We can use the laws of physics to solve this problem. The key here is to recognize that the initial kinetic energy of the car (due to its speed) is converted into potential energy as it goes up the ramp, and then back into kinetic energy as it falls back down.
Using conservation of energy, we can set the initial kinetic energy equal to the final potential energy at the maximum height:
(1/2) * m * v^2 = m * g * h
where m is the mass of the car, v is its initial speed (30 m/s), g is the acceleration due to gravity (9.81 m/s^2), and h is the maximum height.
Simplifying the equation and solving for h, we get:
h = (v^2 * sin^2(theta)) / (2 * g)
where theta is the angle of the ramp (37 degrees in this case).
Plugging in the known values, we get:
h = (30^2 * sin^2(37)) / (2 * 9.81) ≈ 79.1 meters
So the car will reach a height of approximately 79.1 meters off the ground.
To find the maximum height of the gorge that the car could clear, we need to look at the horizontal distance that the car travels while in the air. This distance is given by:
d = v * t
where t is the time that the car spends in the air. We can find t by setting the initial potential energy (at the maximum height) equal to the final kinetic energy (at the end of the jump):
m * g * h = (1/2) * m * (v_f^2)
where v_f is the final velocity of the car at the end of the jump (when it lands back on the ground). We can solve for v_f using:
v_f = sqrt(2gh)
where h is the maximum height (found earlier). Plugging this into the conservation of energy equation, we get:
t = sqrt(2h/g)
Now we can plug in the known values for v and theta to get:
d = 30 * sqrt(2h/g) * cos(theta)
Plugging in the values for h and theta, we get:
d = 30 * sqrt(2*79.1/9.81) * cos(37) ≈ 187.2 meters
So the maximum height of the gorge that the car could clear is approximately 187.2 meters.
OK, once again we have a pendulum, this time of length 1.06 m, which you release from rest at an angle of 41.2 degrees to the vertical. What will be the speed of the pendulum at the instant it reaches an angle of 20.6 degrees above the vertical?
The speed of the pendulum at the instant it reaches an angle of 20.6 degrees above the vertical is 3.02 m/s.
A pendulum is a weight suspended from a fixed point that swings back and forth due to the force of gravity.
Based on the given information, we have a pendulum of length 1.06 m and it is released from rest at an angle of 41.2 degrees to the vertical. We need to find the speed of the pendulum at the instant it reaches an angle of 20.6 degrees above the vertical.
To solve this problem, we can use the conservation of mechanical energy. At the highest point of the pendulum's swing, all of its energy is in the form of potential energy, and at the lowest point of its swing, all of its energy is in the form of kinetic energy. Therefore, we can write:
PE_max = KE_min
where PE_max is the potential energy at the maximum height and KE_min is the kinetic energy at the lowest point.
The potential energy of a pendulum is given by:
PE = mgh
where m is the mass of the pendulum, g is the acceleration due to gravity, and h is the height above some reference point.
The kinetic energy of a pendulum is given by:
KE = (1/2)mv^2
where v is the speed of the pendulum.
First, we need to find the vertical height difference between the pendulum's highest and lowest points. To do this, we can use trigonometry:
h = L(1 - cosθ)
where L is the length of the pendulum and θ is the initial angle to the vertical. Substituting the given values, we get:
h = 1.06(1 - cos(41.2°)) = 0.654 m
Next, we can use the conservation of mechanical energy to find the speed of the pendulum at the lowest point of its swing. At this point, all of the potential energy has been converted into kinetic energy, so we can write:
PE_max = KE_min
mgh = (1/2)mv^2
Canceling out the mass, we get:
gh = (1/2)v^2
Solving for v, we get:
v = sqrt(2gh)
where g is the acceleration due to gravity. Substituting the given values, we get:
v = sqrt(2(9.81 m/s^2)(0.654 m)) = 3.78 m/s
Finally, we need to find the speed of the pendulum when it reaches an angle of 20.6 degrees above the vertical. At this point, the pendulum has a potential energy of:
PE = mgh' = mgh cos(20.6°)
where h' is the height of the pendulum at this point. To find h', we can use trigonometry:
h' = L(1 - cosθ')
where θ' is the angle above the vertical. Substituting the given values, we get:
h' = 1.06(1 - cos(20.6°)) = 0.242 m
Substituting the values for h' and solving for the kinetic energy, we get:
KE = PE_max - mgh' = mgh - mgh'
Substituting the known values, we get:
KE = (1 kg)(9.81 m/s^2)(0.654 m) - (1 kg)(9.81 m/s^2)(0.242 m) = 5.11 J
Now, we can solve for the speed at this point:
KE = (1/2)mv^2
5.11 J = (1/2)(1 kg)v^2
v = sqrt((2)(5.11 J)/(1 kg)) = 3.02 m/s
Therefore, The pendulum is moving at a speed of 3.02 m/s when it reaches an angle of 20.6 degrees above vertical.
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Example 9:
3.
Figure 5.24 shows a barrel of weight 1500 N
and radius 0.5 m that rests against a step of
height 0.2 m.
0.2 m
0.5 m
▲ Figure 5.24
(
What is the smallest horizontal force F th
the centre O needed to push the barrel over
the step
To drive the barrel over the step, the least horizontal force F necessary is 2366.16 N.
How to find horizontal force?To push the barrel over the step, the minimum force required should overcome the force of gravity acting on the barrel and the force of static friction between the barrel and the surface.
The perpendicular component of the weight is given as N = mgcosθ, where m = mass of the barrel,
g = acceleration due to gravity, and
θ = angle between the weight and normal to the surface.
In this case, θ as the inverse tangent of the ratio of the height of step to the distance from the edge of the step to the center of the barrel:
θ = tan⁻¹(0.2/0.5) = 0.39 radians
Therefore, the normal force is N = (1500)(9.81)cos(0.39) = 1443.6 N.
The force of static friction can be found as f = μsN,
where μs = coefficient of static friction.
Assume the coefficient of static friction between the barrel and the surface is 0.6.
f = (0.6)(1443.6) = 866.16 N.
The minimum force required to push the barrel over the step should overcome both these forces. Then, the smallest horizontal force that can push the barrel over the step is:
F = force of gravity + force of static friction
F = 1500 + 866.16
F = 2366.16 N.
Therefore, the smallest horizontal force F required to push the barrel over the step is 2366.16 N.
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When a massive star, much bigger than our sun, reaches the end of its life cycle, it will expand into a red supergiant and then:
A. lose its corona and enter a second stable phase as a star again.
B. explode into a supernova.
C. collapse into a protostar.
D. collapse into a white dwarf.
Answer:
When a massive star, much larger than our sun, reaches the end of its life cycle, it will undergo a series of fusion reactions in its core until it forms iron, which cannot undergo further fusion. Without fusion to counteract the force of gravity, the core collapses in on itself, causing the outer layers of the star to rapidly expand and creating a red supergiant. Eventually, the outer layers of the star will be expelled in a supernova explosion, leaving behind either a neutron star or a black hole, depending on the mass of the original star. So, the correct answer is B.
Problem 2.3. (5 pts) A 0.500-kg cart connected to a light spring for which the force constant is 20.0 N/m oscillates on a frictionless, horizontal air track. (a) Calculate the maximum speed of the cart if the amplitude of the motion is 3.00 cm. (b) What is the velocity of the cart when the position is 2.00 cm? (c) Compute the kinetic and potential energies of the system when the position of the cart is 2.00 cm
The maximum speed of the cart if the amplitude of the motion is 3.00 cm is 0.036 m/s
The velocity of the cart when the position is 2.00 cm is 0.1414 m/s.
The kinetic and potential energies of the system when the position of the cart is 2.00 cm is 5×10⁻³ J and 4×10⁻³ J resp.
a) To find maximum speed potential energy of the spring gets converted into kinetic energy of the cart in the oscillator motion, Hence,
1/2mv² =1/2kA²
1/2×0.5×v² = 1/2 ×20× 0.03²
v² = 40×9×10⁻⁴
v = 0.036 m/s
b) For the velocity at a given point is calculated by the formula,
v = ±ω√(A² - x²)
v = ±ω√(0.03² - 0.02²)
v = ±ω × 0.0223
v = ±√k/m × 0.0223
v = ±√20/0.5× 0.0223
v = 0.1414 m/s
c)
kinetic energy of the system,
K = 1/2 mv²
K = 1/2 ×0.5×0.1414²
K = 5×10⁻³ J
Potential energy of the system
P = 1/2 kx²
P = 1/2 × 20× 0.02²
P = 4×10⁻³ J
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14. Neglecting air resistance, what maximum height will be reached by a stone thrown straight up with an initial speed of 35 m/s?
(a) 98 m
(b) 18 m
(c) 160 m
(d) 63 m
Answer:
D
Explanation:
The maximum height reached by a stone thrown straight up with an initial speed of 35 m/s can be found using the kinematic equation:
v^2f = v^2i - 2gh
where vf is the final velocity (0 m/s at the maximum height), vi is the initial velocity (35 m/s, the magnitude of the velocity with which the stone is thrown upwards), g is the acceleration due to gravity (-9.8 m/s^2), and h is the maximum height reached by the stone.
Rearranging the equation, we get:
h = (vi^2)/(2g)
Substituting the given values, we have:
h = (35 m/s)^2 / (2 * 9.8 m/s^2)
= 62.6 m
Therefore, the maximum height reached by the stone is approximately 63 m.
The answer is (d).
If I want to convert 6,785 mg into grams. I would need to move the decimal places to the
You divide by 1000 so it would be 6.785g
A ball (0.410 kg) is kicked at an angle of 44.0° above the horizontal axis (above the +x-axis). The initial speed of the ball is 24.2 m/s. Ignoring air resistance, determine the momentum of the ball just before it hits the ground.
Answer:
9.922
Explanation:
When ignoring the air resistance and the ball is kicked and fell into the same plane. the initial speed is the speed when it is stricken on the ground. the initial momentum magnitude equals the final momentum magnitude only the direction is changed.
∴ Momentum= mass*velocity
= o.41kg*24.2
= 9.922[tex]kgms^{-1[/tex]
lucy finished 1/4 of her homework at an average speed of 15 questions per hour then she finished the remaining 45 questions at another speed. if the total time spent on the homework was 2.5 hours, what was the amount of time she spent on the remaining 45 questions
Therefore, Lucy spent 1.875 hours on the remaining 45 questions.
How is the amount of time she spent on the remaining 45 questions?Let's start by finding the total number of questions in Lucy's homework. If she finished 1/4 of her homework at an average speed of 15 questions per hour, then the total number of questions must be:
1/4 x Total number of questions = Number of questions finished at 15 questions per hour
1/4 x Total number of questions = 15 questions per hour
Solving for the total number of questions, we get:
Total number of questions = 60 questions
Now we know that Lucy finished 60 - 45 = 15 questions at an average speed of x questions per hour. Let's use the formula:
time = distance / speed
to find the amount of time she spent on the remaining 45 questions.
For the first part of the homework, Lucy spent:
time = distance / speed
time = 15 questions / hour
time = 1/4 x 2.5 hours
time = 0.625 hours
So, she spent 0.625 hours on the first 15 questions.
For the remaining 45 questions, we have:
time = distance / speed
time = 45 questions / x questions per hour
We know that the total time spent on the homework was 2.5 hours, so:
0.625 hours + 45 questions / x questions per hour = 2.5 hours
Solving for x, we get:
x = 18 questions per hour
Now we can use this speed to find the time spent on the remaining 45 questions:
time = distance / speed
time = 45 questions / 18 questions per hour
time = 2.5 hours - 0.625 hours
time = 1.875 hours
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String 1 in the figure has linear density 2.60 g/m and string 2 has linear density 3.30 g/m. A student sends pulses in both directions by quickly pulling up on the knot, then releasing it. She wants both pulses to reach the ends of the strings simultaneously.
What should the string length L1 be?
What should the string length L2 be?
Answer:
Here is your answer please change up some words to remain plagraism free.
Explanation:
To determine the required lengths of strings 1 and 2 so that pulses sent in both directions reach the ends of the strings simultaneously, we need to apply the principle that the time it takes for a wave pulse to travel a distance on a string is equal to the distance divided by the wave speed.
The wave speed, in turn, is determined by the tension in the string and the linear density of the string according to the formula:
v = sqrt(T/μ),
where v is the wave speed, T is the tension, and μ is the linear density.
Let L1 be the length of string 1 and L2 be the length of string 2. Since the wave speed is the same for both strings, we can set up the following equations:
L1/v = L2/v
sqrt(T1/μ1)*L1 = sqrt(T2/μ2)*L2
where T1 and T2 are the tensions in strings 1 and 2, respectively.
We can solve for L1 and L2 by combining these two equations and solving for each variable. Substituting the given linear densities of strings 1 and 2, we get:
sqrt(T1/2.60)*L1 = sqrt(T2/3.30)*L2
Squaring both sides and simplifying, we get:
(T1/T2) = (3.30/2.60) * (L1/L2)^2
Substituting the condition that the pulses reach the ends of the strings simultaneously, we know that the total time for a pulse to travel down string 1 and back up to the knot is equal to the time for a pulse to travel down string 2 and back up to the knot. This condition implies that the total length of string 1 (2L1) is equal to the total length of string 2 (2L2):
2L1 = 2L2
Solving this equation for L2 and substituting it into the expression for T1/T2 derived above, we get:
T1/T2 = (3.30/2.60) * (L1/2L1)^2 = 1.25
Solving for L1, we obtain:
L1 = sqrt(T1/μ1) * (2L2/v) = sqrt((1.25)*(2.60/3.30)) * (2L2)
Simplifying this expression, we get:
L1 = (2/3) * sqrt(2.60/3.30) * L2
Therefore, the required length of string 1 is (2/3) * sqrt(2.60/3.30) times the length of string 2. We can substitute the given length of string 2, say L2 = 1 meter, into this expression to obtain the required length of string 1:
L1 = (2/3) * sqrt(2.60/3.30) * 1 meter ≈ 0.693 meter.
Therefore, the required length of string 1 is approximately 0.693 meter and the required length of string 2 is 1 meter.
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What month is the speed of Earth the fastest?
January
March
June
Decemember
Answer:
July 29, Earth completed a full spin in about 1.59 milliseconds shorter than its standard timeframe ( 23 hours and 56 minutes).
Penguins in Gold Harbour love to communicate with other members of their penguin family. Here in Sanford, the speed of sound in air is about 344.0 m/s. Calculate the speed of sound (in m/s) in Gold Harbour, on a day when the air temperature is -2.7 °C.
Round to the nearest hundredth.
Please show all work!!!!
The speed of sound in Gold Harbour, on a day when the air temperature is -2.7 °C, is 331.5 + 0.6 * (-2.7) = 328.65 m/s.
What is Gold Harbour?Gold Harbour is a small settlement located on Antarctic's King George Island. It is home to a Chilean research base, which is operated by the Chilean Antarctic Institute. It also acts as a summer base for the Chilean Navy, and provides support for the scientific research conducted by other countries, including the United States, United Kingdom, and Russia. The area is known for its stunning natural beauty, with mountains, glaciers, and icebergs all in close proximity. It is also an important habitat for several species of wildlife, including penguins, seals, and sea birds.
The speed of sound in air is affected by temperature, and the formula for calculating the speed of sound in air is v = 331.5 + 0.6 * (air temperature in °C).
Therefore, the speed of sound in Gold Harbour, on a day when the air temperature is -2.7 °C, is 331.5 + 0.6 * (-2.7)
= 328.65 m/s.
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Which one of the following is the longest length?
(a) 100 m
(b) 104 µm
(c) 107 nm
(d) 102 mm
Okay, let's convert all the lengths to the same unit to compare:
(a) 100 m = 100 meters
(b) 104 μm = 104 micrometers = 104 × 10^-6 meters = 0.000104 meters
(c) 107 nm = 107 nanometers = 107 × 10^-9 meters = 0.000000000997 meters
(d) 102 mm = 102 millimeters = 102 × 10^-3 meters = 0.0102 meters
The longest length is:
(a) 100 m = 100 meters
The answer is option (a).
Answer: 100 m
Explanation:
1 μm = [tex]10^{-6}[/tex] m = 0,000001 m
1 nm = [tex]10^{-9}[/tex] m = 0,000000001 m
1 mm = [tex]10^{-3}[/tex] m = 0,001 m
∴ 100 m es la mayor longitud
Diagram A shows a negatively charged conducting rod placed near a light polystyrene ball that is suspended from the ceiling by an insulating thread .Diagram B shows what happens when the ball touches the rod. (a) Explain why the ball is displaced vertically in Diagram A (b)Explain what happens after the ball has been allowed to touch the rod (c)Give a reason why the ball has to be coated with a conducting material such as graphite (d) Explain why the polystyrene ball is suspended by an insulated thread and not by a conducting wire
When a negatively charged conductive rod is brought close to a lightweight Styrofoam ball suspended from the ceiling by insulating threads (see Figure A), the electrons in the ball are repelled by the rod's negative charge.
What happens when the ball hits the pole?As a result, the electrons in the sphere move away from the rod and spread unevenly over the surface of the sphere. This makes the side of the sphere closest to the stick positively charged and the opposite side negatively charged.When the ball touches the negatively charged wand (see Figure B), the negatively charged electrons in the wand repel the negatively charged electrons in the ball and move away from the contact point. This creates a charge imbalance on the surface of the ball, with excess positive charge on one side and negative charge on the other.As a result of this charge imbalance, the ball experiences an electrostatic force and moves vertically away from the rod. The direction of displacement depends on the relative magnitudes of the electrostatic force and the weight of the sphere. If the electrostatic force is stronger than the weight of the ball, the ball will move up. If the weight of the ball is stronger than the electrostatic force, the ball will move down. In both cases the ball moves vertically.For more information on electrostatic force kindly visit to
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Iodine-131 has a half-life of 8 days. How many grams of a 256 g sample would remain at the end of 56 days?
Answer:
Explanation:
The decay of a radioactive substance is governed by the formula:N(t) = N₀ e^(-λt)where N₀ is the initial amount of the substance, N(t) is the amount remaining after time t, and λ is the decay constant.The half-life of Iodine-131 is 8 days, which means that after each 8-day period, the amount remaining will be reduced by half. We can use this fact to calculate the amount remaining after 56 days.First, we need to find the decay constant λ, which is related to the half-life by the formula:λ = ln(2) / t½where ln(2) is the natural logarithm of 2, and t½ is the half-life.Substituting the values we have:λ = ln(2) / 8 days ≈ 0.08664 day^(-1)Next, we can use the formula for N(t) to calculate the amount remaining after 56 days:N(56) = N₀ e^(-λt) = 256 g e^(-0.08664 day^(-1) × 56 days) ≈ 22.6 gTherefore, approximately 22.6 grams of the original 256 gram sample would remain after 56 days.
A displacement vector is 23 km in length and directed 65° south of east. What are the components of this vector?
Eastward
Component Southward. Component
(a) 21 km 9.7 km
(b) 23 km 23 km
(c) 23 km 0 km
(d) 9.7 km 21 km
If a displacement vector is 23 km in length and directed 65° south of east. the components of this vector is: (a) 21 km 9.7 km.
What is the components of this vector?The displacement vector can be resolved into its eastward and southward components using trigonometry. Let's call the eastward component "x" and the southward component "y".
From the given information, we know that the displacement vector makes an angle of 65° south of east. This means that the angle between the vector and the eastward axis is 90° - 65° = 25°.
Using trigonometry, we can relate the length of the vector to its components:
cos 25° = x / 23
sin 25° = y / 23
Solving for x and y, we get:
x = 23 cos 25° ≈ 21 km
y = 23 sin 25° ≈ 9.7 km
Therefore, the answer is (a) 21 km eastward component and 9.7 km southward component.
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The cylinder of a heat engine is filled with an air-fuel mixture. Which property of gases is essential to heat engines ability to do work?
Answer: Pressure of gases.
Explanation:
In a heat engine, the air-fuel mixture is ignited, which causes an increase in pressure of the gases inside the cylinder. This pressure pushes the piston, which is connected to a crankshaft, causing it to rotate and do work. Therefore, pressure is an essential property of gases for the ability of a heat engine to do work.
Sanjay and Ting, each with a mass of 25 kg, are riding opposite each other on the edge of a 150 kg, 3.0-m-diameter playground merry-go-round that's rotating at 15 rpm. Each walks straight inward and stops 35 cm from the center.
What is the new angular velocity, in rpm?
Express your answer in revolutions per minute.
The merry-go-round's new angular velocity is 0.321 rpm.
Calculation-
the system's overall angular momentum is:L = Iω
The moment of inertia of a solid disk rotating about its centre is given by:
[tex]I = (1/2)mr^2I = (1/2)(150 kg)(1.5 m)^2 = 168.75 kg·m^2[/tex]
initial angular momentum of the system
L = Iω = [tex](168.75 kg·m^2)(15 rpm)(2π/60 s) = 52.36 kg·m^2/s[/tex]
The new angular velocity is determined by:
L = I'ω'
where L represents the system's initial angular momentum.
The system's new moment of inertia is:
[tex]I' = I - 2mr^2[/tex]
we get:
[tex]I' = 25 kg - 2(168.75 kg/m2)(0.35 m)^2 = 162.88 kg·m^2[/tex]
We get the following by substituting into the conservation of angular momentum equation:
L = I'ω'
[tex](162.88 kg/m2) / 52.36 kg/m2[/tex]
ω' = 0.321 rpm
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which system best illustrates attractive forces
Answer:
A
Opposite poles will attract
A car starting from rest has an acceleration 0.5m/s after 1 minute then what will be the final velocity of the car
Answer:
The final velocity is 30 m/s
Explanation:
Use the formula:
[tex]V_{f} = V_{i} +a*t\\V_{f} = 0 + 0.5*60\\V_{f} = 30 m/s[/tex]
22. How do we correct the issue of flipped imagery caused by mirrors?
To correct the issue of flipped imagery caused by mirrors you can use a technique called "mirror flipping". Mirror flipping involves using a second mirror to reflect the reflected image from the first mirror which then flips it back to its original orientation
Alternatively you can use a prism to correct the orientation of the image. A prism is a transparent object that can bend light. By placing it infront of the mirror you can reflect the twice, effectively helping in the correction of the orientation of the image
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energy transferred from one thing to another when the gulf balls collide?
This is based on the principle of conservation of energy and momentum.
When the collision of golf balls takes place the energy gets transferred from one ball to the other.
The golf balls experience a force that causes them to change their form and also direction. During the collision, the mechanical energy is converted into heat, sound, and other forms of energy. The rest of energy is used to move in a new direction.
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ray of light exits from a metal with a refractive index of 1.75 travelling to the air the angle of refraction is 25°. what is the angle of deviation
Answer:
Explanation:
To find the angle of deviation, we can use the formula:
angle of deviation = (refractive index of metal - refractive index of air) x angle of incidence
The angle of incidence can be calculated using Snell's law:
sin(angle of incidence) / sin(angle of refraction) = refractive index of air / refractive index of metal
sin(angle of incidence) / sin(25°) = 1 / 1.75
sin(angle of incidence) = 0.5714
angle of incidence = 34°
Now we can substitute this value into the formula for angle of deviation:
angle of deviation = (1.75 - 1) x 34°
angle of deviation = 21°
Therefore, the angle of deviation is 21°.
An overtone is louder than the fundamental tone.
O True
False
The given statement, "An overtone is louder than the fundamental tone" is False.
An overtone is a higher-frequency vibration that occurs simultaneously with the fundamental frequency of a sound wave. These vibrations are multiples of the fundamental frequency and contribute to the overall timbre or tone quality of a sound.
In some cases, overtones can be louder than the fundamental tone, depending on the specific harmonic series present in the sound wave. This phenomenon is known as overtone prominence, and it can be heard in certain musical instruments like bells or cymbals, where the higher harmonics of the sound are emphasized.
However, in most cases, the fundamental tone is perceived as the strongest and most dominant sound in a given sound wave.
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A car is traveling at a speed of 30m/s and leaves the ramp at a 37 degree angle. What is the total hang time of the car?
Explanation:
INITIAL Vertical velocity is given by
30 m/s * sin 37
then gravity begins to slow it down
30 sin37 - (9.81) t = vertical velocity at t
when v = 0 , the car is at its apex and will fall back down in the same amount of time
0 = 30 sin37 - 9.81 t shows t = 1.84 seconds to peak
then another 1.84 seconds to fall to the ground total = 3.7 seconds
Which of the following sets of two charges is experiencing the strongest
attraction?
Charges of +2 C and -2 C, separated by 1 m.
Charges of +1 C and -3 C, separated by 1 m.
Charges of +2 C and +2 C, separated by 1 m.
Charges of +1 C and +3 C, separated by 1 m.
The force of attraction between two charges is proportional to the product of the charges and inversely proportional to the square of the distance between them.
What is a square ?A square is a two-dimensional geometric shape that has four sides of equal length and four right angles (90-degree angles) between them. The sides of a square are parallel to each other and perpendicular to its adjacent sides. All four corners of a square are also known as vertices, and the diagonals of a square bisect each other at right angles.
The area of a square is calculated by multiplying the length of one of its sides by itself. The perimeter of a square is calculated by adding up the lengths of all four sides. The properties of a square make it useful in various applications, such as in geometry, architecture, and construction.
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As part of a movie stunt, a full-size remote-controlled car is driven horizontally off a 9.00 m tall cliff at 24.40 m/s. How far (Δx) from the bottom of the cliff does the car land?
Explanation:
Find the time it takes to hit the bottom....then multiply this time by the horizontal velocity .......
Time to hit bottom :
d = 1/2 at^2
9 m = 1/2 (9.81 m/s^2) (t^2) shows t = 1.35 s
Now the horizontal displacement is
x = rate * time = 24.40 m/s * 1.35 s = 33.1 m
15 A car of mass 750 kg is accelerating up a slope of a certain angle to the horizontal where sin theta = 1/70 at 1.5 m/ s². Ignoring any road resistance, find the tractive force of the engine.
The tractive force of the engine is 1020 N.
Mass of the car, m = 750 kg
Acceleration of the car, a = 1.5 m/s²
Traction, also known as tractive force, is the force used to produce or create motion by using dry friction between a body and an inclined surface.
Weight of the car, W = mg
W = 750 x 9.8
W = 7350
The coefficient of friction is the ratio or percentage of the opposing frictional force to the normal force pressing the two surfaces into contact and motion.
The tractive force of the engine,
F' = ma - mg sinθ = m(a - g sinθ)
F' = 750[1.5 - (9.8/70)]
F' = 750 x 1.36
F' = 1020 N
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What are weird properties of quasars that made them difficult for astronomers to understand?
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Supermassive black holes that are devouring gas at the center of far-off galaxies are known as quasars.
Since, the quasars were initially identified by astronomers in 1963 as objects that resembled stars but gave off radio waves instead, the term quasar is an abbreviation for quasi-stellar radio source.
Quasars are so bright that they drown out the light from all other stars in the same galaxy. Quasars give off radio waves, X-rays, gamma-rays, ultraviolet rays, and visible light across the entire electromagnetic spectrum. Most of them are larger than our solar system.
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What happens as a protostar contracts?
A. More hydrogen is produced to become fuel for the star.
B. The temperature rises.
C. The hydrogen fuses into iron.
D. The temperature drops.
Answer:
As a protostar contract, the temperature rises. This increase in temperature leads to the initiation of nuclear fusion reactions, where hydrogen atoms fuse together to form helium, releasing energy in the process. This energy causes the protostar to heat up and begin to emit light, eventually becoming a stable star.
The answer is B. The temperature rises.
A ball of mass 2 kg is moving with a velocity of 12m/s collides with a stationary ball of mass 6 kg and comes to rest. Calculate velocity of ball of mass 6kg after collision.
Answer: 4 m/s
Explanation:
To solve this problem, we can use the principle of conservation of momentum, which states that the total momentum of a closed system remains constant before and after a collision.
Let's denote the initial velocity of the 2 kg ball as "v1i", the initial velocity of the 6 kg ball as "v2i", the final velocity of the 2 kg ball as "v1f", and the final velocity of the 6 kg ball as "v2f".
According to the principle of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. Mathematically, this can be expressed as:
m1 * v1i + m2 * v2i = m1 * v1f + m2 * v2f
where m1 is the mass of the 2 kg ball, m2 is the mass of the 6 kg ball, v1i and v2i are the initial velocities, and v1f and v2f are the final velocities.
Given:
m1 = 2 kg
m2 = 6 kg
v1i = 12 m/s (initial velocity of the 2 kg ball)
v2i = 0 m/s (initial velocity of the 6 kg ball, as it is stationary)
v1f = 0 m/s (final velocity of the 2 kg ball, as it comes to rest)
Plugging in the given values into the conservation of momentum equation:
2 * 12 + 6 * 0 = 2 * 0 + 6 * v2f
24 = 6 * v2f
Dividing both sides by 6:
v2f = 24 / 6 = 4 m/s
So, the velocity of the 6 kg ball after the collision is 4 m/s.