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zlopas [31]
3 years ago
8

Someone please help me!

Physics
1 answer:
kirill [66]3 years ago
4 0

I think it’s the first one

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light traveling through air encounters a second medium which slows the light to 100,000 miles/second. What is the index of the s
IceJOKER [234]

Answer:

300000 hope it is help full

Explanation:

300000

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Answer:-2

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a ball is whirled on a string and then the string breaks. what causes the ball to move off in a straight line ?
katrin [286]
It would be called Inertia
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3 years ago
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A 0.500 kg bullet is fired from a gun at 25.0 m/s, how much kinetic energy does it have?
slamgirl [31]

Considering the definition of kinetic energy, the bullet has a kinetic energy of 156.25 J.

<h3>Kinetic energy</h3>

Kinetic energy is a form of energy. It is defined as the energy associated with bodies that are in motion and this energy depends on the mass and speed of the body.

Kinetic energy is defined as the amount of work necessary to accelerate a body of a given mass and in a rest position, until it reaches a given speed. Once this point is reached, the amount of accumulated kinetic energy will remain the same unless there is a change in speed or the body returns to its rest state by applying a force to it.

The kinetic energy is represented by the following expression:

Ec= ½ mv²

Where:

  • Ec is the kinetic energy, which is measured in Joules (J).
  • m is the mass measured in kilograms (kg).
  • v is the speed measured in meters over seconds (m/s).

<h3>Kinetic energy of a bullet</h3>

In this case, you know:

  • m= 0.500 kg
  • v= 25 m/s

Replacing in the definition of kinetic energy:

Ec= ½ ×0.500 kg× (25 m/s)²

Solving:

<u><em>Ec= 156.25 J</em></u>

Finally, the bullet has a kinetic energy of 156.25 J.

Learn more about kinetic energy:

brainly.com/question/25959744

brainly.com/question/14028892

#SPJ1

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2 years ago
In the equation vx^2=v0x^2+2ax(x-x0) what does the terms vx, v0x, x, and x0 stand for respectively?
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B. velocity at position x, velocity at position x=0, position x, and the original position

In the equation

v_{x}^{2} = v_{ox}^{2} +2 a x (x - x₀)

v_{x} = velocity at position "x"

v_{ox} = velocity at position "x = 0 "

x = final position

x_{o} = initial position of the object at the start of the motion

6 0
3 years ago
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