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jeka94
3 years ago
11

How long will your trip take if you travel 4000 m at an average speed of 8m/s

Physics
2 answers:
SpyIntel [72]3 years ago
8 0
  • Answer:

<em>500 sec</em>

<em>8 min 20 sec</em>

  • Explanation:

<em>Hi there !</em>

<em />

<em>8 m ................ 1 s </em>

<em>4000 m ........ x s</em>

<em>x = 4000m×1s/8m = 500 sec = 8 min 20 sec</em>

<em />

<em>Good luck ! </em>

Andrews [41]3 years ago
8 0

Answer:8.3mins

Explanation:

S=Vt

where V=average speed

S=4000m V=8m/s

4000=8×t

t=4000/8

t=500s

t=500/60

t=8.3min

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Taking upward motion as positive and downward motion as negative.

Downward motion:

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Mass of ball (m) = 0.150 kg

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Initial velocity (u) = 0 m/s

Acceleration is due to gravity (g) = -9.8 m/s²

Using equation of motion, we have:

v_d^2=u^2+2aS\\\\v=\pm\sqrt{u^2+2aS}\\\\v_d=\pm\sqrt{0+2\times -9.8\times -1.25}\\\\v_d=\pm\sqrt{24.5}=\pm4.95\ m/s

Since, the motion is downward, final velocity must be negative. So,

v_d=-4.95\ m/s

Upward motion:

Given:

Displacement of ball (S) = 0.665 m

Initial velocity (v_d) = 4.95 m/s(Upward direction)

Acceleration is due to gravity (g) = -9.8 m/s²

Using equation of motion, we have:

v_{up}^2=v_d^2+2aS\\\\v_{up}=\pm\sqrt{v_d^2+2aS}\\\\v_{up}=\pm\sqrt{24.5+2\times -9.8\times 0.665}\\\\v_{up}=\pm\sqrt{10.966}=\pm3.31\ m/s

Since, the motion is upward, final velocity must be positive. So,

v_{up}=3.31\ m/s

Now, impulse is equal to change in momentum. So,

Impulse = Final momentum - Initial momentum

J=m(v_{up}-v_d)\\\\J=(0.150\ kg)(3.31-(-4.95))\ m/s\\\\J=0.150\ kg\times 8.26\ m/s\\\\J=1.239\ Ns

Therefore, the impulse given to the ball by the floor is 1.239 kg.m/s in upward direction.

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3 years ago
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