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LUCKY_DIMON [66]
2 years ago
8

A karate master strikes a board with an initial velocity of 10.0 m/s, decreasing to 1.0 m/s as his hand passes through the board

. If the time of contact with the board is 0.002 0 s, and the mass of the coordinated hand and arm is 1.0 kg, what is the force exerted on the board
Physics
1 answer:
kenny6666 [7]2 years ago
6 0

The force exerted on the board by the karate master given the data is -4500 N

<h3>Data obtained from the question </h3>
  • Initial velocity (u) = 10 m/s
  • Final velocity (v) = 1 m/s
  • Time (t) = 0.002 s
  • Mass (m) = 1 Kg
  • Force (F) = ?
<h3>How to determine the force</h3>

The force exerted can be obtained as illustrated below:

F = m(v - u) / t

F = 1 (1 - 10) / 0.002

F = (1 × -9) / 0.002

F = -4500 N

Learn more about momentum:

brainly.com/question/250648

#SPJ1

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They have thick body coverings
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1. Synthesize Information You push your
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2 years ago
A 1300-kg car initially has a velocity of 22.2 m/s due south. It brakes to a stop over a 180 m distance.
Vaselesa [24]
The acceleration of the object which moves from an initial step to a full halt given the distance traveled can be calculated through the equation,
                                     d = v² / 2a
where d is distance, v is the velocity, and a is acceleration
Substituting the known values,
                                     180 = (22.2 m/s)² / 2(a)
The value of a is equal to 1.369 m/s²
The force needed for the object to be stopped is equal to the product of the mass and the acceleration.
                                      F = (1300 kg)(1.369 m/s²) 
                                            F = 1779.7 N
4 0
3 years ago
The gravitational force, F, between an object and the Earth is inversely proportional to the square of the distance from the obj
andreev551 [17]
<h2>Weight of astronaut 2450 miles above the Earth is 80.38 pounds</h2>

Explanation:

Given that gravitational force, F, between an object and the Earth is inversely proportional to the square of the distance from the object and the center of the Earth.

             F=\frac{k}{r^2}

Where F is gravitational force  between an object and the Earth, r is the distance from the object and the center of the Earth and k is a constant.

Radius of Earth = 4000 miles

In case 1 an astronaut weighs 209 pounds on the surface of the Earth,

              209=\frac{k}{4000^2}\\\\k=3.344\times 10^9

Now we need to find weight of astronaut 2450 miles above the Earth    

              r = 4000 + 2450 = 6450 miles

               F=\frac{k}{6450^2}\\\\F=\frac{3.344\times 10^9}{6450^2}=80.38pounds

Weight of astronaut 2450 miles above the Earth is 80.38 pounds  

3 0
3 years ago
A small 1.0 kg steel ball rolls west at 3.0 m/s collides with a large 3.0 kg ball at rest. After the collision, the small ball m
77julia77 [94]

Answer:

The direction of the momentum of the large ball after the collision with respect to east is 146.58°.

Explanation:

Given that,

Mass of large ball = 3.0 kg

Mass of steel ball = 1.0 kg

Velocity = 3.0 kg

After collision,

Velocity = 2.0 m/s

Using conservation of momentum

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}

3.0\times0+1.0\times(3.0)(-i)=1.0\times2(-j)+3.0\times v_{2}

-3i+2j=3.0\times v_{2}

v_{2}=-i+0.66j

The direction of the momentum

tan\theta=\dfrac{0.66}{-1}

\theta=tan^{-1}\dfrac{0.66}{-1}

\theta=-33.42^{\circ}

The direction of the momentum with respect to east

\theta=180-33.42=146.58^{\circ}

Hence, The direction of the momentum of the large ball after the collision with respect to east is 146.58°.

7 0
2 years ago
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