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faust18 [17]
2 years ago
11

Lucy is cruising through space in her new spaceship. As she coasts along, a tiny spacebug drifts into her path and bounces off t

he window. Consider several statements concerning this scenario. Evaluate each statement according to the law of momentum conservation and match it to the appropriate category.
Physics
1 answer:
Elden [556K]2 years ago
4 0

Your question is not complete, please let me assume this to be your complete question:

Lucy is cruising through space in her new spaceship. As she coasts along, a spacebug drifts into her path and bounces off the window. Consider several statements concerning this scenario. Evaluate each statement according to the law of momentum conservation and match it to the appropriate category.

Below are several statements concerning this scenario. Evaluate each statement and decide if it is true, false, or undetermined by the principle of momentum conservation.

i) The total change in momentum for this interaction is zero.

ii) The change in the space bug's momentum is greater than the change in the spaceship's momentum.

iii) If the space bug had stuck to the spaceship instead of bouncing off, momentum would not have been conserved for this interaction.

Answer:

I is true

II is false

III is undetermined

Explanation:

STATEMENT I : This statement is TRUE because because they was no change in velocity and mass of the spaceship and spacebug, after the collision, as the bug bounced outside the window and the spaceship retains it's velocity. Therefore the total momentum in the system before the collision, is equal to the total momentum of the system after the collision.

Where;

momentum = Mass × velocity

M1V1 = M2V2

Therefore;

M2V2 - M1V1 = 0

STATEMENT II : This statement is FALSE, because the change in the momentum of the spaceship and bug are equal, as the spaceship and bug remains in constant motion after collision. The collision did not have any effect in their velocity nor mass.

Ps = Pb

Ps = Momentum of spaceship

Pb = Momentum of spacebug

STATEMENT III : The statement is UNDETERMINE, because it will depend on the momentum we are considering. If the spaceship is still in constant motion, that means moment of the spaceship is conserved, while that of the spacebug is not conserved.

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The gravitational field strength at a distance R from the center of moon is gR. The satellite is moved to a new circular orbit t
3241004551 [841]

Answer:

g'=\frac{g__R}{4}

Explanation:

Given:

  • gravitational field strength of moon at a distance R from its center, g__R
  • Distance of the satellite from the center of the moon, h=2R

<u>Now as we know that the value of gravity of any heavenly body is at height h is given as:</u>

g'=g__{R}} \times \frac{R^2}{(2R)^2}

g'=\frac{g__R}{4}

∴The gravitational field strength will become one-fourth of what it is at the surface of the moon.

6 0
2 years ago
(a) A 15.0 kg block is released from rest at point A in the figure below. The track is frictionless except for the portion betwe
castortr0y [4]

Answer:

(a) coefficient of friction = 0.451

This was calculated by the application of energy conservation principle (the total sum of energy in a closed system is conserved)

(b) No, it comes to a stop 5.35m short of point B. This is so because the spring on expanding only does a work of 43 J on the block which is not enough to meet up the workdone of 398 J against friction.

Explanation:

The detailed step by step solution to this problems can be found in the attachment below. The solution for part (a) was divided into two: the motion of the body from point A to point B and from point B to point C. The total energy in the system is gotten from the initial gravitational potential energy. This energy becomes transformed into the work done against friction and the work done in compression the spring. A work of 398J was done in overcoming friction over a distance of 6.00m. The energy used in doing so is lost as friction is not a conservative force. This leaves only 43J of energy which compresses the spring. On expansion the spring does a work of 43J back on the block is only enough to push it over a distance of 0.65m stopping short of 5.35m from point B.

Thank you for reading and I hope this is helpful to you.

4 0
2 years ago
A heat engine (Power Cycle) with a thermal efficiency of 35 percent efficiency produces 750 kJ of work. Heat transfer to the eng
frosja888 [35]

Answer:

a) The schematic illustrating is attached

b) The heat transfer to the heat engine is 2142.86 kJ, the heat transfer from the heat engine is 1392.86 kJ

c) The heat transfer to the heat engine is 1648.35 kJ, the heat transfer from the heat engine is 898.35 kJ

Explanation:

b) The heat transfer to the engine and the heat transfer from the engine to the air is:

Q_{1} =\frac{W}{n}

Where

W = 750 kJ

n = 35% = 0.25

Replacing:

Q_{1} =\frac{750}{0.35} =2142.86kJ

Q_{2} =Q_{1} -W=2142.86-750=1392.86kJ

c) The efficiency of Carnot engine is:

n=1-\frac{300K}{550K} =0.455

The heat transfer to the heat engine is:

Q_{1c} =\frac{750}{0.455} =1648.35kJ

The heat transfer from the heat engine is:

Q_{2c} =1648.35-750=898.35kJ

4 0
2 years ago
If you find an igneous rock which has 450 radioactive isotopes and 3,150 stable daughter isotopes, how many half-lifes of this i
slavikrds [6]

Answer:

3t_{1/2}  

Explanation:

To find the half-lifes of the isotope we need to use the following equation:

N_{t} = N_{0}2^{-\frac{t}{t_{1/2}}}     (1)

<em>where Nt: is the amount of the isotope that has not yet decayed after a time t, N₀: is the initial amount of the isotope, t: is the time and </em>t_{1/2}<em>: is the half-lifes.</em>

By solving equation (1) for t we have:

\frac{t}{t_{1/2}} = - \frac{Ln(Nt/N_{0})}{Ln(2)}

<u>Having that:</u>

Nt = 450

N₀ = 3150 + 450 = 3600,

The half-lifes of the isotope is:

t = - \frac{Ln(450/3600)}{Ln(2)} \cdot t_{1/2} = 3t_{1/2}

Therefore, 3 half-lives of the isotope passed since the rock was formed.

I hope it helps you!

3 0
2 years ago
La luz pasa del medio A al medio B formando un ángulo de 35° con la frontera horizontal entre ambos. Si el ángulo de refracción
zaharov [31]

Answer:

Índice de refracción entre los dos medios = 1,43

Refractive index between the two media = 1.43

Explanation:

El índice de refracción entre dos medios se explica mejor entendiendo primero la refracción.

Cuando las olas se mueven de un medio a otro, a menudo experimentan un cambio de dirección con respecto al medio en el que viajan.

Por lo tanto, el índice de refracción se expresa como el seno del ángulo de incidencia dividido por el seno del ángulo de refracción.

El seno del ángulo de incidencia y la refracción utilizados en esta fórmula de índice de refracción se miden respectivamente con respecto a la vertical.

En esta pregunta Ángulo de incidencia = 35° a la horizontal = (90° - 35°) a la vertical = 55° a la vertical.

Ángulo de refracción = 35°

Índice de refracción entre los dos medios

= (Sin 55°) ÷ (Sin 35°)

= 0.8192 ÷ 0.5736

= 1.428 = 1.43 a 2 d.p.

¡¡¡Espero que esto ayude!!!

English Translation

The light passes from medium A to medium B at an angle of 35 ° with the horizontal border between the two. If the angle of refraction is also 35 °, what is the relative refractive index between the two media?

Solution

The refractive index between two media is best explained by first understanding refraction.

When waves move from one medium to another, they often experience a change in direction with respect to the medium in which they are travelling.

Hence, refractive index is expressed as the sine of angle of incidence dibided by the sine of angle of refraction.

The sine of angle of incidence and refraction used in this refractive index formula are both respectively measured with respect to the vertical.

In this question,

Angle of incidence = 35° to the horizontal = (90° - 35°) to the vertical = 55° to the vertical.

Angle of refraction = 35°

Refractive index between the two media

= (Sin 55°) ÷ (Sin 35°)

= 0.8192 ÷ 0.5736

= 1.428 = 1.43 to 2 d.p.

Hope this Helps!!!

3 0
2 years ago
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