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4 years ago

As you brake your bicycle, your velocity changes from 20 m/s east to 10 m/s east in 5 seconds. What’s your acceleration?

Physics

1 answer:

charle [14.2K]4 years ago

8 0

Hello!

$\large\boxed{-2m/s^{2} }$

Find the acceleration using the formula a = (vf - vi) / t where:

vf = final velocity

vi = initial velocity

t = time

Plug in the points above:

vf = +10 m/s

vi = +20 m/s

t = 5 sec

(10 - 20) / 5 = -10 / 5 = -2 m/s²

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3 years ago

A horizontal 791-N merry-go-round is a solid disk of radius 1.56 m and is started from rest by a constant horizontal force of 49

ICE Princess25 [194]

Answer:

$K.E=273.5J$

Explanation:

Given data

$F_{g}=791N\\F_{h}=49N\\r=1.56m\\t=3.03s$

To find

Kinetic Energy

Solution

The moment of inertia is given as:

$I=(\frac{1}{2} )MR^2\\I=\frac{1}{2}(F_{g}/g)R^2\\ I=\frac{1}{2}(\frac{791N}{9.8m/s^2} )(1.56m)^2\\ I=98.21kg.m^2$

The angular acceleration is given as:

$\alpha =\frac{T}{I}\\\alpha =\frac{F_{y}R}{I}\\ \alpha =\frac{(49N)(1.56m)}{98.2kg.m^2}\\\alpha =0.778rad/s^2$

Now the angular velocity is given by:

$w=\alpha t\\w=(0.778rad/s^2)(3.03s)\\w=2.36rad/s$

So the kinetic energy given as:

$K.E=(\frac{1}{2} )Iw^2\\K.E=\frac{1}{2}(98.21kg.m^2)(2.36rad/s)^2\\ K.E=273.5J$

8 0

3 years ago

A Ping-Pong ball moving East at a speed of 4 m/s collides with a stationary bowling ball. The Ping-Pong ball bounces back to the

In-s [12.5K]

Answer:

They experience the same magnitude impulse

Explanation:

We have a ping-pong ball colliding with a stationary bowling ball. According to the law of conservation of momentum, we have that the total momentum before and after the collision must be conserved:

$p_i = p_f\\p_p + p_b = p'_p+p'_b$

where

$p_p$ is the initial momentum of the ping-poll ball

$p_b$ is the initial momentum of the bowling ball (which is zero, since the ball is stationary)

$p'_p$ is the final momentum of the ping-poll ball

$p'_f$ is the final momentum of the bowling ball

We can re-arrange the equation as follows

$p_p - p'_p = p_b'-p_b$

$-\Delta p_p = \Delta p_b$

which means

$|\Delta p_p | = |\Delta p_b|$ (1)

so the magnitude of the change in momentum of the ping-pong ball is equal to the magnitude of the change in momentum of the bowling ball.

However, we also know that the magnitude of the impulse on an object is equal to the change of momentum of the object:

$I=\Delta p$ (2)

Therefore, (1)+(2) tells us that the ping-pong ball and the bowling ball experiences the same magnitude impulse:

$|I_p| = |I_b|$

4 0

4 years ago