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Scilla [17]
2 years ago
3

A ball weighing 10 kg traveling at 1.4 m/s hits and rebounds back off a wall and now travels 1.4 m/s in the opposite direction.

What is the work done on it?
Physics
1 answer:
Marizza181 [45]2 years ago
4 0

<em>NO work is done on the ball.</em>

Its momentum changes, but its kinetic energy doesn't.

<u>Before it hits the wall:</u>

Momentum = m v = (10 kg) x (1.4 m/s) = +14 kg-m/s

Kinetic energy = (1/2)(m)(v²) = (1/2)(10kg)(1.4)² = 9.8 Joules

<u>After it hits the wall:</u>

Momentum = m v = (10 kg) x (-1.4 m/s) = -14 kg-m/s

Kinetic energy = (1/2)(m)(v²) = (1/2)(10kg)(-1.4)² = 9.8 Joules

No change in energy  ==>  no work done on the ball.

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Spills out 10cm³ of wather.

Explanation:

The ball being submerged halfway in the full glass of water will spill a volume of water equal to the volume of the ball that is being submerged, because the entire volume of the submerged ball was previously occupied by water and is not compressed should move out of the glass.This volume will be 10cm³ since the total of the ball is 20cm³ and it is submerged halfway.

8 0
2 years ago
Henrietta is going off to her physics class, jogging down the sidewalk at a speed of 3.20 m/s. Her husband Bruce suddenly realiz
Zigmanuir [339]

Answer:

u_x=8.5454\ m.s^{-1}

d=9.5782+16=25.5782\ m was the her position ahead from her window when she caught the object.

Explanation:

Given:

  • speed of walking of Henrietta, v_w=3.2\ m.s^{-1}
  • height of projection of projectile, h=43.9\ m
  • Horizontal distance between the window and Henrietta when the projectile was launched:

r=3.2\times 5

r=16\ m

  • Since the projectile was thrown at the time when she had passed below the window 5 seconds ago.

Since the projectile was thrown horizontally therefore the vertical component of velocity is zero.

<u>Now the time taken for the object to reach to Henrietta ignoring the height where she catches:</u>

h=u_x.t+\frac{1}{2} \times g.t^2

43.9=0+\frac{1}{2} \times 9.8\times t^2

t=2.9932\ s is the time taken by the projectile to reach Henrietta.

<u>Now the distance further walked by Henrietta in the above time from the point where she was when the projectile was launched:</u>

\Delta d=v_w\times t

\Delta d=9.5782\ m

<u>Now the total  horizontal distance from the window to Henrietta when the projectile reached her:</u>

d=\Delta d+r

d=9.5782+16=25.5782\ m was the her position ahead from her window when she caught the object.

<u>Now the initial horizontal velocity of launch of the projectile:</u>

u_x=\frac{d}{t}

u_x=\frac{25.5782}{2.9932}

u_x=8.5454\ m.s^{-1}

8 0
2 years ago
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1 year ago
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Answer:

v = 172 km/h

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So, the average speed of the whole journey is 172 km/h.

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