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2 years ago

Changes of state

Physics

2 answers:

andrew11 [14]2 years ago

5 0

Answer: first statement, "At segment T-U, the substance changes from liquid to gas and does not change temperature"

Explanation:

if the segment T-U is flat (not inclined), means that the temperature is not changing. In this case, the heat provided to the substance is being used to overcome the intermolecucular attraction forces, which permits the particles to get more separated and, in such way, the liquid pass to gas phase.

Mrrafil [7]2 years ago

4 0

Answer:

Option A

Explanation:

At segment T-U, the substance changes from a liquid to a gas and does not change temperature.

The reason is because latent heat of vaporisation allows for the absorption of heat in the change of state and temperature remains constant until it has fully changed state.

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Can you tell from the coefficient of restitution whether a collision has added kinetic energy to a system, taken some away, or l

Maurinko [17]

For an inelastic collision where coefficient of restitution,e, is equal to 0, the momentum is conserved but not the kinetic energy. So, there is addition or elimination of kinetic energy.

On the otherhand, when e = 1, like for an elastic collision, kinetic energy and momentum is conserved. Thus, the system's kinetic energy is unchanged.

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3 years ago

Sean, whose mass is 60 kg, is riding on a 5.0 kg sled initially traveling at 8.0 m/s. He brakes the sled with a constant force,

alexgriva [62]

Answer:

Explanation:

according to Newtons second law;

F = ma

F = m(v-u)/t

m is the mass = 60+5 = 65kg

v is the final velocity = 0m/s

initial velocity = 8m/s

time t = 4.0s

F = 65(8)/4

F = 65*2

F = 130N

Hence he applied a force of 130N

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3 years ago

Si se aplica una fuerza de 3n sobre un sistema se genera 15000 cal de calor generandose a su vez un trabajo de 300 j ¿en cuanto

Marizza181 [45]

Answer: +/- 71,65 Calorías

Explanation: Se deduce que la fuerza aplica para aumentar o disminuir la energía del generador.

+/- x Calorías

X = Equivalente de 300 Joules en calorías

Para poder pasar Joules a calorías.

Hacemos una regla de 3 simple

1 Calorías = 4,18 Julios

0,2392 calorías = 1 Julio

( 1 / 4,18 = 0,2392 calorías)

300 Julios = 71,65 calorías

(0,2392 calorías x 300 julios = +/- 71,65 calorías)

5 0

3 years ago

A 620 kg moose is standing in the middle of a train track. A 10,000 kg train moving at 10 m/s is unable to stop and the moose en

ddd [48]

Answer:

This is an example of Inelastic colission

Explanation:

Step one:

given:

mass of moose m1 = 620 kg

mass of train m2= 10,000kg

Initial velocity of moose u1= 0 m/s

Initial velocity of train v1 = 10m/s

combined velocity of the system is given as v

Applying the conservation of momentum equation we have

m1u1+ m2u1= (m1+m2)V

substitutting we have

620*0+10000*10= (620+10000)V

100000= 10620V

divide both sides by 10620

V = 100000/10620

V=9.41m/s

The velocity of the moose after impact is 9.41m/s

4 0

3 years ago

Two 22.7 kg ice sleds initially at rest, are placed a short distance apart, one directly behind the other, as shown in Fig. 1. A

boyakko [2]

Newton's third law of motion sates that force is directly proportional to the rate of change of momentum produced

(a) The final speeds of the ice sleds is approximately 0.49 m/s each

(b) The impulse on the cat is 11.0715 kg·m/s

The reason for arriving at the above values is as follows:

The given parameters are;

The masses of the two ice sleds, m₁ = m₂ = 22.7 kg

The initial speed of the ice, v₁ = v₂ = 0

The mass of the cat, m₃ = 3.63 kg

The initial speed of the cat, v₃ = 0

The horizontal speed of the cat, v₃ = 3.05 m/s

(a) The required parameter:

The final speed of the two sleds

For the first jump to the right, we have;

By the law of conservation of momentum

Initial momentum = Final momentum

∴ m₁ × v₁ + m₃ × v₃ = m₁ × v₁' + m₃ × v₃'

Where;

v₁' = The final velocity of the ice sled on the left

v₃' = The final velocity of the cat

Plugging in the values gives;

22.7 kg × 0 + 3.63 × 0 = 22.7 × v₁' + 3.63 × 3.05

∴ 22.7 × v₁' = -3.63 × 3.05

v₁' = -3.63 × 3.05/22.7 ≈ -0.49

The final velocity of the ice sled on the left, v₁' ≈ -0.49 m/s (opposite to the direction to the motion of the cat)

The final speed ≈ 0.49 m/s

For the second jump to the left, we have;

By conservation of momentum law, m₂ × v₂ + m₃ × v₃ = m₂ × v₂' + m₃ × v₃'

Where;

v₂' = The final velocity of the ice sled on the right

v₃' = The final velocity of the cat

Plugging in the values gives;

22.7 kg × 0 + 3.63 × 0 = 22.7 × v₂' + 3.63 × 3.05

∴ 22.7 × v₂' = -3.63 × 3.05

v₂' = -3.63 × 3.05/22.7 ≈ -0.49

The final velocity of the ice sled on the right = -0.49 m/s (opposite to the direction to the motion of the cat)

The final speed ≈ 0.49 m/s

(b) The required parameter;

The impulse of the force

The impulse on the cat = Mass of the cat × Change in velocity

The change in velocity, Δv = Initial velocity - Final velocity

Where;

The initial velocity = The velocity of the cat before it lands = 3.05 m/s

The final velocity = The velocity of the cat after coming to rest =

∴ Δv = 3.05 m/s - 0 = 3.05 m/s

The impulse on the cat = 3.63 kg × 3.05 m/s = 11.0715 kg·m/s

(c) The required information

The average velocity

Impulse = $F_{average}$ × Δt

Where;

Δt = The time of collision = The time it takes the cat to finish landing = 12 ms

12 ms = 12/1000 s = 0.012 s

We get;

$F_{average} = \mathbf{\dfrac{Impulse}{\Delta \ t}}$

∴ $F_{average} = \dfrac{11.0715 \ kg \cdot m/s}{0.012 \ s} = 922.625 \ kg\cdot m/s^2 = 922.625 \ N$

The average force on the right sled applied by the cat while landing, $\mathbf{F_{average}}$ = 922.625 N

Learn more about conservation of momentum here:

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8 0

3 years ago