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harkovskaia [24]
3 years ago
14

What is the displacement of the car between t=1s and t=4s

Physics
1 answer:
tensa zangetsu [6.8K]3 years ago
4 0

Answer:

Option C. 30 m

Explanation:

From the graph given in the question above,

At t = 1 s,

The displacement of the car is 10 m

At t = 4 s

The displacement of the car is 40 m

Thus, we can simply calculate the displacement of the car between t = 1 and t = 4 by calculating the difference in the displacement at the various time. This is illustrated below:

Displacement at t = 1 s (d1) = 10 m

Displacement at t= 4 s (d2) = 40

Displacement between t = 1 and t = 4 (ΔD) =?

ΔD = d2 – d1

ΔD = 40 – 10

ΔD = 30 m.

Therefore, the displacement of the car between t = 1 and t = 4 is 30 m.

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Xamine the graph. Select the statement that best describes the energy change in the particles of a substance during melting.
lesya692 [45]

Answer:

  • Option B) Absorbed energy results in the change in potential energy.

Explanation:

Please, find attached the graph that accompanies this question.

The<em> melting</em> proces is the change from solid phase to liquid phase. It is represented with the lower flat line with the symbol ΔHfus over it.

The line is flat because the temperature remains constant during this process. Thus, you know the option "C) As the temperature increases during melting, the kinetic energy also increases" is FALSE.

What happens during this process is:

  • Most of the energy received by the particles from heating, during the melting process, goes to overcome the intermolecular bonds between the particles. This results in increasing the distance between the particles, so the internal potential energy increases. This is what the option <em>"B) Absorbed energy results in the change in potential energy" correctly describes.</em> Hence, option B) is TRUE.

Althoug most of the heat energy received is transformed into potential energy, yet a small part of the heat energy increases a bit the kinetic energy of the particles, because the particles will vibrate faster around their relatively fixed positions. Hence, the option "<em>A) The kinetic energy of the particles remains unchanged</em>" is FALSE.

As for option D) it is not reasonable at all: none chemical or physical priciple can be used to state that <em>the kinetic energy decreases as the particles move farther apart</em>. Thus, this is FALSE.

6 0
3 years ago
A 6 kg mass collides with a body at rest .After collision they travel together with a velocity equal
Mars2501 [29]

Answer:

12 kg

Explanation:

Momentum before collision = momentum after collision

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

After the collision, they have the same velocity, so v₁ = v₂ = v:

m₁ u₁ + m₂ u₂ = m₁ v + m₂ v

m₁ u₁ + m₂ u₂ = (m₁ + m₂) v

We know that m₁ = 6 kg, u₂ = 0 m/s, and v = u₁ / 3.

(6 kg) u₁ + m₂ (0 m/s) = (6 kg + m₂) (u₁ / 3)

(6 kg) u₁ = (6 kg + m₂) (u₁ / 3)

6 kg = (6 kg + m₂) (1/3)

18 kg = 6 kg + m₂

m₂ = 12 kg

3 0
4 years ago
Light is reflected from a crystal of table salt with an index of refraction of 1.544. An analyser is placed to intercept the ref
Vitek1552 [10]

Answer:

hola me llamo bruno y tu?

Explanation:

pero yo soy de mexico

5 0
3 years ago
In the first stage of a two-stage rocket, the rocket is fired from the launch pad starting from rest but with a constant acceler
AnnZ [28]

Answer:

3091.56

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s² (positive downward and negative upward)

s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times 25+\frac{1}{2}\times 3.5\times 25^2\\\Rightarrow s=1093.75\ m

Distance traveled in the first stage is 1093.75 m

v=u+at\\\Rightarrow v=0+3.5\times 25\\\Rightarrow v=87.5\ m/s

Velocity at the end of first stage is 87.5 m/s

v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{132.5-87.5}{10}\\\Rightarrow a=4.5\ m/s^2

Acceleration of the second stage is 4.5 m/s²

s=ut+\frac{1}{2}at^2\\\Rightarrow s=87.5\times 10+\frac{1}{2}\times 4.5\times 10^2\\\Rightarrow s=1100\ m

Distance traveled in the second stage is 1100 m

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-132.5^2}{2\times -9.81}\\\Rightarrow s=894.81\ m

Distance traveled after the second stage has stopped firing is 894.81 m

Total height the rocket reached = 1093.75+1100+897.81 = 3091.56 m

3 0
4 years ago
If the 50 kg objects slows down to a velocity of 1 m/s how much kinetic energy does it have?
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It has 50kg with a velocity of 1 m/s times the speed of the cart divided by 2 and multiplied by kinectic x plus 5

3 0
3 years ago
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