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2 years ago

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A cylinder is given a push and then rolls up an inclined plane. If the origin is the starting point, sketch the position, veloci

ty, and acceleration of the cylinder vs. time as it goes up and then down the plane.

1 answer:

Damm [24]2 years ago

7 0

It will take a shorter amount of time for the cylinder to go down the plane down off the plane Because more pressure is applied one going up then going down there’s no pressure at all it’s the gravity is helping

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Assume that the stopping distance of a van varies directly with the square of the speed. A van traveling 40 miles per hour can s

Daniel [21]

Answer:

d = 100.8 ft

Explanation:

As we know that initial speed of the van is 40 miles then the stopping distance is given as 70 feet

here we know that

$v_f^2 - v_i^2 = 2 ad$

so here we have

$0^2 - 40^2 = 2 a (70 feet)$

now again if the speed is increased to 48 mph then let say the stopping distance is "d"

so we will have

$0^2 - 48^2 = 2 a (d)$

now divide the above two equations

$\frac{40^2}{48^2} = \frac{70 feet}{d}$

$d = 100.8 ft$

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2 years ago

When monochromatic light shines perpendicularly on a soap film (n = 1.33) with air on each side, the second smallest nonzero fil

Anika [276]

Let us start from considering monochromatic light as an incidence on the film of a thickness t whose material has an index of refraction n determined by their respective properties.

From this point of view part of the light will be reflated and the other will be transmitted to the thin film. That additional distance traveled by the ray that was reflected from the bottom will be twice the thickness of the thin film at the point where the light strikes. Therefore, this relation of phase differences and additional distance can be expressed mathematically as

$2t + \frac{1}{2} \lambda_{film} = (m+\frac{1}{2})\lambda_{film}$

We are given the second smallest nonzero thickness at which destructive interference occurs.

This corresponds to, m = 2, therefore

$2t = 2\lambda_{film}$

$t = \lambda_{film}$

The index of refraction of soap is given, then

$\lambda_{film} = \frac{\lambda_{vacuum}}{n}$

Combining the results of all steps we get

$t = \frac{\lambda_{vacuum}}{n}$

Rearranging, we find

$\lambda_{vacuum} = tn$

$\lambda_{vacuum} = (278)(1.33)$

$\lambda_{vacuum} = 369.74nm$

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2 years ago

The critical angle for a beam of light passing from water into air is 48.8°. this means that all light rays with an angle of inc

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totally internally reflected

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2 years ago

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Kinetic energy is the energy of . It increases with and .Potential energy is the energy of or . It increases with , and

lorasvet [3.4K]

Answer:

k is the energy of motion it increase with four of load you carry

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2 years ago

A bus contains a 1440 kg flywheel (a disk that has a 0.63 m radius) and has a total mass of 10200 kg. Calculate the angular velo

CaHeK987 [17]

Answer: $\omega =93.51 rad/s$

Explanation:

Given

mass of Flywheel $m_1=1440 kg$

mass of bus $m_b=10200 kg$

radius of Flywheel $r=0.63 m$

final speed of bus $v=21 m/s$

Conserving Energy i.e.

0.9(Rotational Energy of Flywheel)= change in Kinetic Energy of bus

Let $\omega$ be the angular velocity of Flywheel

$0.9\cdot \frac{I\omega ^2}{2}=\frac{m_bv^2}{2}$

$I=moment\ of\ Inertia =mr^2=1440\cdot 0.63^2=571.536 kg-m^2$

$0.9\cdot \frac{571.536\cdot \omega ^2}{2}=\frac{10200\cdot 21^2}{2}$

$\omega ^2=21^2\times \frac{10200}{0.9\times 571.536}$

$\omega =21\times 4.45=93.51 rad/s$

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2 years ago