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adelina 88 [10]
4 years ago
9

You throw a 3.00-N rock vertically into the air from ground level. You observe that when it is 14.9 m above the ground, it is tr

aveling at 24.9 m/s upward. Part A Use the work-energy theorem to find the rock's speed just as it left the ground.
Physics
1 answer:
Liula [17]4 years ago
3 0

Answer:

rock's speed just as it left the ground = 30.2 m/s

Explanation:

The work energy theorem explains that the change in kinetic energy between two points = workdone between the two points.

Mass of the rock = 3/9.8 = 0.306 kg

Kinetic energy as it leaves the ground = mv₀²/2 = 0.153 v₀²

Kinetic energy at the height it reached = m(24.9²)/2 = 0.306 × 24.9²/2 = 94.862 J

Work done by the force of gravity in between those two points = (-mgh) = - 3 × 14.9 = - 44.7 J

94.862 - 0.153 v₀² = - 44.7

0.153 v₀² = 94.862 + 44.7 = 139.562

v₀ = 30.2 m/s

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Neutrons and protons are located in the nucleus of the atom.

Explanation:

And electrons are in the electron cloud.

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What do you need to know to establish motion
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3 years ago
A ball is thrown horizontally from the top of a building 14.9 m high. The ball strikes the ground at a point 107 m from the base
umka2103 [35]

Answer:

1) t=1.743 sec

2)Vo=61.388  m/sec

3)the x component of its velocity just be- fore it strikes the ground is the same as the  initial velocity of the ball that is=61.388  m/sec

4)Vf=17.08 m/s

Explanation:

1)From second equation of motion we get

h=Vit+(1/2)gt^2

here in case(a): Vi=0 m/s,h=14.9m,,put these values in above equation to find the time the ball is in motion

14.9=(0)*t+(1/2)(9.8)t^2

t^2=14.9/4.9

t^2=3.040 sec

t=1.743 sec

2) s=Vo*t

Putting values we get

107=Vo*1.743

Vo=61.388  m/sec

3)the x component of its velocity just be- fore it strikes the ground is the same as the  initial velocity of the ball that is=61.388  m/sec

4)From third equation of motion we know that

Vf^2-Vi^2=2gh

here Vi=0 m/s,h=14.9 m

Vf^2=Vi^2+2gh=0+2(9.8)(14.9)

Vf^2=292.04

Vf=17.08 m/s

8 0
4 years ago
A child on a bridge throws a rock straight down to the water below. The point where the child released the rock is 82 m above th
eimsori [14]

Answer:

40 m/s

Explanation:

given,

height of the fall, h = 82 m

time taken to fall, t = 1.3 s

rock velocity, v = ?

acceleration due to gravity, g = 9.8 m/s²

rock is released initial velocity, u = 0 m/s

using equation of motion

v² = u² + 2 a s

v² = 0 + 2 x 9.8 x 82

v² = 1607.2

v = 40 m/s

hence, rock's velocity is equal to 40 m/s

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3 years ago
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RoseWind [281]
To the picture the answer is A. I can’t answer the typed question because I need the picture for the box
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