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emmainna [20.7K]
3 years ago
7

If you were to walk at a constant speed 20m/s for 30 seconds, how far would you walk?

Physics
1 answer:
lana [24]3 years ago
6 0

Answer:

600m

Explanation:

30×20 at a constant speed is 600m.

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A horizontal uniform bar of mass 2.7 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to t
Jlenok [28]

Answer:

14.36 N

Explanation:

T_{1} = Tension in string 1

T_{2} = Tension in string 2

m_b = mass of the bar = 2.7 kg

W_b = weight of the bar

weight of the bar is given as

W_b = m_{b} g = (2.7) (9.8) = 26.46N

m_m = mass of the bar = 1.35 kg

W_m = weight of the monkey

weight of the monkey is given as

W_m = m_{m} g = (1.35) (9.8) = 13.23N

Using equilibrium of torque about left end

W_{m} (AB) + W_{b} (AB) = T_{2} (AC)\\W_{m} (AB) + W_{b} (AB) = T_{2} (AD - CD)\\(13.23) (1.5) + (26.46)(1.5) = T_{2} (3 - 0.65)\\\\T_{2} = 25.33 N

Using equilibrium of force in vertical direction

T_{1} + T_{2} = W_{b} + W_{m}\\T_{1} + 25.33 = 26.46 + 13.23\\T_{1} = 14.36 N

7 0
3 years ago
A 0.05-kg car starts from rest at a height of 0.95 m. Assuming no friction, what is the kinetic energy of the car when it reache
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The kinetic energy of the car is A.) 0.466 J
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A clock moves past you at a speed of 0.9 c. How much time passes for you for each second that elapses on the moving clock?
Zigmanuir [339]

Answer:

Explanation:

Time dilation formula is

T = T₀ / √ 1-v²/c²

T₀ is time elapsed in moving reference , T time elapsed in stationary reference.

Here T₀ = 1 second

T = 1/√ 1-0.9² = 1/.4358 = 2.3 second

So 2.3 second will pass for each second on moving reference.

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The force between charged objects decreases when their separation ..... A)increase<br> B)decrease
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Inertia is the force in play here
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