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gavmur [86]
3 years ago
10

2. Before 0.5-kg ball is dropped from a third story window 20 m above the sidewalk. What is the potential

Physics
2 answers:
timofeeve [1]3 years ago
5 0
(C). Pe=mgh
Pe=(.5kg)(10m/s^2)(20m)
Pe= 100J≈ 98J using 9.8 for G
Flauer [41]3 years ago
4 0

Answer:

C

Explanation:

C

20*9.8*0.5

98

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73 Newton is the correct answer
6 0
3 years ago
A cue ball initially moving at 3.4 m/s strikes a stationary eight ball of the same size and mass. After the collision, the cue b
romanna [79]

Answer:

speed of eight ball speed after the collision is 3.27 m/s

Explanation:

given data

initially moving v1i = 3.4 m/s

final speed is v1f = 0.94 m/s

angle = θ w.r.t. original line of motion

solution

we assume elastic collision

so here using conservation of energy

initial kinetic energy = final kinetic energy .............1

before collision kinetic energy = 0.5 × m× (v1i)²

and

after collision kinetic energy =  0.5 × m× (v1f)²  + 0.5 × m× (v2f)²

put in equation 1

0.5 × m× (v1i)² =  0.5 × m× (v1f)²  + 0.5 × m× (v2f)²

(v2f)² = (v1i)² - (v1f)²

(v2f)² = 3.4² - 0.94²

(v2f)² = 10.68

taking the square root both

v2f = 3.27 m/s

speed of eight ball speed after the collision is 3.27 m/s

5 0
3 years ago
Which of the following is an example of the transformation of gravitational potential energy into motion energy (kinetic energy
nikklg [1K]

Answer:

A drop of water falling from a faucet into a sink

5 0
3 years ago
What is velocity in science
katovenus [111]

Answer:

Velocity is the rate at which the position changes. The average velocity is the displacement or position change (a vector quantity) per time ratio.

Hope this helps! ^-^

8 0
3 years ago
Read 2 more answers
You throw a rock straight up from the edge of a cliff. It leaves your hand at time t = 0 moving at 13.0 m/s. Air resistance can
Zarrin [17]

Answer:

0.36s, 2.3s

Explanation:

Let gravitational acceleration g = 9.81 m/s2. And let the throwing point as the ground 0 for the upward motion. The equation of motion for the rock leaving your hand can be written as the following:

s = v_0t + gt^2/2

where s = 4 m is the position at 4m above your hand. v_0 = 13 m/s is the initial speed of the rock when it leaves your hand. g = -9.81m/s2 is the deceleration because it's in the downward direction. And t it the time(s) it take to get to 4m, which we are looking for

4 = 13t - 9.81t^2/2

4.905 t^2 - 13t + 4 = 0

t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

t= \frac{13\pm \sqrt{(-13)^2 - 4*(4.905)*(4)}}{2*(4.905)}

t= \frac{13\pm9.51}{9.81}

t = 2.3 or t = 0.36

7 0
4 years ago
Read 2 more answers
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