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3 years ago

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Spaceship 1 and Spaceship 2 have equal masses of 200 kg. Spaceship 1 has a speed of 0 m/s, and Spaceship 2 has a speed of 6 m/s.

They collide and stick together. What is their speed?

1 answer:

rusak2 [61]3 years ago

5 0

Answer:

3 ms⁻¹

Explanation:

<em><u>Theory</u></em>

<u>The law of conservation of linear momentum</u>

The sum of linear momentum of a system under no external force remains constant.

In this scenario although different forces act on two spaceship by each other when we consider the total system of two spaceship no external unbalance force act on them. So we can apply this law.

As law says,

The sum of linear momentum before collision = The sum of linear momentum after collision.

Momentum = mass × velocity.

So we get, 200×0 +200×6 = 400×v where v = velocity of the locked spaceships,

v = 3 ms⁻¹

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A heat engine accepts 200,000 Btu of heat from a source at 1500 R and rejects 100,000 Btu of heat to a sink at 600 R. Calculate

diamong [38]

To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.

By definition we know that the change in entropy is given by

$\Delta S = \frac{Q}{T}$

Where,

Q = Heat transfer

T = Temperature

On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is

$W = Q_{source}-Q_{sink}$

According to the data given we have to,

$Q_{source} = 200000Btu$

$T_{source} = 1500R$

$Q_{sink} = 100000Btu$

$T_{sink} = 600R$

PART A) The total change in entropy, would be given by the changes that exist in the source and sink, that is

$\Delta S_{sink} = \frac{Q_{sink}}{T_{sink}}$

$\Delta S_{sink} = \frac{100000}{600}$

$\Delta S_{sink} = 166.67Btu/R$

On the other hand,

$\Delta S_{source} = \frac{Q_{source}}{T_{source}}$

$\Delta S_{source} = \frac{-200000}{1500}$

$\Delta S_{source} = -133.33Btu/R$

The total change of entropy would be,

$S = \Delta S_{source}+\Delta S_{sink}$

$S = -133.33+166.67$

$S = 33.34Btu/R$

Since $S\neq 0$ the heat engine is not reversible.

PART B)

Work done by heat engine is given by

$W=Q_{source}-Q_{sink}$

$W = 200000-100000$

$W = 100000 Btu$

Therefore the work in the system is 100000Btu

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3 years ago

When a object is at rest, what is it’s speed??

nadya68 [22]

Answer:

0 (there is no speed)

Explanation:

If an object is at rest, it is not moving, and it doesn't have a speed, so the speed is zero.

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3 years ago

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marishachu [46]

Answer:

the last one: weight force

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3 years ago

Urgente!!!!! <br> Necesito ayuda con esto!!!!

ahrayia [7]

media.discordapp.net/attachments/782414373888458783/826224189828366377/video0.mp4

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3 years ago

A 3.5 kg object moving in two dimensions initially has a velocity v1 = (12.0 i^ + 22.0 j^) m/s. A net force F then acts on the o

lys-0071 [83]

Answer:

The work done by the force is 820.745 joules.

Explanation:

Let suppose that changes in potential energy can be neglected. According to the Work-Energy Theorem, an external conservative force generates a change in the state of motion of the object, that is a change in kinetic energy. This phenomenon is describe by the following mathematical model:

$K_{1} + W_{F} = K_{2}$

Where:

$W_{F}$ - Work done by the external force, measured in joules.

$K_{1}$ , $K_{2}$ - Translational potential energy, measured in joules.

The work done by the external force is now cleared within:

$W_{F} = K_{2} - K_{1}$

After using the definition of translational kinetic energy, the previous expression is now expanded as a function of mass and initial and final speeds of the object:

$W_{F} = \frac{1}{2}\cdot m \cdot (v_{2}^{2}-v_{1}^{2})$

Where:

$m$ - Mass of the object, measured in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final speeds of the object, measured in meters per second.

Now, each speed is the magnitude of respective velocity vector:

Initial velocity

$v_{1} = \sqrt{v_{1,x}^{2}+v_{1,y}^{2}}$

$v_{1} = \sqrt{\left(12\,\frac{m}{s} \right)^{2}+\left(22\,\frac{m}{s} \right)^{2}}$

$v_{1} \approx 25.060\,\frac{m}{s}$

Final velocity

$v_{2} = \sqrt{v_{2,x}^{2}+v_{2,y}^{2}}$

$v_{2} = \sqrt{\left(16\,\frac{m}{s} \right)^{2}+\left(29\,\frac{m}{s} \right)^{2}}$

$v_{2} \approx 33.121\,\frac{m}{s}$

Finally, if $m = 3.5\,kg$ , $v_{1} \approx 25.060\,\frac{m}{s}$ and $v_{2} \approx 33.121\,\frac{m}{s}$ , then the work done by the force is:

$W_{F} = \frac{1}{2}\cdot (3.5\,kg)\cdot \left[\left(33.121\,\frac{m}{s} \right)^{2}-\left(25.060\,\frac{m}{s} \right)^{2}\right]$

$W_{F} = 820.745\,J$

The work done by the force is 820.745 joules.

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3 years ago