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3 years ago

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Arunner has a speed of 23 m/s. They see the finish line and speed up to 27 m/s. This happens in 5 seconds. If the runner has a m

ass of 78 kg, with what force did the
runner cross the finish line? Show all work to receive full credit.
Type your answer.

1 answer:

sp2606 [1]3 years ago

6 0

The answer is 100

Have a great day

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the earth's moon has a gravitational field strength of about 1.6 n/kg near its surface. the moon has a mass of 7.35x10^22 kg. wh

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A camcorder has a power rating of 13 watts. If the output voltage from its battery is 6 volts, what current does it use?

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2 years ago

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A hockey player uses her hockey stick to exert a force of 6.81 N on a stationary hockey puck. The hockey puck has a mass of 165

Anna007 [38]

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41.3 m/s^2 option (e)

Explanation:

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3 years ago

A playground merry-go-round has a mass of 115 kg and a radius of 2.50 m and it is rotating with an angular velocity of 0.520 rev

tatuchka [14]

Answer:

$W_f = 2.319 rad/s$

Explanation:

For answer this we will use the law of the conservation of the angular momentum.

$L_i = L_f$

so:

$I_mW_m = I_sW_f$

where $I_m$ is the moment of inertia of the merry-go-round, $W_m$ is the initial angular velocity of the merry-go-round, $I_s$ is the moment of inertia of the merry-go-round and the child together and $W_f$ is the final angular velocity.

First, we will find the moment of inertia of the merry-go-round using:

I = $\frac{1}{2}M_mR^2$

I = $\frac{1}{2}(115 kg)(2.5m)^2$

I = 359.375 kg*m^2

Where $M_m$ is the mass and R is the radio of the merry-go-round

Second, we will change the initial angular velocity to rad/s as:

W = 0.520*2 $\pi$ rad/s

W = 3.2672 rad/s

Third, we will find the moment of inertia of both after the collision:

$I_s = \frac{1}{2}M_mR^2+mR^2$

$I_s = \frac{1}{2}(115kg)(2.5m)^2+(23.5kg)(2.5m)^2$

$I_s = 506.25kg*m^2$

Finally we replace all the data:

$(359.375)(3.2672) = (506.25)W_f$

Solving for $W_f$ :

$W_f = 2.319 rad/s$

7 0

3 years ago