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Tems11 [23]
3 years ago
12

When landing from a jump, a basketball player of mass 82 kg has a velocity of 1.2 m/s right before they hit the ground. The play

er then lands on the floor with their feet and is quickly brought to a stop. If the the floor deforms by 0.025 m while the player lands on it, what was the average force under the player's feet during the landing
Physics
1 answer:
saw5 [17]3 years ago
6 0

Answer:

2361.6N

Explanation:

Mass of player = 82kg

Velocity = 1.2m/s

Kinetic energy of player:

= 1/2mv²

= 1/2*82*1.2²

= 41x1.44

= 59.04J

Final kinetic energy = 0

Change in kinetic energy

|∆k| = |0-59.04|

= 59.04

Workdone by the feet = fd

d = 0.025

Fd = 59.04

F = 59.04/0.025

= 2361.6N

This is his average force.

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_____________ is the study of movement in athletes. A. Sports biomechanics B. Posture C. Dynamics D. Anatomy
liubo4ka [24]

Answer:

I believe its A: Sports biomechanics.

3 0
2 years ago
The prop blades of an airplane spin with a linear velocity of 875 m/s and have a centripetal acceleration on the farthest edge o
storchak [24]

The radius of the prop blade of an airplane is determined as 4.25 m.

<h3>Radius of the prop blade</h3>

The radius of the prop blade of an airplane is calculated as follows;

a = v²/r

where;

  • v is the linear speed
  • r is the radius of the prop blade
  • a is the centripetal acceleration

r = v²/a

r = (875²)/(180,000)

r = 4.25 m

Thus, the radius of the prop blade of an airplane is determined as 4.25 m.

Learn more about centripetal acceleration here: brainly.com/question/79801

#SPJ1

4 0
2 years ago
If a nearsighted person has a far point df that is 3.50 m from the eye, what is the focal length f1 of the contact lenses that t
Anastaziya [24]

Answer:

f1 = -3.50 m

Explanation:

For a nearsighted person an object at infinity must be made to  appear  to be at his far point which is 3.50 m away. The image of an object at infinity must be formed on the same side of the lens as the object.

∴ v = -3.5 m

Using mirror formula,

i/f1 = 1/v + 1/u

Where f1 = focal length of the contact lens, v = image distance = -3.5 m, u =         object distance = at infinity(∞) = 1/0

∴ 1/f1 = (1/-3.5) + 1/infinity

  Note that, 1/infinity = 1/(1/0) = 0/1 =0.

∴ 1/f1 = 1/(-3.5) + 0

  1/f1 = 1/(-3.5)

Solving the equation by finding the inverse of both side of the equation.

∴ f1 = -3.50 m

 Therefore a converging lens of focal length  f1 = -3.50 m

would be needed by the person to see an object at infinity clearly

8 0
3 years ago
A 200g piece of iron is heated at 100C. It is than dropped into water to bring its temperature down to 22C. What is the amount o
gizmo_the_mogwai [7]

Answer:

6926.4J

Explanation:

Given parameters:

Mass of iron  = 200g

Initial temperature  = 100°C

Final temperature  = 22°C

Unknown:

Amount of heat transferred to the water  = ?

Solution:

The quantity of heat transferred to the water is a function of mass and temperature of the iron;

 H  = m c Ф

m is the mass of the iron

Ф is the change in temperature

C is the specific heat capacity of iron = 0.444 J/g°C

Now;

 insert the parameters and solve;

      H  = 200 x 0.444 x (100-22)

      H = 6926.4J

8 0
3 years ago
Two identical bowling balls are moving down a bowling alley so that their centers of mass have the same velocity, but one just s
tamaranim1 [39]

Answer:

Rolling Ball

Explanation

7 0
3 years ago
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