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3 years ago

g A ball is thrown against the wall and bounces back with the same velocity. What type of collision is this

Physics

1 answer:

Softa [21]3 years ago

8 0

Answer:

Perfectly elastic collision

Explanation:

In a closed system, an elastic collision is a type of collision between two bodies, where the total momentum and kinetic energy are conserved.

We are told from the problem that the ball bounces back with its original velocity. For the ball to bounce back to the thrower in the first place, this is our first hint that the collision is elastic. If the collision was inelastic, the ball would most likely have stuck to the wall.

In addition to that, the velocity of the ball remains fairly unchanged even after the collision. This confirms that the kinetic energy it had before the collision is the same as the kinetic energy it has after the collision.

As a result of this, the collision is perfectly elastic

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What is the first law of conservation of energy? How is the law applicable to the earths energy balance

Vlad1618 [11]

Answer:

First law of thermodynamic state that energy is always conserved, in other word, energy cannot be created or destroyed.

This means that energy can only transfer from one source to another.

For example, the electricity we used to power our phone didn't come from nowhere.

We charge our phone using the energy from the electricity from the outlet.

The outlet gets its energy from the power plant, who generated that energy from burning coal.

The coal they burned got their energy while they were alive as plants.

The plants got their energy from the sun.

The sun got its energy from other cosmic stuff and so on...

My point is that the energy we use will always cycle around. It doesn't come from thin air and it doesn't disappear into thin air either. This help keeps things balance because if energy is created out of thin air then the earth might just blow up. If energy gets erased into thin air then there'll be nothing to fuel life.

5 0

3 years ago

An automobile light has a 1.0-a current when it is connected to a 12-v battery. determine the resistance of the light.

Elis [28]

<span>Resistance of automobile light is equal to 12 ohms.
</span>Resistance = Voltage / Current
where
Voltage = 12 volts
Current = 1.0 Amperes
Resistance = 12 volts/ 1.0 amperes
Resistance = 12 ohms.
Automobile light has 12 ohms resistance when it is connected to 12volts battery with 1.0 ampere current.

3 0

3 years ago

A racquet ball with mass m = 0.221 kg is moving toward the wall at v = 13.9 m/s and at an angle of θ = 25° with respect to the h

Salsk061 [2.6K]

Answer:

1) 3.07kgm/s

2) 5.56kgm/s

3) 76.16N

4) 4.33kgm/s

5) 0.57s

6) -8.66J

Explanation:

Given

m = 0.221kg

v = 13.9m/s

θ = 25°

t = 0.073s

1) to get the magnitude of the initial momentum is the racquet ball. We use the formula,

P(i) = mv(i)

P(i) = 0.221 * 13.9

P(i) = 3.07kgm/s

2) Magnitude of the change in momentum of the ball,

P(i,x) = P(i) cos θ

P(i,x) = 3.07 * cos25

P(i,x) = 3.07 * 0.9063

P(i,x) = 2.78

ΔP = 2P(i,x)

ΔP = 2 * 2.78 = 5.56kgm/s

3) magnitude of the average force exerted by the wall,

F(ave) = ΔP/Δt

F(ave) = 5.56/0.073

F(ave) = 76.16N

4) ΔP(z) = mv(f) - mv(i)

ΔP(z) = 0.221*-7.8 - 0.221*11.8

ΔP(z) = -1.72 - 2.61

ΔP(z) = 4.33kgm/s

5) F(ave) = ΔP/Δt

Δt = ΔP/F(ave)

Δt = 4.33 / 76.16

Δt = 0.57s

6) KE(i) = 0.5mv(i)²

KE(f) = 0.5mv(f)²

ΔKE = 0.5m[v(f)² - v(i)²]

ΔKE = 0.5 * 0.221 [(-7.8)² - 11.8²]

ΔKE = 0.1105 ( 60.84 - 139.24 )

ΔKE = 0.1105 * -78.4

ΔKE = -8.66J

6 0

3 years ago

Most applications will ask you to provide all of the following information

Colt1911 [192]

its A chief your welcome

4 0

3 years ago

Read 2 more answers

The physician changed the medication order because the infection was refractory, which means that the infection was

Alexeev081 [22]

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8 0

3 years ago