The minimum slit width possible is 1.15 mm
How to find what minimum slit width this is possible?To find this double slit experiment we will use the equation:
sin(theta) = m × lambda / d
where:
theta = angle of the dark fringem = order of the dark fringe (1 for the first dark fringe, 2 for the second dark fringe, etc.)lambda = wavelength of the lightd = distance between the slitWe are given that theta = 1.72 mrad, m = 1 for 600 nm light, and m = 2 for 500 nm light. We can solve for d in each case:
d = 600 nm × sin(1.72 mrad) / 1 = 2.44 mm
d = 500 nm × sin(1.72 mrad) / 2 = 1.15 mm
We can see that the minimum slit width possible is 1.15 mm
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An optical fiber uses flint glass surrounded by a crown glass cladding.
What is the critical angle for total internal reflection in degrees?
Given: Glass, crown n = 1.52; glass, flint n = 1.66
air n = 1.00
The critical angle for total internal reflection in an optical fiber composed of flint glass surrounded by a crown glass cladding can be determined using the given refractive indices.
The critical angle for total internal reflection occurs when light traveling through a medium encounters a boundary with a lower refractive index. In this case, the light travels from the flint glass (n = 1.66) to the crown glass cladding (n = 1.52), with the surrounding medium being air (n = 1.00).
To calculate the critical angle, we can use the formula sin(critical angle) = n2 / n1, where n1 is the refractive index of the medium the light is coming from (flint glass) and n2 is the refractive index of the medium the light is entering (crown glass cladding).
Plugging in the values, sin(critical angle) = 1.52 / 1.66. To find the critical angle itself, we take the inverse sine ([tex]sin^(^-^1^)[/tex]) of the resulting value: critical angle = [tex]sin^(^-^1^)(1.52 / 1.66)[/tex]. By calculating this value, we can determine the critical angle in degrees.
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A 1. 5-kilogram cart initially moves at 2. 0 meters
per second. It is brought to rest by a constant net
force in 0. 30 second. What is the magnitude of
the net force?
(1) 0. 40 N (3) 10. N
(2) 0. 90 N (4) 15 N
The initial velocity of a 1.5-kilogram cart is 2.0 meters per second. It is brought to a stop in 0.30 seconds by a constant net force.To solve this problem, you must first recall the formula F = ma, where F is force, m is mass, and a is acceleration.
We can rearrange the formula to solve for force as follows:F = maWhere F is force, m is mass, and a is acceleration.We can use the formula to solve for force since we know the mass of the cart and its acceleration.First, we must calculate the acceleration, which can be found using the formula a = Δv / Δt, where a is acceleration, Δv is the change in velocity, and Δt is the time taken. We can substitute the values that we know into the equation:
a = Δv / Δt= (0 m/s - 2 m/s) / 0.30 sa = -2/0.3a = -6.67 m/s²
We obtained a negative acceleration because the velocity of the cart decreases during the time interval of 0.30 seconds.To determine the net force on the cart, we can now use the formula F = ma:F = ma= 1.5 kg x -6.67 m/s²= -10.0 N (approximately)Therefore, the magnitude of the net force on the cart is 10 N. Answer (3) is correct.
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When you eat food, not all of the food can be broken down into the basic building blocks and why?
Answer:
cause you crazy..
Explanation:
The device shown below contains 2 kg of water. When the cylinder is allowed to fall 250 m, the temperature of the water increases by 1.4°C. Suppose 2 kg of water are added to the container and the cylinder is allowed to fall 750 m. What would the increase in temperature be in this case? Gizmo image A. 0.7°C B. 1.4°C C. 2.1°C D. 2.8°C
Answer:c. 2.1°C
Explanation:
I just did it
A design team is developing a prototype CO2 cartridge for a manufacturer of rubber rafts. This cartridge will allow a user to quickly inflate a raft. A typical raft is shown in the sketch. Assume a raft inflation pressure of 3 p, which means that the absolute pressure is 3 p greater than local atmospheric pressure). Estimate the volume of the raft and the mass of CO2 in grams in the prototype cartridge.
Length: 6 m
Width: 2 m
Tube diameter: 0.4 m
The estimated volume of the raft is 4.8 cubic meters (m³). Without knowing the pressure and volume of the CO2 cartridge, we cannot accurately estimate the mass of CO2 in grams.
To estimate the volume of the raft, we can assume it has a simple rectangular shape. The volume (V) of a rectangular object is calculated by multiplying its length (L), width (W), and height (H):
V = L * W * H
Given:
Length (L) = 6 m
Width (W) = 2 m
Tube diameter (H) = 0.4 m (assuming it represents the height of the raft)
V = 6 m * 2 m * 0.4 m = 4.8 m³
Next, let's estimate the mass of CO2 in grams in the prototype cartridge. To do this, we need to know the pressure and volume of the CO2 gas. However, the provided information does not specify the pressure or volume of the cartridge. Without these values, it is not possible to accurately estimate the mass of CO2 in grams.
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find the speed the block has as it passes through equilibrium (for the first time) if the coefficient of friction between block and surface is k = 0.350.
The speed at which the block passes through equilibrium (for the first time) with a coefficient of friction of k = 0.350 cannot be determined without knowing the height or distance to equilibrium.
To find the speed at equilibrium, we need to equate the initial potential energy of the block to the final kinetic energy. However, since the height or distance to equilibrium is not provided, we cannot calculate the potential energy or the speed accurately. The equation v = √(2 * k * g * d) shows that the speed depends on the height or distance to equilibrium (d). Without this information, we cannot determine the speed. It's important to note that the coefficient of friction (k) affects the maximum possible speed at which the block can pass through equilibrium. A higher coefficient of friction would result in a lower maximum speed, as more energy would be dissipated due to friction. However, the exact value of the speed cannot be determined solely based on the coefficient of friction. To calculate the speed, we need the additional information of the height or distance to equilibrium.
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GIVING BRAINLIEST PLEASE HELP!!
-if you answer correctly ill give you brainliest which will give you 27pts-
Answer:
C. The lever applies three times more force than you hand can apply.
Explanation:
Since it's advantage is 3, that means you'll have to multiply the input of it by 3, making this apply 3x more force than your hand.
Hope this helped! <3
Source(s): Me and a bit of g*ogle for clarification
Light is polarized by using:
Answer:
Polaroid fliter
Explanation:
light can be polarized by using Polaroid filters
Polaroid fliter are made of special material that is capable of blocking one of the two planes of vibration of an electromagnetic wave
hope this is useful--(have a good day)
Energy that comes from the heat inside the Earth is called ________ energy.
Answer:
Geothermal Energy.
Explanation:
The 500-N force F is applied to the vertical pole as shown(1) Determine the scalar components of the force vector F along the x'- and y'-axes. (2) Determine the scalar components of F along the x- and y'-axes.
Solution :
Given :
Force, F = 500 N
Let [tex]$ \vec F = F_x\ \hat i + F_y\ \hat j$[/tex]
[tex]$|\vec F|=\sqrt{F_x^2+F_y^2}$[/tex]
∴ [tex]$F_x=F \cos 60^\circ = 500 \ \cos 60^\circ = 250 \ N$[/tex]
[tex]$F_y=-F \cos 30^\circ = -500 \ \cos 30^\circ = -433.01 \ N$[/tex] (since [tex]$F_y$[/tex] direction is in negative y-axis)
[tex]$F=250 \ \hat i - 433.01 \ \hat j$[/tex]
So scalar components are : 250 N and 433.01 N
vector components are : [tex]$250 \ \hat i$[/tex] and [tex]$-433.01\ \hat j$[/tex]
1. Scalar components along :
x' axis = 500 N, since the force is in this direction.
[tex]$F_{x'}= F \ \cos \theta = 500\ \cos \theta$[/tex]
Here, θ = 0° , since force and axis in the same direction.
So, cos θ = cos 0° = 1
∴ [tex]$F_{x'}=500 \times 1=500\ N$[/tex]
[tex]$F_{y'}= F \ \sin \theta = 500\ \sin 0^\circ=500 \times 0=0$[/tex]
[tex]$F_{y'}=F\ cos \theta$[/tex] but here θ is 90°. So the force ad axis are perpendicular to each other.
[tex]$F_{y'}=F\ \cos 90^\circ= 500 \ \cos 90^\circ = 500 \times 0=0$[/tex]
∴ [tex]$F_{x'}= 500\ N \text{ and}\ F_{y'}=0\ N$[/tex]
2. Scalar components of F along:
x-axis :
[tex]$F_x=F\ \cos \theta$[/tex], here θ is the angle between x-axis and F = 60°.
[tex]$F_x=500 \times \cos60^\circ=250\ N$[/tex]
y'-axis :
[tex]$F_{y'}=F\ \cos \theta$[/tex], here θ is the angle between y'-axis and F = 90°.
[tex]$F_{y'}=500 \times \cos90^\circ=500\times 0=0\ N$[/tex]
∴ [tex]$F_{x}= 250\ N \text{ and}\ F_{y'}=0\ N$[/tex]
you hear the sound of a distance cannon 2.0 s after seeing a flash of light from it. how far you from the cannon?
You are approximately 686 meters away from the cannon.
To determine the distance between you and the cannon, we can use the speed of sound as a reference.
The speed of sound in air at room temperature is approximately 343 meters per second (m/s).
Given that you hear the sound of the cannon 2.0 seconds after seeing the flash of light, this time delay represents the time it took for the sound to travel from the cannon to your location.
We can use the formula:
Distance = Speed * Time
Distance = 343 m/s * 2.0 s
Distance = 686 meters
Therefore, you are approximately 686 meters away from the cannon.
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Why might a scientist use a magnet and small strainer when investigating the physical properties of a substance?
Answer:
In order to check its magnetic properties and for removing impurities.
Explanation:
A scientist use a magnet and small strainer while investigating the physical properties of a substance in order to check the magnetic properties of the substance as well as to separate the substance from the other impurities. Magnet attract substances that is made of metals or magnetic characteristics while on the other hand, small strainer is used to separate impurities from the investigating substance so that's why the scientist use a magnet and small strainer during investigation of a substance.
A scientist use a magnet to check magnetic properties of the substance and use small strainer to separate the substance from the other impurities.
The magnet have Attractive Property , Magnet attracts ferromagnetic materials like iron, cobalt, and nickel.Magnet attract substances that is made of metals or magnetic characteristics .The small strainer is used to separate impurities from the investigating substanceLearn more:
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A woman uses a pulley and a rope to raise a 12 weight to a height of 3 . If it takes 4 to do this, about how much power is she supplying? a. 90 b. 190 c. 290 d. 390
The woman is supplying approximately 290 units of power to raise the weight using a pulley and a rope.
Power is calculated using the formula: Power = Work/Time. In this case, the work done is equal to the weight lifted multiplied by the height gained, which is 12 units * 3 units = 36 units of work. The time taken to perform this work is given as 4 units of time.
Therefore, the power supplied can be calculated as 36 units of work divided by 4 units of time, resulting in 9 units of power. However, the answer options provided do not match this calculation.
To determine the correct answer, we need to convert the given units to match the answer options. Since the units of work and time are not specified, we can assume they are arbitrary units. Given that, we can multiply the calculated power (9 units) by a conversion factor to match the answer options. The closest option is 290, so the correct answer is option c. The woman is supplying approximately 290 units of power to raise the weight.
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Solved Exa
Example 1. An iron ball of mass 3 kg is
released from a height of 125 m and falls
freely to the ground. Assuming that the
value of g is 10 m/s2, calculate
(i) time taken by the ball to reach the
ground
(ii) velocity of the ball on reaching the
ground
(iii) the height of the ball at half the time it
takes to reach the ground.
According to the equations of motion, the time taken to reach the ground is 5 seconds.
Using;
s = ut + 1/2gt^2
s = distance
u = initial velocity
t = time taken
g = acceleration due to gravity
Note that u = 0 m/s since the object was dropped from a height
Substituting values;
125 = 1/2 × 10 × t^2
125 = 5t^2
t^2 = 125/5
t^2 = 25
t = 5 secs
Velocity on reaching the ground is obtained from
v = u + gt
Where u = 0 m/s
v = gt
v = 10 × 5
v = 50 m/s
At half the time it takes to reach the ground;
s = ut + 1/2gt^2
Where u = 0 m/s
s = 1/2gt^2
s = 1/2 × 10 × (2.5)^2
s = 31.25 m
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Answer:
(i) time taken by the ball to reach the
ground is 5 sec.
(ii) velocity of the ball on reaching the
ground is 50 m/s.
(iii) the height of the ball at half the time it
takes to reach the ground is 31.25 m.
Step-by-step explanation:
Solution :(i) time taken by the ball to reach the
ground
[tex]\longrightarrow{\sf{ \: \: s= ut + \dfrac{1}{2} a{(t)}^2}}[/tex]
[tex]\longrightarrow{\sf{ \: \: 125= 0 \times t + \dfrac{1}{2} \times 10 \times {(t)}^2}}[/tex]
[tex]\longrightarrow{\sf{ \: \: 125= 0 + \dfrac{10}{2} \times {(t)}^2}}[/tex]
[tex]\longrightarrow{\sf{ \: \: 125= 0 + 5\times {(t)}^2}}[/tex]
[tex]\longrightarrow{\sf{ \: \: 125= 5\times {(t)}^2}}[/tex]
[tex]\longrightarrow{\sf{ \: \: {(t)}^2 = \dfrac{125}{5}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: {(t)}^2 = \dfrac{ \cancel{125}}{\cancel{5}}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: {(t)}^2 = 25}}[/tex]
[tex]\longrightarrow{\sf{ \: \: t = \sqrt{25} }}[/tex]
[tex]\longrightarrow \: \: {\sf{\underline{\underline{\red{ t = 5 \: sec}}}}}[/tex]
Hence, the ball taken 5 sec to reach the ground.
[tex]\begin{gathered}\end{gathered}[/tex]
(ii) velocity of the ball on reaching the
ground
[tex]\longrightarrow{\sf{ \: \: {v}^{2} - {u}^{2} = 2as}}[/tex]
[tex]\longrightarrow{\sf{ \: \: {v}^{2} - {0}^{2} = 2 \times 10 \times 125}}[/tex]
[tex]\longrightarrow{\sf{ \: \: {v}^{2} = 20 \times 125}}[/tex]
[tex]\longrightarrow{\sf{ \: \: {v}^{2} = 2500}}[/tex]
[tex]\longrightarrow{\sf{ \: \: {v} = \sqrt{2500} }}[/tex]
[tex]\longrightarrow{\sf{ \: \: \underline{\underline{ \red{{v} = 50 \: m/s }}}}}[/tex]
Hence, the velocity of ball is 50 m/s.
[tex]\begin{gathered}\end{gathered}[/tex]
(iii) the height of the ball at half the time it
takes to reach the ground.
[tex]\longrightarrow{\sf{ \: \: s= ut + \dfrac{1}{2} a{(t)}^2}}[/tex]
[tex]\longrightarrow{\sf{ \: \: s= 0 \times \dfrac{5}{2} + \dfrac{1}{2} \times 10 \times { \left( \dfrac{5}{2} \right)}^2}}[/tex]
[tex]\longrightarrow{\sf{ \: \: s= 0 + \dfrac{10}{2} \times { \left( \dfrac{5}{2} \times \dfrac{5}{2} \right)}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: s= \dfrac{10}{2} \times { \left( \dfrac{5 \times 5}{2 \times 2} \right)}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: s= \dfrac{10}{2} \times { \left( \dfrac{25}{4} \right)}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: s= \dfrac{10}{2} \times \dfrac{25}{4}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: s= \dfrac{10 \times 25}{2 \times 4}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: s= \dfrac{250}{8}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: s= \dfrac{\cancel{250}}{\cancel{8}}}}[/tex]
[tex]\longrightarrow{\sf{ \: \: {\underline{\underline{\red{s= 31.25 \: m}}}}}}[/tex]
Hence, the height of the ball to reach the ground is 31.25 m.
[tex]\underline{\rule{220pt}{3.5pt}}[/tex]
After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 53.0 cmcm . The explorer finds that the pendulum completes 102 full swing cycles in a time of 138 ss .
What is the magnitude of the gravitational acceleration on this planet?
The magnitude of the gravitational acceleration on this unfamiliar planet is approximately 18.31 m/s².
First, we need to calculate the period of one swing cycle (t) using the formula:
[tex]t = T / N \\t = 138 s / 102[/tex]
Next, we can calculate the time for one full swing (T_full) by dividing t by 2:
[tex]T_{full} = t / 2[/tex]
Now, we can use the formula for the period of a pendulum to solve for the gravitational acceleration (g):
T_full = 2π √(L / g)
Rearranging the formula to solve for g, we have:
g = (4π² L) / T_full²
Substituting the values:
g = (4π² * 53.0 cm) / (T_full)²
There are 100 cm in 1 m, so we divide the length by 100:
g = (4π² * 0.53 m) / (T_full)²
Now, we substitute the value of T_full obtained earlier:
g = (4π² * 0.53 m) / (t/2)²
Calculating t/2 and simplifying further, we have:
g = (4π² * 0.53 m) / [(138 s / 102) / 2]²
g = (4π² * 0.53 m) / [(138 s / 204)²]
g = (4π² * 0.53 m) / (0.676 s)²
Calculating the numerator, we get:
4π² * 0.53 m ≈ 8.365 m²/s²
Substituting this value, we have:
g ≈ 8.365 m²/s² / (0.676 s)²
Calculating the denominator, we get:
(0.676 s)² ≈ 0.456 s²
Finally, substituting the values and calculating, we find:
g ≈ 8.365 m²/s² / 0.456 s²
g ≈ 18.31 m/s²
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A snail is traveling along a straight path. The snail's velocity can be modeled by v(t) = 1.4 In (1 +r²) inches per minute for 0 ≤ 1 ≤ 15 minutes. (a) Find the acceleration of the snail at time t = 5 minutes. (b) What is the displacement of the snail over the interval 0 ≤ 1 ≤ 15 minutes?
The snail's acceleration at t = 5 minutes is approximately 0.079 inches per minute squared. Over the interval 0 ≤ t ≤ 15 minutes, the snail's displacement is approximately 15.405 inches.
To find the acceleration at t = 5 minutes, we need to differentiate the velocity function with respect to time. The given velocity function is v(t) = 1.4 ln(1 + r²), where ln denotes the natural logarithm. Let's differentiate v(t) with respect to t to find the acceleration function a(t):
a(t) = d/dt (1.4 ln(1 + r²))
To differentiate ln(1 + r²), we use the chain rule:
a(t) = 1.4 * d/dt (ln(1 + r²))
The derivative of ln(1 + r²) with respect to r² is 1 / (1 + r²), so we can rewrite the acceleration function as:
a(t) = 1.4 * (1 / (1 + r²)) * d/dt (1 + r²)
The derivative of 1 + r² with respect to t is 2r dr/dt. Substituting this back into the acceleration function, we get:
a(t) = 1.4 * (1 / (1 + r²)) * 2r dr/dt
Since we're evaluating the acceleration at t = 5 minutes, we substitute t = 5 into the expression and solve for the corresponding values of r and dr/dt. Then, we calculate the acceleration.
Now, to find the displacement over the interval 0 ≤ t ≤ 15 minutes, we integrate the velocity function with respect to time over that interval:
∫[0,15] (1.4 ln(1 + r²)) dt
By evaluating this definite integral, we obtain the displacement of the snail over the given time interval.
Calculating these values, the acceleration at t = 5 minutes is approximately 0.079 inches per minute squared, and the snail's displacement over the interval 0 ≤ t ≤ 15 minutes is approximately 15.405 inches.
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PLEASE HELP ME!!! THIS IS DO TODAY. AND NO LINKS PLEASE!!!!!
Answer:
to late
Explanation:
if 100 cm 3 of a gas with a density of 0.025 g/cm 3 condenses into 4.5 cm 3 of liquid, what is the density of the liquid? A. 1,125 g/mc3 B. 2,5 g/mc3 C. 0,56 g/mc3 D. 180 g/mc3
The required density of the liquid is 0.56 g/cm³. Option C is correct.
To find the density of the liquid, we can use the formula:
Density = Mass / Volume
Given that the initial gas has a density of 0.025 g/cm³ and condenses into 4.5 cm³ of liquid, we need to find the mass of the liquid to determine its density.
The initial volume of the gas is 100 cm³, and its density is 0.025 g/cm³. Therefore, the initial mass of the gas can be calculated as:
Mass of gas = Density of gas * Volume of gas
= 0.025 g/cm³ * 100 cm³
= 2.5 g
Since the gas condenses into 4.5 cm of liquid, the volume of the liquid is 4.5 cm³.
Now we can find the density of the liquid:
The density of liquid = Mass of liquid / Volume of liquid
= 2.5/4.5 = 0.56 g/mc³
Therefore, the required density of the liquid is 0.56 g/cm³.
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Measurement of density contrasts the mass of an object with its volume. High density refers to the amount of matter in a given volume of an object. Here the density of the liquid is 0.55 g/cm³. The correct option is C.
The density of a substance indicates how dense it is in a given area. Mass per unit volume is the definition of a material's density. In essence, density is a measurement of how closely stuff is packed. It is a particular physical characteristic of a specific thing.
The amount of space occupied by matter is measured in volume. It is common practice to measure liquids in liters (L) or milliliters (mL).
The equation connecting density and volume is:
V₁D₁ = V₂D₂
D₂ = V₁D₁ / V₂
D₂ = 100 × 0.025 / 4.5 = 0.55 g/cm³
Thus the correct option is C.
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A drill bit in a hand drill is turning at 1200 revolutions per minute (1200 rpm). Express this angular speed in radians per second (rad/s). a. 2.1 rad/s b. 19 rad/s c. 0.67 rad/s d. 126 rad/s e. 39 rad/s
The angular speed of 1200 rpm is option d. 126 rad/s.
What is angular speed?
Angular speed refers to the rate at which an object rotates or moves in a circular path. It measures how quickly an object is changing its angular position with respect to time.
To convert the angular speed from revolutions per minute (rpm) to radians per second (rad/s), we need to use the following conversion factor:
1 revolution = 2π radians
First, we convert the given angular speed of 1200 rpm to revolutions per second:
1200 rpm = 1200 revolutions/minute * (1 minute / 60 seconds) = 20 revolutions/second
Then, we can convert the revolutions to radians by multiplying by 2π:
20 revolutions/second * 2π radians/revolution = 40π radians/second
Since the answer options are given in decimal form, we can approximate the value of 40π:
40π ≈ 40 * 3.14 ≈ 125.6 rad/s
Therefore, the angular speed of 1200 rpm is d. 126 rad/s.
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state Newton's law of gravitation.
Explanation:
Newton’s law of gravitation, statement that any particle of matter in the universe attracts any other with a force varying directly as the product of the masses and inversely as the square of the distance between them.
Hope this helps!
Please find attached photograph for your answer
A child on a sled starts from rest at the top of a 15.0 degree slope. If the trip to the bottom takes 15.2s, how long is the slope? Assume that frictional forces may be neglected.
A child on a sled starting from rest at the top of a 15.0-degree slope takes 15.2 seconds to reach the bottom, with the slope's length of 5.823 meters, neglecting frictional forces.
To find the length of the slope, we can use the equations of motion for motion along an inclined plane.
Given:
The angle of the slope: θ = 15.0 degrees
Time is taken to reach the bottom: t = 15.2 seconds
Initial velocity: u = 0 (since the child starts from rest)
We can use the equation of motion for displacement along an inclined plane:
s = ut + (1/2)at²
In this case, since the child starts from rest, the initial velocity u is 0, and we can simplify the equation to:
s = (1/2)at²
To find the acceleration a, we can use the equation for acceleration along an inclined plane:
a = g * sin(θ)
where g is the acceleration due to gravity (approximately 9.8 m/s²).
Plugging in the values, we have:
a = 9.8 m/s² * sin(15.0 degrees)
Calculating the value of a, we get:
a ≈ 2.529 m/s²
Now, we can use the equation s = (1/2)at² to find the length of the slope s:
s = (1/2) * (2.529 m/s²) * (15.2 s)²
Calculating the value of s, we get:
s ≈ 5.823 meters
Therefore, the length of the slope is approximately 5.823 meters.
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According to Newton's first law of motion when will an object at rest begin to move
Answer:
When acted upon by a force.
Explanation:
"If a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force."
a small remote-control car with a mass of 1.61 kg moves at a constant speed of v = 12.0 m/s in a vertical circle inside a hollow metal cylinder that has a radius of 5.00 m
The tension in the string is approximately 46.7 N. To find the tension in the string, we can analyze the forces acting on the remote-control car at the top and bottom of the vertical circle.
To find the tension in the string, we can analyze the forces acting on the remote-control car at the top and bottom of the vertical circle.
At the top of the circle:
The downward gravitational force (mg) and the tension in the string (T) act downward.
The net force in the upward direction is provided by the centripetal force (Fc).
At the bottom of the circle:
The downward gravitational force (mg) and the tension in the string (T) act downward.
The net force in the upward direction is the sum of the centripetal force (Fc) and the car's weight (mg).
We can set up the following equations of motion at the top and bottom of the circle:
At the top:
T - mg = Fc ...(1)
At the bottom:
T + mg = Fc + mg ...(2)
We can substitute the expression for the centripetal force (Fc = mv^2 / r) into the equations:
At the top:
T - mg = mv^2 / r ...(3)
At the bottom:
T + mg = mv^2 / r + mg ...(4)
Now we can solve these equations to find the tension in the string.
At the top:
T - mg = mv^2 / r
T = mv^2 / r + mg ...(5)
At the bottom:
T + mg = mv^2 / r + mg
From equation (5), we can substitute the expression for T:
mv^2 / r + mg + mg = mv^2 / r + mg
2mg = mv^2 / r
Now we can solve for the tension (T):
T = mv^2 / r - mg
T = (1.61 kg)(12.0 m/s)^2 / 5.00 m - (1.61 kg)(9.8 m/s^2)
T ≈ 46.7 N
Therefore, the tension in the string is approximately 46.7 N.
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The drag bucket for Laminar Flow airfoils is in what AOA regime?
A. all AOA regimes
B. low AOA regimes
C. high AOA regimes
D. No drag bucket exists with laminar flow wings
the correct answer is option (B): low AOA regimes. The drag bucket for Laminar Flow airfoils is primarily found in the low angle of attack (AOA) regime.
The term "drag bucket" refers to a range of angles of attack where the drag coefficient of an airfoil remains relatively low compared to surrounding angles. Laminar Flow airfoils are designed to maintain laminar boundary layer flow over a significant portion of their upper surface, which helps reduce drag.
Option A: All AOA regimes - This option is incorrect because the drag bucket for Laminar Flow airfoils is not present across all angles of attack. The purpose of Laminar Flow airfoils is to delay the onset of turbulent flow, and this effect is most prominent in a specific range of low angles of attack.
Option B: Low AOA regimes - This option is correct. Laminar Flow airfoils exhibit a drag bucket in the low AOA regime, typically from near zero AOA up to a specific critical AOA. In this range, the laminar boundary layer remains attached, resulting in lower drag compared to higher angles of attack.
Option C: High AOA regimes - This option is incorrect because at high angles of attack, the boundary layer on a Laminar Flow airfoil typically transitions to turbulent flow. Consequently, the laminar flow advantages are lost, and the drag increases significantly.
Option D: No drag bucket exists with laminar flow wings - This option is incorrect because the drag bucket is indeed a characteristic feature of Laminar Flow airfoils, allowing for improved aerodynamic performance in the low AOA regime.
In summary, the correct answer is B: low AOA regimes, as this is where the drag bucket is typically observed for Laminar Flow airfoils.
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The displacement of the tip of the 10 cm long minute hand of aclock between 12:15 A.M. and 12:45 P.M. is:
Question 4 answers
10 cm,90°
10 cm,180°
10 cm,4500°
20 cm,180°
20 cm,540°
The displacement of the tip of the 10 cm long minute hand of a clock between 12:15 A.M. and 12:45 P.M. is 10 cm and 180°.
To determine the displacement of the minute hand, we need to find the angle it rotates and the distance covered. Between 12:15 A.M. and 12:45 P.M., there are 12 hours and 30 minutes. The minute hand of a clock completes a full revolution (360°) in 60 minutes.
First, let's find the angle covered by the minute hand. Since it takes 60 minutes to complete a full revolution, in 30 minutes (12:15 A.M. to 12:45 P.M.), the minute hand will cover half of that angle, which is 180°.
Next, let's calculate the distance covered by the minute hand. The length of the minute hand is given as 10 cm. Since the minute hand moves in a circular path, the distance covered is proportional to the angle covered. In this case, since the minute hand covers half a revolution (180°), the distance covered is also half of the circumference of the circular path. Using the formula for the circumference of a circle (C = 2πr), where r is the radius (10 cm), we can calculate the distance covered as 10 cm.
Therefore, the displacement of the tip of the 10 cm long minute hand of a clock between 12:15 A.M. and 12:45 P.M. is 10 cm and 180°.
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Why do you fall forward when you stub your toe on a chair? Explain in terms.
Answer & Explanation:
This can be explained with Newton's first law, Inertia - if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force.
When you walk, your entire body is in a forwarding motion. If your toe hits an object (in this case, a chair), only this unfortunate toe will stop while the rest of your body continues its forwarding motion, resulting in you falling forward.
consider that on the interior of a motor 33 a travels through a 250 turn circular loop that is 12.5 cm in radius. What is the magnetic field strength created at its center?
The magnetic field strength created at the center of a circular loop with 250 turns and a radius of 12.5 cm, through which a current of 33 A flows is 0.208 T.
The magnetic field strength created at the center of the circular loop carrying the current can be found by using the formula: B=μ0IN/2R where B is the magnetic field strength, μ0 is the permeability of free space, I is the current passing through the loop, N is the number of turns in the coil, and R is the radius of the coil. By substituting the given values, we get: B = (4π×10^-7) × 33 × 250 / (2 × 0.125)B = 0.208 T Therefore, the magnetic field strength at the center of the circular loop is 0.208 T.
The intensity of a magnetic field in a specific area is measured by its magnetic field strength. Addressed as H, attractive field strength is ordinarily estimated in amperes per meter (A/m), as characterized by the Global Arrangement of Units (SI).
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Six identical resistors, each with resistance R, are connected to an emf E Part A What is the current I through each of the resistors if they are connected in parallel?
Part B If they are connected in series? Express your answer in terms of the variables E and R.
An emf E is connected to six identical resistors, each with resistance R.
(A) When identical resistors are connected in parallel, the current through each resistor is the same and is given by [tex]\begin{equation}I = \frac{E}{R}[/tex].
(B) When identical resistors are connected in series, the total resistance is 6R, and the current through each resistor is given by [tex]\begin{equation}I = \frac{E}{6R}[/tex].
Part A: When the identical resistors are connected in parallel, the current (I) through each resistor is the same. To calculate the current, we can use Ohm's Law, which states that the current (I) flowing through a resistor is equal to the voltage (V) across the resistor divided by its resistance (R):
[tex]\begin{equation}I = \frac{V}{R}[/tex]
In this case, the voltage across each resistor is the same, and it is equal to the emf (E). Therefore, the current through each resistor connected in parallel is:
[tex]\begin{equation}I = \frac{E}{R}[/tex]
Part B: When the identical resistors are connected in series, the total resistance ([tex]R_total[/tex]) is the sum of the individual resistances. Therefore, the current (I) flowing through the resistors in series is given by Ohm's Law:
[tex]\begin{equation}\I = \frac{E}{R_\text{total}}[/tex]
Since the resistors are identical, the total resistance can be calculated as:
[tex]R_total[/tex] = R + R + R + R + R + R = 6R
Substituting this value into the equation for the current, we get:
[tex]\begin{equation}I = \frac{E}{6R}[/tex]
So, when the resistors are connected in series, the current through each resistor is given by [tex]\begin{equation}I = \frac{E}{6R}[/tex].
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(a) An insulating sphere with radiusa has a uniform charge density rho. The sphere isnot centered at the origin but at.
r=b
Show that the electric field inside thesphere is given by
e=p(r - b)/3E0
To show that the electric field inside the insulating sphere is given by E = ρ(r - b)/(3ε₀), where ρ is the charge density, r is the distance from the centre of the sphere, b is the displacement of the centre from the origin, and ε₀ is the permittivity of free space, we can use Gauss's law.
Gauss's law states that the electric flux through a closed surface is proportional to the charge enclosed by that surface. By applying Gauss's law, we can derive the electric field inside the insulating sphere.
Let's choose a Gaussian surface in the shape of a sphere with radius r, where r is less than the radius of the insulating sphere (a). Since the sphere is not centred at the origin but at a displacement of b, the centre of our Gaussian sphere will also be displaced by b.
According to Gauss's law, the electric flux through this Gaussian surface is given by:
Φ = E * A
where Φ is the electric flux, E is the electric field, and A is the area of the Gaussian surface.
Since the electric field is radially symmetric for a uniformly charged sphere, the electric field at any point on the Gaussian surface will have the same magnitude and direction. Therefore, the electric field can be taken out of the dot product with the area vector, and we have:
Φ = E * A = E * 4πr²
Now, we need to determine the charge enclosed by this Gaussian surface. Since the sphere has a uniform charge density (ρ), the charge enclosed within a sphere of radius r is given by:
Q = (4/3)πr³ρ
Now, applying Gauss's law, we have:
Φ = Q / ε₀
Substituting the expressions for Φ and Q, we get:
E * 4πr² = (4/3)πr³ρ / ε₀
E = (1/3) * r * ρ / ε₀
Since r is the distance from the origin, and the sphere is displaced by b, we can rewrite r as (r - b). Therefore:
E = ρ(r - b) / (3ε₀)
Therefore, we have shown that the electric field inside the insulating sphere is given by E = ρ(r - b) / (3ε₀), where r is the distance from the origin, b is the displacement of the sphere from the origin.
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175 g of water was heated from 15 to 88 celsius how many kilocalories were absorbed by the water
In the given condition, approximately 12.85 kilocalories were absorbed by the water.
To calculate the amount of heat absorbed by water, we can use the formula:
Q = m × c × ΔT
where Q is the amount of heat absorbed by water, m is the mass of water, c is the specific heat capacity of water, and ΔT is the change in temperature.
In this case, we have 175 g of water that was heated from 15 to 88 degrees Celsius. The change in temperature is:
ΔT = 88 - 15 = 73 °C
The specific heat capacity of water is approximately 4.18 J/g°C. Therefore, we can calculate the amount of heat absorbed by water as follows:
Q = m * c * ΔT Q = 175 g * 4.18 J/g°C * 73 °C Q = 53,765 J
To convert this to kilocalories, we can divide by 4.184 J/cal:
Q = 53,765 J / 4.184 J/cal Q = 12.85 kcal
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