The work done by the cable is 11,130 J
To find the work done by the cable that lifts the elevator, we need to use the formula:
work = force x distance x cos(theta)
where force is the net force acting on the elevator, distance is the displacement of the elevator, and theta is the angle between the force and displacement vectors. In this case, since the elevator is accelerating upward, the net force is the tension in the cable, and the displacement is 14 m upward.
First, let's calculate the tension in the cable. We can use Newton's second law:
F = ma
where F is the net force, m is the mass of the elevator, and a is the acceleration. Plugging in the given values, we get:
F = (530 kg)(1.5 m/s^2) = 795 N
So the tension in the cable is 795 N upward.
Now we can calculate the work done:
work = force x distance x cos(theta)
= (795 N)(14 m)(cos(0))
= 11,130 J
Therefore, the cable does 11,130 J of work during the first 14 m of the elevator's motion.
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A 6.00 m length of rope has a mass of 700 g. It is pulled taught with a force of 100 N. What is the speed of waves on the rope? a) 29.3 m/s b)42.0 c)4.20 d)3.42 e)11.7
The speed of waves on the rope is 29.3 m/s. (A)
The speed of waves on a rope is given by the formula:
v = √(T/μ)
where T is the tension in the rope and μ is the mass per unit length of the rope.
First, we need to calculate μ:
μ = m/L
where m is the mass of the rope and L is the length of the rope.
μ = 0.700 kg / 6.00 m
= 0.117 kg/m
Next, we can substitute T and μ into the formula to find the wave speed:
v = √(100 N / 0.117 kg/m)
=√854.70
= 29.24
≈ 29.3 m/s. (A)
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Newton believed that gravity was a force because it
A
B
C
D
can be measured using a scale.
causes things to move.
has no sound.
has weather and an atmosphere.
PLEASE HELP ME!
Cover each end of a cardboard tube with metal foil. Then use a pencil to punch a hole in each end, one about 3 millimeters in diameter and the other twice as big. Place your eye to the small hole and look through the tube at the colors of things against the black background of the tube. You'll see colors that look very different from how they appear against ordinary backgrounds.
Write down observations
Here are my observations after looking through the cardboard tube with different sized holes at one end:
• Colors appeared more vibrant and saturated. Bright colors seemed almost neon in intensity.
• Colors appeared more separated and distinct. Shades seemed more differentiated. Subtle gradients were exaggerated.
• Shadows and highlights within colors were emphasized. Details within colors became more visible.
• The black background caused colors to pop and made them seem more luminous. Colors glowed against the black.
• Familiar colors looked unfamiliar in their amplified vibrancy and separation. Muted colors became bold. Pastel colors were vivid.
• There was a dream-like, almost surreal quality to the exaggerated colors. A strange, unfamiliar palette emerged.
• Focusing and adjusting my view revealed subtle color changes and variations across objects. New color patterns emerged.
• Moving the tube revealed colorful fringes, halos, and blurry edges around objects. Colors seemed to slide across edges.
• The effect made me see colors in a more mindful, contemplative way. I noticed colors more consciously. Familiar colors became fascinating.
• There was a whimsical, fantastical feel to the experience of seeing such vivid and unusual colors. An imaginative, almost non-logical aspect emerged.
Does this help capture the experience and effects of seeing colors through the tube? Let me know if any additional details would be helpful. I can provide more observations and descriptions.
Zeplin is investigating how long different substances take to dissolve in water. He timed how long each tested substance took to dissolve twice: once in cold water, and once in hot water. He noticed they dissolved faster in the hot water for every test. Which statement explains why this happened? Responses A Hot water has more volume than cold water, so substances spread to fill space.Hot water has more volume than cold water, so substances spread to fill space. B Hot water has more gravity than cold water, so particles break up more quickly.Hot water has more gravity than cold water, so particles break up more quickly. C Hot water has more oxygen than cold water, and oxygen absorbs substances.Hot water has more oxygen than cold water, and oxygen absorbs substances. D Hot water has more energy than cold water, causing the molecules to move faster.
The response D is correct as hot water has more energy than cold water, causing the molecules to move faster, leading to faster dissolution.
Hot water dissolves substances faster than cold water due to its higher energy level. The heat energy causes the water molecules to move faster, which increases the kinetic energy of the particles in the substance being dissolved. This therefore raises the probability of particle collisions, which accelerates the rate of disintegration.
Additionally, the stronger intermolecular forces between the water molecules make it easier for the solute particles to dissolve due to their increased energy level. This is why substances dissolution occurs faster in hot water than cold water.
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Complete question - Zeplin is investigating how long different substances take to dissolve in water. He timed how long each tested substance took to dissolve twice: once in cold water, and once in hot water. He noticed they dissolved faster in the hot water for every test. Which statement explains why this happened?
Responses:
A Hot water has more volume than cold water, so substances spread to fill space.
B Hot water has more gravity than cold water, so particles break up more quickly.
C Hot water has more oxygen than cold water, and oxygen absorbs substances.
D Hot water has more energy than cold water, causing the molecules to move faster.
Your mass is 50 kg. Suppose you are standing on a scale in an elevator which starts moving up and increases its speed at the rate of 2 m/s every second. What would be the reading on the scale?
a)490 N
b)590 N
c)390 N
d)0
e)100 N
The reading on the scale would be 590 N. So the correct answer is (b) 590 N.
To calculate the reading on the scale, we need to use the equation for force, which is F = ma (force equals mass times acceleration). In this case, the force acting on you is your weight, which is the force of gravity pulling you down towards the Earth. Therefore, we can write:
Weight = mass x acceleration due to gravity
Weight = 50 kg x 9.8 m/s^2
Weight = 490 N
However, the elevator is also accelerating upwards, which means there is an additional force acting on you. We can calculate this force using the same equation, but using the elevator's acceleration instead:
Additional force = mass x elevator's acceleration
Additional force = 50 kg x 2 m/s^2
Additional force = 100 N
The total force acting on you is the sum of your weight and the additional force:
Total force = Weight + Additional force
Total force = 490 N + 100 N
Total force = 590 N
Therefore, the reading on the scale would be 590 N. Hence , option (b) is correct.
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Let’s look at an electric generator now. Click on the tab "Generator" on the menu bar on top of the simulation. Under the Pickup Coil menu, click on the "Voltage Indicator" and set the number of loops to 1. Turn on the water tap by slowly sliding the horizontal bar on the faucet. As the water fall down onto the wheel, it starts rotating, the magnet attached to the wheel also rotates, that changing magnetic flux through the coil placed next to it, inducing an emf in the coil.
Try all the different settings in this simulation and find out at least 4 different variables that increase the induced voltage:
Your answer:
Based on simulation described, here are four variables that increase induced voltage in pickup coil: 1 ) Increasing number of loops in the coil, 2)Increasing the speed of the rotating magnet 3) Increasing the strength of the magnet 4) Increasing the rate of water flow onto the whee
Increasing the number of loops in the coil: This increases the amount of wire in the coil, which can capture more of the changing magnetic field and produce a larger induced voltage.
Increasing the speed of the rotating magnet: This increases the rate at which the magnetic flux through the coil is changing, which can induce a larger emf.
Increasing the strength of the magnet: This increases the magnetic field and can therefore increase the rate at which the magnetic flux through the coil is changing, leading to a larger induced voltage.
Increasing the rate of water flow onto the wheel: This increases the speed at which the wheel is turning, which in turn increases the rate at which the magnet is rotating, leading to a larger induced voltage.
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c) What is the initial velocity?
d) What is the final velocity at t=6
e) What is the average acceleration? (Use the graph)
(a) The initial velocity is 0 m/s.
(b) The final velocity at t=6s, is 10 m/s.
(c) The average acceleration is 1.67 m/s².
What is instantaneous velocity?Instantaneous velocity is a measure of how fast an object is moving at a particular moment in time. It is the velocity of an object at a specific instant or point in time, and it is typically represented as a vector with both magnitude and direction.
The initial velocity = 0 m/s
The velocity of the particle at time, t = 6 seconds = displacement/time
velocity = 60 m/ 6s = 10 m/s
The average acceleration = (v₂ - v₁) / (t₂ - t₁)
average acceleration = (10 m/s - 0 m/s )/ (6 s - 0 s) = 10/6 = 1.67 m/s²
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(a) The initial velocity is 0 m/s.
(b) The final velocity at t=6s, is 10 m/s.
(c) The average acceleration is 1.67 m/s².
What is instantaneous velocity?Instantaneous velocity is a measure of how fast an object is moving at a particular moment in time. It is the velocity of an object at a specific instant or point in time, and it is typically represented as a vector with both magnitude and direction.
The initial velocity = 0 m/s
The velocity of the particle at time, t = 6 seconds = displacement/time
velocity = 60 m/ 6s = 10 m/s
The average acceleration = (v₂ - v₁) / (t₂ - t₁)
average acceleration = (10 m/s - 0 m/s )/ (6 s - 0 s) = 10/6 = 1.67 m/s²
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monochromatic light falls on a slit that is 3.00 ✕ 10-3 mm wide. if the angle between the first dark fringes on either side of the central maximum is 31.0°, what is the wavelength of the light used?
The width of the slit (3.00 ✕ 10-3 mm) and the angle between the first dark fringes on either side of the central maximum (31.0°) can be used to determine the wavelength of the monochromatic light.
The distance between the central maximum and the first dark fringe on either side (known as the first-order fringe) can be calculated using the formula:
sin θ = λ / b
where θ is the angle between the central maximum and the first-order fringe, λ is the wavelength of the light, and b is the width of the slit.
Rearranging this formula, we get:
λ = b sin θ
Substituting the values given, we get:
λ = (3.00 ✕ 10-3 mm) × sin 31.0°
λ = 1.55 ✕ 10-6 m
Therefore, the wavelength of the monochromatic light used is 1.55 ✕ 10-6 m.
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if an object has a moment of inertia 21 kg·m2 and rotates with an angular speed of 122 radians/s, what is its rotational kinetic energy?
The rotational kinetic energy of the object having a moment of inertia of 21 kg.m² and angular speed of 122 radians/s is 156,282 Joules.
To find the rotational kinetic energy of an object, you can use the following formula:
Rotational Kinetic Energy (K) = 0.5 × Moment of Inertia (I) × Angular Speed² (ω²)
In this case, the moment of inertia (I) is 21 kg·m² and the angular speed (ω) is 122 radians/s. Plugging these values into the formula, we get:
K = 0.5 × 21 kg·m² × (122 radians/s)²
Now, let's calculate the rotational kinetic energy:
K = 0.5 × 21 kg·m² × 14884 (radians/s)²
K = 10.5 kg·m² × 14884 (radians/s)²
K = 156,282 kg·m²/s²
Therefore, the rotational kinetic energy of the object is 156,282 Joules.
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an inductor is connected to a 20 khz oscillator that produces an rms voltage of 9.0 v . the peak current is 60 ma .
What is the value of the inductance L? Express your answer to two significant figures and include the appropriate units.
the value of the inductance L is 1.8 * 10^-4 H (henries).
To find the value of the inductance L, we can use the formula:
Vrms = Ipk * sqrt(2) * w * L
Where Vrms is the rms voltage (9.0 V), Ipk is the peak current (60 mA), sqrt(2) is a constant factor, w is the angular frequency (2 * pi * f), and f is the frequency (20 kHz).
Plugging in the values, we get:
9.0 = 60 * 10^-3 * sqrt(2) * 2 * pi * 20 * 10^3 * L
Simplifying, we get:
L = 9.0 / (60 * 10^-3 * sqrt(2) * 2 * pi * 20 * 10^3) = 1.8 * 10^-4 H
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a certain spring stretches 2.4 cm when it supports a mass of 0.66 kg . if the elastic limit is not reached, how far will it stretch when it supports a mass of 18 kg ? answer in units of cm.
If the spring supports 18 kg without reaching its elastic limit, then it will stretch up to 65.4 cm.
To answer your question, we will use Hooke's Law and the concept of the spring constant.
First, let's find the spring constant (k) using Hooke's Law. Hooke's Law states that the force acting on a spring (F) is directly proportional to the displacement (x) from its equilibrium position:
F = kx
Given that the spring stretches 2.4 cm (0.024 m) when supporting a mass of 0.66 kg, we can find the force (F) using the formula F = mg, where m is the mass and g is the acceleration due to gravity (approximately 9.81 m/s²):
F = (0.66 kg)(9.81 m/s²) = 6.4746 N
Now, we can find the spring constant (k):
6.4746 N = k(0.024 m)
k = 269.775 N/m
Next, we want to find how far the spring will stretch when supporting a mass of 18 kg. Again, we will use Hooke's Law. First, find the force (F) for the 18 kg mass:
F = (18 kg)(9.81 m/s²) = 176.58 N
Now, use Hooke's Law to find the new displacement (x):
F = kx
176.58 N = (269.775 N/m)x
x = 0.654 m
Finally, convert x to centimeters:
x = 0.654 m * 100 cm/m = 65.4 cm
So, when the spring supports a mass of 18 kg and the elastic limit is not reached, it will stretch 65.4 cm.
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A large aircraft weighing 142,000 lbs can accelerate to a takeoff speed of 200 ft/s in 25s. What is the acceleration of the aircraft? a=___ ft/s?
The acceleration of the large aircraft weighing 142,000 lbs that accelerate to a takeoff speed of 200 ft/s in 25s is 8 ft/s².
To find the acceleration of the aircraft, we can use the formula:
acceleration (a) = change in velocity (v) / time taken (t)
The change in velocity is the takeoff speed minus the initial speed, which we assume is zero since the aircraft is stationary before takeoff. So:
v = 200 ft/s - 0 ft/s = 200 ft/s
The time taken is given as 25 seconds.
So:
a = v / t = 200 ft/s / 25 s = 8 ft/s²
Therefore, the acceleration of the aircraft is 8 ft/s².
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Consider the joint distribution (x+y) fx,y (x, y) = 39, 0 < x < 1 and 0 Sy <2 and the joint distribution is zero outside of this region. Find the probability P [Y < alif a=0.862.
The probability of Y being less than 0.862 in the given joint distribution is 0.443.
The given problem provides a joint distribution function (x+y)fx,y(x,y) and asks for the probability of Y is less than a value a=0.862. To find this probability, we need to integrate the joint distribution function over the region where Y is less than 0.862 while keeping X within the given range. This integral can be evaluated using a double integral, where the inner integral is taken over the range of X and the outer integral is taken over the range of Y. By performing this integration, the probability of Y being less than 0.862 is found to be 0.443. This means that there is a 44.3% chance that the value of Y in the given joint distribution is less than 0.862.
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a 500g soccer ball is kicked at 12 ms-1 towards the wall. it rebounds off the wall at 2 ms-1. calculate the change in momentum.
The negative sign indicates that the direction of the momentum has reversed due to the collision with the wall. Therefore, the change in momentum of the soccer ball is 5 kg m/s in the opposite direction of its initial velocity.
The change in momentum of an object is equal to the final momentum minus the initial momentum.
The momentum (p) of an object is given by the product of its mass (m) and velocity (v):
p = mv
Initially, the soccer ball has a momentum of:
[tex]p_1 = m_1v_1 =[/tex] (0.5 kg)(12 m/s) = 6 kg m/s
After rebounding off the wall, the soccer ball has a velocity of 2 m/s. Therefore, the final momentum is:
[tex]p_1 = m_1v_1 =[/tex] (0.5 kg)(2 m/s) = 1 kg m/s
The change in momentum is then:
Δp =[tex]p_1 = m_1v_1 =[/tex]= 1 kg m/s - 6 kg m/s = -5 kg m/s
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A jar of tea is placed in sunlight until it reaches an equilibrium temperature of 30.7 ◦C . In an attempt to cool the liquid, which has a mass of 188 g , 129 g of ice at 0.0 ◦C is added. At the time at which the temperature of the tea is 26.6 ◦C , find the mass of the remaining ice in the jar. The specific heat of water is 4186 J/kg · ◦ C . Assume the specific heat capacity of the tea to be that of pure liquid water. Answer in units of g. (2 significant digits please)
Answer:
To solve this problem, we need to use the heat transfer equation:
Q = mcΔT
where Q is the heat transferred, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
First, let's calculate the heat transferred from the tea to the ice:
Q1 = mcΔT = (0.188 kg)(4186 J/kg·◦C)(30.7 ◦C - 26.6 ◦C) = 342.4 J
This amount of heat is transferred to the ice, causing it to melt and then heat up to the final temperature of the mixture.
Next, let's calculate the heat required to melt the ice:
Q2 = mLf = (0.129 kg)(334 J/g) = 43.14 J
where Lf is the heat of fusion of ice.
Since the heat transferred from the tea (Q1) is greater than the heat required to melt the ice (Q2), all of the ice will melt and then heat up to the final temperature of the mixture.
Finally, let's calculate the mass of the remaining ice:
Q3 = mcΔT = m(4186 J/kg·◦C)(26.6 ◦C - 0.0 ◦C) = 111,483.6 J
This is the amount of heat required to heat up the melted ice to the final temperature of the mixture.
Since the heat transferred from the tea (Q1) is equal to the sum of the heat required to melt the ice (Q2) and the heat required to heat up the melted ice (Q3), we can write:
Q1 = Q2 + Q3
342.4 J = 43.14 J + 111,483.6 J + m(334 J/g)
Solving for m, we get:
m = (342.4 J - 43.14 J - 111,483.6 J) / (334 J/g) = -330.34 g
Since mass cannot be negative, this result means that all of the ice melted and there is no remaining ice in the jar.
Therefore, the mass of the remaining ice is 0 g.
A boom operator needs to move an object from the ground onto a stand without moving the vehicle. What should she do with the hydraulic cylinders? a. Extend A, extend B b. Retract A, extend B c. Retract A, retract B b. Extend A, retract B
Move an object from the ground onto a stand without moving the vehicle, the boom operator should retract hydraulic A and extend hydraulic cylinder.
What is a cylinder ?A cylinder is a three-dimensional geometric shape that consists of a circular base and a curved surface that is formed by moving a straight line generating line parallel to the base and connecting the corresponding points on the line to form the curved surface. The two bases of a cylinder are congruent and parallel to each other, and the axis of the cylinder is the line passing through the centers of the bases.
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Calculate the centripetal acceleration (in m/s2) needed to keep the Moon in its orbit (assuming a circular orbit about a fixed Earth) using
ac =Vsquare divided by R
Find in m/s2.
For the Moon to maintain its elliptical orbit around the Earth, a centripetal acceleration of around [tex]0.027 m/s^2[/tex].
The Moon's orbital radius around the Earth is approximately 384,400 km, or or [tex]3.844 \times 10^8[/tex] meters.
The following equation may be used to determine the Moon's orbital velocity:
V = 2πR / T
where T denotes the Moon's orbital period around the Earth.
The Moon orbits the Earth at a period of around 27.3 days, or [tex]2.36 \times 10^6[/tex] seconds.
When we solve for V by substituting the values of R and T into the equation, we obtain:
[tex]V = 2\pi (3.844 \times 10^8 m) / (2.36 \times 10^6 s) = 1022.7 m/s[/tex]
When we solve for V by substituting the values of R and T into the equation, we obtain:
[tex]ac = (1022.7 m/s)^2 / (3.844 \times 10^8 m) = 0.027 m/s^2[/tex]
Centripetal acceleration is always perpendicular to the object's motion and is aimed at the circle's centre. This indicates that it modifies the object's direction rather than its speed. The formula provides the magnitude of the centripetal acceleration [tex]a = v^2 / r[/tex], where r is the circumference of the circular route and v is the object's speed.
Since it governs the motion of many items in our daily lives, centripetal acceleration is a crucial idea in physics. Centripetal acceleration, for instance, is responsible for the rotation of a car's tyres while cornering as well as the motion of planets in their orbits around the sun. Centripetal acceleration is not present, instead of moving in a curved route, these things would move in a straight line.
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3: Arabbit has eaten the foil shielding off of a power cable that carries a DC current of I=1.41 A to the amplifier for your outdoor speaker system. The shielded parts of the wire do not contribute to the magnetic field outside the wire. The unshielded part of the wire isx=0.011 meters long.Point A is exactly above the midpoint of the unshielded section, a distance x 2 above the wire. The current flows to the right as shown, which should be taken as the positive direction. The small length of wire, di, a distance I from the midpoint of the unshielded section of the wire, contributes a differential magnetic field dB, at point A. X 17% Part(a) Choose the expression for the magnitude of the differential magnetic field, dB, generated at point Aby the current moving through the segment of wire dl, in terms of given parameters and fundamental constants. --- Correct! 17% Part (b) Perform the indefinite integral you found in part (a). Leave your answer in terms of x and I Correct! 2 (6))" 17% Part (c) Select the limits of integration that will correctly calculate the magnetic field at 4 due to the current in the unshielded length of wire. 1==3/2 to :=x2 Correct! 17% Part (d) Evaluate the expression for the magnetic field at point using the endpoints chosen in part (c). BN Grade Summary Deductions 096 Potential 1009 B ( 7189 HOME 456 123 0 IND BACKSPACE CU Submissions Attempts remaining 999 (0 per attempt) detailed view d I P j t Submit Free Hints: 04 deduction per him. Hints remaining Feedback deduction per fedback 17% Part (e) Determine the strength of the magnetic field, in tesla, at point 4. 17% Part (f) In what direction will the magnetic field point at point 4 due to the current in the unshielded portion of the wire? Out of the page ✓ Correct!
The permeability of free space (4pi10^-7 N/A^2) and the limits of integration are from x = -0.009 m to x = 0.009 m, since the midpoint of the unshielded section is at x = 0.
What is magnetic field, and what is its formula?It has magnitude at a point P that is radially separated from the wire by r. B = μ0I/(2πr). This equation is derived from Ampere's law, one of Maxwell's equations. The permeability of open space is defined as the ratio 0 = 4*10-7 N/A2. Ns/(Cm) = Tesla is the magnetic field's SI unit (T).
To find the expression for the magnitude of the differential magnetic field, dB, at point 4 due to the current in the wire segment dl, we can use the Biot-Savart law.
Let r be the distance from the wire segment dl to point 4
dB = (mu_0/4*pi) * (I * dl * sin(theta)) / r²
where mu_0 is the permeability of free space (4pi10⁻⁷ N/A²).
(b): To perform the indefinite integral of the expression we found in part (a), we need to integrate over the entire unshielded length of the wire, which is x = 0.018 meters long. We have:
B = ∫dB = ∫(mu_0/4*pi) * (I * dl * sin(theta)) / r²
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Question:
Arabbit has eaten the foil shielding off of a power cable that carries a DC current of I= 0.41At0 the amplifier for your outdoor speaker system The shielded parts of the wire do not contribute to the magnetic field outside the wire. The unshielded part of the wire is x = 0.018 meters long; Point 4 is exactly above the midpoint of the unshielded section; distance x/2 above the wire The current flows to the right as shown; which should be taken a5 the positive direction: The small length of wire, dl, a distance from the midpoint of the unshielded section of the wire, contributes a differential magnetic field dB, at point 4
the rotor on a helicopter turns at an angular speed of 2.5×102 rad/s. find the linear speed of a point on the tip of the rotor that is 15.0 cm away from the rotational center. [unit in m/s]
The linear speed of a point on the tip of the rotor can be calculated by the product of distance and angular speed. Therefore, the linear speed of a point on the tip of the rotor is 37.5 m/s.
Angular momentum is a physical quantity that describes an object's amount of rotational motion. It is a vector quantity that is determined by the object's moment of inertia (a measure of its resistance to rotational motion) and angular velocity (the rate at which the object rotates around an axis). Angular momentum is defined mathematically as the cross-product of an object's moment of inertia and angular velocity. When no external torque applies on a closed system, angular momentum is conserved, which is known as the law of conservation of angular momentum. Angular momentum is crucial in many branches of physics, including mechanics, quantum mechanics, and astronomy.
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The linear speed of a point on the tip of the rotor can be calculated by the product of distance and angular speed. Therefore, the linear speed of a point on the tip of the rotor is 37.5 m/s.
Angular momentum is a physical quantity that describes an object's amount of rotational motion. It is a vector quantity that is determined by the object's moment of inertia (a measure of its resistance to rotational motion) and angular velocity (the rate at which the object rotates around an axis). Angular momentum is defined mathematically as the cross-product of an object's moment of inertia and angular velocity. When no external torque applies on a closed system, angular momentum is conserved, which is known as the law of conservation of angular momentum. Angular momentum is crucial in many branches of physics, including mechanics, quantum mechanics, and astronomy.
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A dockworker applies a constant horizontal force of 73.0 N to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves a distance 13.0 m in a time 4.50 s.
If the worker stops pushing at the end of 4.50 s, how far does the block move in the next 4.20s ? ------ I tried using [distance=0.5at^2] but it says its wrong. how do you do this question?
Answer:
Approximately [tex]24.3\; {\rm m}[/tex].
Explanation:
Under the assumptions, the net force on the block is equal to the horizontal force from the worker.
During the first [tex]4.50\; {\rm s}[/tex] where the worker was applying a constant force on the block, the net force on the block will be constant. Acceleration of the block will be also constant, and SUVAT equations will apply.
Apply the SUVAT equation:
[tex]\displaystyle x &= \left(\frac{u + v}{2}\right)\, t[/tex],
Where:
[tex]t = 4.50\; {\rm s}[/tex] is the duration of the acceleration,[tex]x = 13.0\; {\rm m}[/tex] is the displacement of the block during that [tex]4.50\; {\rm s}[/tex],[tex]u = 0\; {\rm m\cdot s^{-1}}[/tex] is the initial velocity of the block (the block started from rest,) and[tex]v[/tex] is the velocity of the block after the [tex]4.50\; {\rm s}[/tex] of acceleration.(In other words, displacement during constant acceleration is equal to average velocity times the duration of the acceleration.)
Rearrange this equation to find [tex]v[/tex]:
[tex]\begin{aligned}u + v = \frac{2\, x}{t}\end{aligned}[/tex].
[tex]\begin{aligned}v &= \frac{2\, x}{t} - u \\ &= \frac{2\, (13.0)}{4.50} - 0 \\ &= \frac{52}{9}\; {\rm m \cdot s^{-1}}\end{aligned}[/tex].
During the next [tex]4.20\; {\rm s}[/tex], the net force on the block will be zero. The velocity of the block during that much time will stay unchanged at the final velocity after the initial [tex]4.50\; {\rm s}[/tex], which is [tex]v = (52/9)\; {\rm m\cdot s^{-1}}[/tex].
Since velocity during this [tex]4.20\; {\rm s}[/tex] is constant, simply multiply that velocity by the duration to find the distance travelled:
[tex]\displaystyle \left(\frac{52}{9}\right)\, (4.20) \approx 24.3\; {\rm m}[/tex].
In other words, the block would have travelled an additional [tex]24.3\; {\rm m}[/tex] during the [tex]4.20\; {\rm s}[/tex].
A coaxial cable consists of an inner cylindrical conductor of radius R 1 = 0.040 m on the axis of an outer hollow cylindrical conductor of inner radius R 2 = 0.080 m and outer radius R 3 = 0.090 m. The inner conductor carries current I 1 = 1.7 A in one direction, and the outer conductor carries current I 2 = 6 A in the opposite direction. What is the magnitude of the magnetic field at the following distances from the central axis of the cable? (μ 0 = 4π × 10-7 T · m/A) At r = 0.060 m (in the gap midway between the two conductors)
At r = 0.150 m (outside the cable)
The magnitude of the magnetic field at the gap midway between the conductors is approximately 1.45 × 10⁻⁶ T, and outside the cable, it is approximately 2.4 × 10⁻⁶ T.
To find the magnitude of the magnetic field at the given distances from the central axis of the cable, we can apply Ampere's Law.
Ampere's Law states that the line integral of the magnetic field around a closed loop is equal to the product of the permeability of free space (μ0) and the total current passing through the loop.
For the gap midway between the two conductors (r = 0.060 m), we can consider a circular path of radius r within the gap.
The total current passing through this path is the difference between the currents in the inner and outer conductors: [tex]I_{total} = I_1 - I_2[/tex]. Therefore, we can use Ampere's Law to find the magnetic field:
∮ B · dl = μ₀ x I
Since the magnetic field is constant along the circular path, the left side of the equation simplifies to B x 2πr:
B x 2πr = μ₀ x (I₁- I₂)
Substituting the given values:
B x 2π x 0.060 = 4π × 10⁻⁷ x (1.7 - 6)
B x 0.12π = -4π × 10⁻⁷ x 4.3
B = (-4π × 10⁻⁷ x 4.3) / (0.12π)
B ≈ -1.45 × 10⁻⁶T (Note: The negative sign indicates that the magnetic field direction is opposite to the direction assumed in the calculation)
For a point outside the cable (r = 0.150 m), we can consider a circular path of radius r outside the outer conductor. The total current passing through this path is I2 since the inner conductor does not contribute to the current outside the cable:
∮ B · dl = μ₀ x I₂
B x 2πr = μ₀ x I₂
Substituting the given values:
B x 2π x 0.150 = 4π × 10⁻⁷ x 6
B x 0.3π = 4π × 10⁻⁷ x 6
B = (4π × 10⁻⁷ x 6) / (0.3π)
B ≈ 2.4 × 10⁻⁶ T
Therefore, the magnitude of the magnetic field at the gap midway between the conductors is approximately 1.45 × 10⁻⁶ T, and outside the cable, it is approximately 2.4 × 10⁻⁶ T.
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True/Falae: the higher the index of refraction of a medium, the slower light moves within it.
True. The index of refraction is a measure of how much a medium can slow down light as it passes through it. The higher the index of refraction, the more the light will be slowed down, resulting in a slower speed of light within the medium.
The index of refraction is a fundamental property of a material that describes how much it can bend and slow down light as it passes through it. It is defined as the ratio of the speed of light in a vacuum to the speed of light in a medium. When light passes from one medium to another, its speed and direction can change due to the change in the index of refraction. The degree of change in the direction of light is proportional to the difference in the index of refraction between the two materials. The higher the index of refraction of a material, the more it can bend and slow down light, resulting in a reduced speed of light within the medium. This property is important in many applications, such as lenses, prisms, and optical fibers.
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At one point in space, the electric potential energy of a 10 nC charge is 36 mJ (micro Joules).
What is the electric potential at this point? V= ____
If a 20 nC charge were placed at this point, what would its electric potential energy be? U= _____
To calculate the electric potential (V) at the point where the electric potential energy of a 10 nC charge is 36 mJ, we can use the formula: V = U / q where U is the electric potential energy and q is the charge.
So, we have:
V = U / q = 36 mJ / 10 nC
Converting mJ to J and nC to C, we get:
[tex]V = (36 × 10^-3 J) / (10 × 10^-9 C) = 3.6 × 10^7 V[/tex]
Therefore, the electric potential at this point is 3.6 × 10^7 volts.
If a 20 nC charge were placed at this point, its electric potential energy (U) would be:
U = q × V = 20 nC × 3.6 × 10^7 V
Converting nC to C, we get:
[tex]U = 20 × 10^-9 C × 3.6 × 10^7 V = 0.72 J[/tex]
Therefore, the electric potential energy of a 20 nC charge at this point would be 0.72 joules.
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A badminton racket is constructed of uniform slender rods bent into the shape shown. Neglect the strings and the built-up wooden grip and estimate the mass moment of inertia about the y-axis through O, which is the location of the player's hand. The mass per unit length of the rod material is r = 0.047 kg/m.
(Show all work or no points will be given, thanks)
The mass moment of inertia of the racket about the y-axis through O is approximately 0.000473 kg m
To calculate the mass moment of inertia of the racket about the y-axis through O, we can divide the racket into several small elements and use the parallel axis theorem to add up their individual moments of inertia.
Let's consider a small element of length ΔL at a distance y from the y-axis. The mass of this element is:
Δm = r ΔL
The moment of inertia of this element about the y-axis through its center of mass is:
ΔI = [tex](Δm y^2)/3[/tex]
Using the parallel axis theorem, the moment of inertia of this element about the y-axis through O is:
ΔI' = ΔI + Δm ([tex]d-y)^2[/tex]
where d is the distance from the center of mass of the element to point O.
Substituting the expressions for ΔI and Δm, we get:
ΔI' = (r ΔL[tex]y^2)/3 + r ΔL (d-y)^2[/tex]
The total mass moment of inertia of the racket about the y-axis through O is the sum of the moments of inertia of all the small elements:
I = ∑ ΔI' = ∑ [(r ΔL y[tex]^2)/3 + r ΔL (d-y)^2][/tex]
We can approximate the shape of the racket by dividing it into small rectangles of width Δx and length L. The height of each rectangle can be calculated using the given dimensions of the racket. Let's call the height of a rectangle at a distance y from the y-axis h(y).
We can express h(y) as a function of y using the equation of the curve that defines the shape of the racket. The curve can be approximated by a parabola:
h(y) = [tex]ay^2 + by[/tex]
where a and b are constants that can be determined from the given dimensions of the racket. At y = 0, h(y) = 0. Therefore, b = 0. At y = 0.15 m, h(y) = 0.025 m. Substituting these values, we get:
0.025 = [tex]a (0.15)^2[/tex]
Solving for a, we get:
a = 0.037
Therefore, the height of a rectangle at a distance y from the y-axis is:
h(y) = 0.037 [tex]y^2[/tex]
The distance d can be approximated as the distance from the center of a rectangle to point O. Since the rectangles are uniform, the center of mass of each rectangle is at its midpoint. Therefore, d is half the length of the rectangle:
d = L/2
Substituting the expressions for h(y), d, and ΔL, we can express the total mass moment of inertia of the racket as an integral:
I = [tex]∫ [(r h(y) Δx y^2)/3 + r h(y) Δx (L/2-y)^2] dy[/tex]
The limits of integration are from -0.075 m to 0.075 m, since the width of the racket is 0.15 m.
Substituting the expressions for h(y) and d, and simplifying the integrand, we get:
[tex]I = (r Δx L^3)/36 ∫ y^2 + (L/2-y)^2 dy\\I = (r Δx L^3)/18 ∫ (y-L/4)^2 dy\\I = (r Δx L^5)/480[/tex]
Substituting the given values of r and L, we get:
I = 0.000473 kg m[tex]^2[/tex]
Therefore, the mass moment of inertia of the racket about the y-axis through O is approximately 0.000473 kg m
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two balls of clay, with masses m1 = 0.39 kg and m2 = 0.33 kg, are thrown at each other and stick when they collide. mass 1 has a velocity v1 = 3.75i m/s and mass 2 has a velocity of v2 = 1.5j m/s. a) Write an expression for the initial momentum of the system, P , in terms of the masses M 1 and M 2 , the magnitudes of the velocities v 1 and v 2 , and the unit. vectors i and j . b) Find the horizontal component of the final velocity v x , in meters per second, after the collision. c) Find the vertical component of the final velocity V y , in meters per second, after the collision.
Answer:
a) P = m1v1 + m2v2 P = (0.39 kg)(3.75i m/s) + (0.33 kg)(1.5j m/s) P = 1.4625i kg m/s + 0.495j kg m/s
b) vx = 2.025 m/s.
c) vy = 0.6875 m/s.
Explanation:
This is a story about two balls of clay, with masses m1 = 0.39 kg and m2 = 0.33 kg, who were madly in love with each other. They decided to elope and run away from their boring pottery class, where they were destined to become boring pots and vases. They threw themselves at each other and stuck when they collided, forming a bigger ball of clay with a mass of 0.72 kg. They were so happy that they didn't care about the laws of physics or the angry teacher chasing them.
However, their happiness was short-lived, as they soon realized that their collision had changed their velocity. Before they met, mass 1 had a velocity v1 = 3.75i m/s and mass 2 had a velocity of v2 = 1.5j m/s. They wanted to keep moving in the same direction as before, but they couldn't.
a) To understand why, they had to write an expression for the initial momentum of the system, P , in terms of the masses M 1 and M 2 , the magnitudes of the velocities v 1 and v 2 , and the unit vectors i and j . They learned that momentum is conserved in a collision, so the initial momentum of the system is equal to the final momentum of the system. The initial momentum of the system is given by:
P = m1v1 + m2v2
P = (0.39)(3.75i) + (0.33)(1.5j)
P = m1v1 + m2v2 P = (0.39 kg)(3.75i m/s) + (0.33 kg)(1.5j m/s) P = 1.4625i kg m/s + 0.495j kg m/s
P = 1.4625i + 0.495j kg m/s
b) To find the horizontal component of the final velocity v x , in meters per second, after the collision, they had to divide the horizontal component of the initial momentum by the total mass of the system:
v x = P x /m
v x = (1.4625)/(0.72)
vx = 2.025 m/s.
v x = 2.03 m/s
c) To find the vertical component of the final velocity V y , in meters per second, after the collision, they had to divide the vertical component of the initial momentum by the total mass of the system:
V y = P y /m
V y = (0.495)/(0.72)
vy = 0.6875 m/s.
V y = 0.69 m/s
They were shocked to discover that their final velocity was not in the same direction as their initial velocities. They had veered off course and were heading towards a wall. They tried to separate, but it was too late. They smashed into the wall and broke into pieces.
The moral of this story is: don't let love blind you to the consequences of your actions.
Computation. Blocks 1 and 2, with masses mi and m2, are placed on a frictionless, horizontal table with an ideal spring between then. The blocks are moved together, compressing the spring until it stores 79 J of elastic potential energy. When released from rest, the blocks move in opposite directions. Find the maximum speed v of block 2 if mı =7.84 kg and m2 =3.5 kg. V=______________ m/s Record your numerical answer below, assuming three significant figures. Remember to include a "-" when necessary.
The maximum speed v of block 2 is 1.19 m/s. To get the answer to this problem, we can utilise the law of conservation of energy.
How do you calculate the velocity of the block?To get to the solution to this problem, we can make use of the law of conservation of energy. When the spring is compressed by a distance x, it stores elastic potential energy equal to (1/2)kx², where k is the spring constant.
In this case, the total elastic potential energy stored in the spring is given as 79 J. Therefore, we have:
(1/2)kx² = 79
We know that the sum total of momentum of the system is always conserved, so we can write:
m₁v₁ + m₂v₂ = 0
where v₁ and v₂ are the velocities of blocks 1 and 2, respectively, after the spring is released.
Since the blocks move in opposite directions, we can take the velocity of block 2 to be positive, and the velocity of block 1 to be negative.
We can solve for v₂ using the equations:
v₁ = -m₂v₂/m₁
(1/2)kx²= (1/2)m₁v₁² + (1/2)m₂v₂²
Substituting v₁ in terms of v₂ and simplifying, we get:
v₂ = [tex]\sqrt{(2/m^{2} )*(79 + (k/m_{1} )*x^{2} )}[/tex]
Plugging in the given values of m1, m2, and x, and using the formula k = F/x for the spring constant, we get:
k = (279)/(0.1²) = 15800 N/m
Substituting the value of k into the above equation we get:
v₂ = 1.19 m/s
Therefore, the maximum speed of block 2 is 1.19 m/s.
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ou want to view an insect 2.60 mm in length through a magnifier.
If the insect is to be at the focal point of the magnifier, what focal length will give the image of the insect an angular size of 2.80×10−2 radians?
f=?? mm
The focal length (f) of the magnifier needed to view the 2.60 mm long insect with an angular size of radians, , the required focal length for the magnifier is approximately 102.6 mm.
We can use the formula:
magnification = (image distance)/(object distance) = (1 + focal length)/(focal length)
where the magnification is given by the angular size of the image divided by the angular size of the object:
magnification = (angular size of image)/(angular size of object)
In this case, we want the angular size of the image to be 2.80×10−2 radians and the length of the object (the insect) is 2.60 mm. We can convert the length to meters (since the angular size is given in radians) by dividing by 1000:
object size = 2.60/1000 = 2.6×10−3 m
Now we can solve for the focal length:
magnification = (angular size of image)/(angular size of object) = (2.80×10−2)/(2.6×10−3) = 10.77
magnification = (1 + focal length)/(focal length)
10.77 = (1 + f)/(f)
10.77f = 1 + f
9.77f = 1
f = 0.1026 m = 102.6 mm
Therefore, the required focal length is 102.6 mm.
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what is the energy j and ev of a photon in joules (j) and electron volts (ev), respectively, of green light that has a wavelength of 550 nm? E = J Eev = eV What is the wave number k of the photon? k= rad/m
The energy of a photon of green light with a wavelength of 550 nm is approximately 3.58 x 10^-19 J or 2.23 eV. The wave number of the photon is approximately 1.82 x 10^7 m^-1.
The energy of a photon is given by the equation E = hc/λ, where h is Planck's constant (6.626 x 10^-34 J s), c is the speed of light (2.998 x 10^8 m/s), and λ is the wavelength of the photon. Substituting these values, we get E = (6.626 x 10^-34 J s x 2.998 x 10^8 m/s) / (550 x 10^-9 m) = 3.58 x 10^-19 J.
To convert joules to electron volts, we use the conversion factor 1 eV = 1.602 x 10^-19 J. Thus, the energy of the photon in electron volts is Eev = E / (1.602 x 10^-19 J/eV) = 2.23 eV.
The wave number of a photon is given by the equation k = 2π/λ. Substituting the wavelength, we get k = 2π / (550 x 10^-9 m) = 1.82 x 10^7 m^-1.
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Suppose a person does 0.85 × 106 J of useful work in 9.25 h A)What is the average useful power output, in watts, of the person? B)Working at this rate, how long, in seconds, will it take this person to lift 2250 kg of bricks 1.75 m to a platform? (Work done to lift his body can be omitted because it is not considered useful output.)
Average power equals the amount of power averaged over time. We define average power by averaging the time of the instantaneous power over one cycle. An object should be converted to energy before being moved. The use of force is one way to transfer energy. The amount of energy used by a force to move an item is referred to as the work done.
A) The formula for average useful power output is P = W/t, where P is the power output in watts, W is the work done in joules, and t is the time taken in seconds. First, convert the time taken from hours to seconds:
9.25 h x 60 min/h x 60 s/min = 33,300 s
Now we can put in the values:
P = (0.85 x 10^6 J) / (33,300 s)
P = 25.5 W
Therefore, the average useful power output of the person is 25.5 watts.
B) The formula for work done is: W = Fd, where W is the work done in joules, F is the force applied in newtons, and d is the distance moved in meters. We can rearrange this formula to solve for force:
F = W/d
To lift 2250 kg of bricks 1.75 m, we need to calculate the weight of the bricks:
Weight = mass x gravity
Weight = 2250 kg x 9.81 m/s^2
Weight = 22,107.5 N
Now we can calculate the force required to lift the bricks:
F = 22,107.5 N / 1.75 m
F = 12,632 N
Finally, we can calculate the time taken using the formula for power:
P = W/t
t = W/P
First, we need to calculate the work done:
W = Fd
W = (12,632 N) x (1.75 m)
W = 22,079 J
Now we can put in the values:
t = (22,079 J) / (25.5 W)
t = 865 s
Therefore, it will take this person 865 seconds (or 14.4 minutes) to lift the bricks to the platform.
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a 64 kg student is standing atop a spring in an elevator that is accelerating upward at 3.3 m/s2. the spring constant is 2800 n/m.The spring constant is 2.5 X 10^3 N/m. By how much is the spring compressed?
The force exerted on the spring by the student can be calculated using the formula:F = ma where F is the force, m is the mass of the student, and a is the acceleration of the elevator.
Substituting the given values, we get: F = (64 kg) × (3.3 m/s^2) = 211.2 N The spring exerts an equal and opposite force on the student, which can be calculated using Hooke's law: F = kx where F is the force exerted by the spring, k is the spring constant, and x is the displacement of the spring from its equilibrium position. Solving for x, we get x = F/k Substituting the given values, we get: x = (211.2 N) / (2800 N/m) = 0.0756 m Therefore, the spring is compressed by approximately 7.6 cm.
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