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Ray Of Light [21]
3 years ago
12

How does kinetic energy affect the stopping distance of a small vehicle compared to a large vehicle?

Physics
2 answers:
Yuliya22 [10]3 years ago
6 0
The more mass the vehicle has, the more that is needed to stop the vehicle in motion.
Nuetrik [128]3 years ago
3 0
Whenever a vehicle is in motion, it has got kinetic energy. Kinetic energy has a direct relationship with the stopping distance. Kinetic energy is dependent on the mass of the vehicle and also the velocity at which it is traveling. In case of a small vehicle, the mass of the vehicle will be small and so the stopping distance will also be less compared to a large vehicle. In the case of the large vehicle traveling at the same velocity as the small vehicle, the stopping distance will be greater becuse the large vehicle has a larger mass. So in case of a small vehicle the kinetic energy will be less and so the distance for stopping will be less than that of the large vehicle. 
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max2010maxim [7]

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A proton in a particle accelerator is traveling at a speed of 0.99c
Gwar [14]

A proton in a particle accelerator is traveling at a speed of 0.99c has a speed magnitude of 2.97 x 10⁸ m/s.

<h3>What is speed of proton?</h3>

The speed of a proton is the rate at which a proton is moving through a given space.

The given speed of the proton is 0.99c

where;

  • c is speed of light

<h3>What is speed of light?</h3>

The speed of light in vacuum, commonly denoted c, is a universal physical constant that is important in many areas of physics.

The value of speed of light in a vacuum is given as 3 x 10⁸ m/s.

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8 0
2 years ago
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alexgriva [62]
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4 0
3 years ago
The water stream strikes the inclined surface of the cart. Determine the power produced by the stream if, due to rolling frictio
levacccp [35]

Answer

given,

constant speed of cart on right side = 2 m/s

diameter of nozzle = 50 mm = 0.05 m

discharge flow through nozzle = 0.04 m³

One-fourth of the discharge flows down the incline

three-fourths flows up the incline

Power = ?

Normal force i.e. Fn acting on the cart

F_n = \rho A (v - u)^2 sin \theta

v is the velocity of jet

Q = A V

v = \dfrac{0.04}{\dfrac{\pi}{4}d^2}

v= \dfrac{0.04}{\dfrac{\pi}{4}\times 0.05^2}

v = 20.37 m/s

u be the speed of cart assuming it to be u = 2 m/s

angle angle of inclination be 60°

now,

F_n = 1000 \times \dfrac{\pi}{4}\times 0.05^2\times (20.37 - 2)^2 sin 60^0

F n = 2295 N

now force along x direction

F_x = F_n sin 60^0

F_x = 2295 \times sin 60^0

F_x = 1987.52\ N

Power of the cart

P = F x v

P = 1987.52 x 20.37

P = 40485 watt

P = 40.5 kW

3 0
3 years ago
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