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VladimirAG [237]
2 years ago
14

Type the correct answer in each box. Round your answers to the nearest hundredth.

Physics
1 answer:
Natali [406]2 years ago
8 0

The velocity of the ball at position A is 3.13 m/s.Since all energy is converted to KE.

<h3>What is potential energy?</h3>

The potential energy is due to the virtue of the position and the height. The unit for the potential energy is the joule.

A person has the most potential energy due to her position. when he is sitting on the highest point.

Given data;

Mass,m = .5 kg

Height,H = 0.5 m

Acceleration due to gravity,g = 9.8 m/s²

H is the height

The potential energy is found as;

The potential energy is mainly dependent upon the height of the object.

Potential energy = mgh

The potential energy at point B;

PE = mgh

PE = ( 1.5 kg) ( 9.8 m/s2) (0.5 m)

PE = 7.35 J

Since all energy is converted to KE, the kinetic energy at point B;

KE = 0.5mv^2

KE = PE

7.35 = 0.53(1.5) v^2

V = 3.13 m/s

Hence, the velocity of the ball at position A is 3.13 m/s.

To learn more about the potential energy, refer to the link;

brainly.com/question/24284560

#SPJ1

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Read 2 more answers
A person slaps her leg with her hand, which results in her hand coming to rest in a time interval of 2.65 ms from an initial spe
dezoksy [38]

Answer:

the magnitude of the average contact force exerted on the leg is  3466.98 N

Explanation:

Given the data in the question;

Initial velocity of hand v₀ = 5.25 m/s

final velocity of hand v = 0 m/s

time interval t = 2.65 ms = 0.00265 s

mass of hand m = 1.75 kg

We calculate force on the hand F_{hand

using equation for impulse in momentum

F_{hand × t = m( v - v₀ )

we substitute

F_{hand × 0.00265  = 1.75( 0 - 5.25 )

F_{hand × 0.00265  = 1.75( - 5.25 )

F_{hand × 0.00265  = -9.1875

F_{hand = -9.1875 / 0.00265  

F_{hand = -3466.98 N

Next we determine force on the leg F_{leg

Using Newton's third law of motion

for every action, there is an equal opposite reaction;

so, F_{leg = - F_{hand

we substitute

F_{leg = - ( -3466.98 N )

F_{leg = 3466.98 N

Therefore, the magnitude of the average contact force exerted on the leg is  3466.98 N

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