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lubasha [3.4K]
3 years ago
11

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.
Physics
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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Answer:

1750km

Explanation:

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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?
leva [86]

Power = (voltage) x (current)

13 watts = (6 volts) x (current)

Divide each side by (6 volts):

Current = (13 watts) / (6 volts)

Current = (13/6) Amperes

<em>Current = 2.17 Amperes</em>

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A driverless car collides head on with a stationary sign. The collision brings the car to a stop, but does not move the sign.
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Answer:

I'm sorry hindi ko po alam ung sagot sorry po

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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]

Answer:

41.3 m/s^2 option (e)

Explanation:

force, F = 6.81 N

mass, m = 165 g = 0.165 kg

Let a be the acceleration of the puck.

Use newtons' second law

Force = mass x acceleration

6.81 = 0.165 x a

a = 41.27 m/s^2

a = 41.3 m/s^2

Thus, the acceleration of the puck is 41.3 m/s^2.

5 0
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
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