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2 years ago

14

A children's roller coaster is released from the top of a track. If its maximum speed at ground level is 3 m/s, find the height

it was released from.

1 answer:

san4es73 [151]2 years ago

4 0

Answer:

h = 0.46 m

Explanation:

According to the law of conservation of energy:

Potential Energy Lost by Roller Coaster = Kinetic Energy Gained by Roller Coaster

$mgh = \frac{1}{2}mv^2\\\\2gh = v^2\\\\h = \frac{v^2}{2g}$

where,

h = height = ?

v = speed at bottom = 3 m/s

g = acceleration due to gravity = 9.81 m/s²

Therefore,

$h = \frac{(3\ m/s)^2}{(2)(9.81\ m/s^2)}$

<u>h = 0.46 m</u>

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Two points are located on a rigid wheel that is rotating with decreasing angular velocity about a fixed axis. Point A is located

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d.Both points have the same instantaneous angular velocity.

Explanation:

Let's analyze each option one by one:

a.the angular velocity at point A is greater than that of point B. --> FALSE. In fact, for a rigid body (such as the rigid wheel in this problem), all the points along the body rotates with the same angular velocity. This is because the angular velocity is defined as

$\omega=\frac{\theta}{t}$

where $\theta$ is the angular displacement and t is the time taken. Therefore, all the points rotate by the same angle $\theta$ in the same time t.

b.Both points have the same centripetal acceleration. --> FALSE. The centripetal acceleration of a point along the disk is given by

$a_c=\omega^2 r$

where $\omega$ is the angular velocity and r is the distance of the point from the axis of rotation. Here A is located farther away from the axis with respect to B, so it will have a greater centripetal acceleration.

c.Both points have the same tangential acceleration. --> FALSE. The tangential acceleration is given by

$a_t = \alpha r$

where $\alpha$ is the angular acceleration. $\alpha$ is the same for every point for a rigid body, but $r$ is not (A is located farther away), so A and B do not have same tangential acceleration.

d.Both points have the same instantaneous angular velocity. --> TRUE. This is true for what we said at point A).

e.Each second, point A turns through a greater angle than point B. --> FALSE. The two points cover the same angular displacement in the same time, since they have same angular velocity.

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3 years ago

The wind is blowing in the direction of a vector pointing at an angle 30 degrees to the east of due north, at 35 miles/hour. An

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3 years ago

Deimos completes one (circular) orbit of Mars in 1.26 days. The distance from Mars to Deimos is 2.35×107m. What is the centripet

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Answer:

The centripetal acceleration of Deimos is $0.077m/s^{2}$ .

Explanation:

The centripetal acceleration is defined as:

$a = \frac{v^{2}}{r}$ (1)

Where v is the velocity of Deimos and r is the orbital distance.

Notice that is necessary to determine the velocity first.

The speed of the Deimos can be found by means of the Universal law of gravity:

$F = G\frac{M \cdot m}{r^{2}}$ (2)

Then, replacing Newton's second law in equation 2 it is gotten:

$m\cdot a = G\frac{M \cdot m}{r^{2}}$ (3)

However, a is the centripetal acceleration since Deimos almost describes a circular motion around Mars:

$a = \frac{v^{2}}{r}$ (4)

Replacing equation 4 in equation 3 it is gotten:

$m\frac{v^{2}}{r} = G\frac{M \cdot m}{r^{2}}$

$m \cdot v^{2} = G \frac{M \cdot m}{r^{2}}r$

$m \cdot v^{2} = G \frac{M \cdot m}{r}$

$v^{2} = G \frac{M \cdot m}{rm}$

$v^{2} = G \frac{M}{r}$

$v = \sqrt{\frac{G M}{r}}$ (5)

Where v is the orbital speed, G is the gravitational constant, M is the mass of Mars, and r is the orbital radius.

$v = \sqrt{\frac{(6.67x10^{-11}N.m^{2}/kg^{2})(6.39x10^{23}kg)}{2.35x10^{7}m}}$

$v = 1346m/s$

Finally, equation 4 can be used:

$a = \frac{(1346m/s)^{2}}{2.35x10^{7}m}$

$a = 0.077m/s^{2}$

Hence, the centripetal acceleration of Deimos is $0.077m/s^{2}$ .

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2 years ago