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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?

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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If element X has 99 protons, how many elctrons does it have?

slega [8]

It has 99 electrons because an element has the same number of protons and electrons

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3 years ago

Read 2 more answers

Dafna11 [192]

Answers:

a) -2.54 m/s

b) -2351.25 J

Explanation:

This problem can be solved by the <u>Conservation of Momentum principle</u>, which establishes that the initial momentum $p_{o}$ must be equal to the final momentum $p_{f}$ :

$p_{o}=p_{f}$ (1)

Where:

$p_{o}=m_{1} V_{o} + m_{2} U_{o}$ (2)

$p_{f}=(m_{1} + m_{2}) V_{f}$ (3)

$m_{1}=110 kg$ is the mass of the first football player

$V{o}=-7 m/s$ is the velocity of the first football player (to the south)

$m_{2}=75 kg$ is the mass of the second football player

$U_{o}=4 m/s$ is the velocity of the second football player (to the north)

$V_{f}$ is the final velocity of both football players

With this in mind, let's begin with the answers:

a) Velocity of the players just after the tackle

Substituting (2) and (3) in (1):

$m_{1} V_{o} + m_{2} U_{o}=(m_{1} + m_{2}) V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{m_{1} V_{o} + m_{2} U_{o}}{m_{1} + m_{2}}$ (5)

$V_{f}=\frac{(110 kg)(-7 m/s) + (75 kg) (4 m/s)}{110 kg + 75 kg}$ (6)

$V_{f}=-2.54 m/s$ (7) The negative sign indicates the direction of the final velocity, to the south

b) Decrease in kinetic energy of the 110kg player

The change in Kinetic energy $\Delta K$ is defined as:

$\Delta K=\frac{1}{2} m_{1}V_{f}^{2} - \frac{1}{2} m_{1}V_{o}^{2}$ (8)

Simplifying:

$\Delta K=\frac{1}{2} m_{1}(V_{f}^{2} - V_{o}^{2})$ (9)

$\Delta K=\frac{1}{2} 110 kg((-2.5 m/s)^{2} - (-7 m/s)^{2})$ (10)

Finally:

$\Delta K=-2351.25 J$ (10) Where the minus sign indicates the player's kinetic energy has decreased due to the perfectly inelastic collision

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3 years ago

Where does every piece of matter begin?

Margarita [4]

Every piece of matter begins “Out of this world”

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It can take some of those large , industrial vehicles up to ____ feet to stop when traveling only 60 MPH, and therefore , you sh

kherson [118]

Answer:

a 200 feet, and trains go a whole mile even after hitting the brakes

Explanation:

4 0

3 years ago

A carriage of 20 kg is pulled with a force of 35 N. How far the carriage will go

Gennadij [26K]

Answer:

2.71 m

Explanation:

Force is the product of mass and acceleration

F=m*a

Work done is the product of force and distance

Work done=F*d

In this case;

F= 35 N

Work done = 95 J

95 =35 * d

95 /35 = d

2.71 m= d

6 0

3 years ago