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Art [367]
3 years ago
8

An astronaut in a spacecraft looks out her window and observes a comet travel in the opposite direction at a relative speed of 2

37 m/s. The velocity of the spacecraft is 114 m/s directly away from the sun. What is the comet's velocity relative to the sun? Assume that motion away from the sun is in the positive direction.
Physics
2 answers:
Sergeu [11.5K]3 years ago
7 0
The velocity of the spacecraft with respect to the Sun is
v_s = +114 m/s
the velocity of the comet in the reference system of the spacecraft is
v_c' = 237 m/s

So, the velocity of the comet in the reference system of the Sun will be
v_c = v_c'-v_s = -237 m/s - 114 m/s=-351 m/s
And the negative sign tells us that the comet is directed toward the Sun.
Umnica [9.8K]3 years ago
4 0

Answer: The relative velocity of the comet with respect to sun is '-351 m/s'.

Explanation:

Let the magnitude of the velocity of the spaceship moving away from the sun be |v_s|=114 m/s

Let the magnitude of the velocity of the comet moving towards the sun be |v_c| m/s

Assuming the direction away from the sun to be positive then :

The velocity of the space ship :v_s=+114 m/s

The  velocity of the comet :-v_c

v_r = relative velocity of spaceship with respect to comet

v_r=-237 m/s=v_s+(-v_c)=114 m/s-v_c

v_c=351 m/s

Velocity of the comet.v_c=351 m/s

v_R = Relative velocity of comet with respect to sun:

v_R=v_s+v_c=0 m/s+(-351 m/s)=-351 m/s

v_s= 0 m/s(Sun is fixed at one point)

The negative sign means that comet is moving towards the sun.The relative velocity of the comet with respect to sun is '-351 m/s'.

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A 3.53-g lead bullet traveling at 428 m/s strikes a target, converting its kinetic energy into thermal energy. Its initial tempe
Taya2010 [7]

Complete question:

A 3.53-g lead bullet traveling at 428 m/s strikes a target, converting its kinetic energy into thermal energy. Its initial temperature is 40.0°C. The specific heat is 128 J/(kg · °C), latent heat of fusion is 24.5 kJ/kg, and the melting point of lead is 327°C.

(a) Find the available kinetic energy of the bullet. J

(b) Find the heat required to melt the bullet. J

Answer:

Part (a) the available kinetic energy of the bullet is 323.32 J

Part (b) the heat required to melt the bullet is 216.17 J

Explanation:

Given;

mass of the bullet = 3.53 g = 0.00353 kg

velocity of the bullet = 428 m/s

initial temperature of the bullet = 40.0°C

final temperature of the bullet =  327°C

specific heat capacity, c= 128 J/(kg · °C)

latent heat of fusion, Hf  = 24.5 kJ/kg

Part (a) the available kinetic energy of the bullet. J

KE = ¹/₂ × mv²

KE = ¹/₂ × 0.00353 × 428²

     = 323.32 J

Part (b) the heat required to melt the bullet. J

This is the thermal energy required to increase the temperature of the bullet and the heat energy required to melt the bullet.

Quantity of heat required to raise the temperature of the bullet:

Q = mcΔT

   = 0.00353 × 128 × (327-40)

   = 0.00353 × 128 × 287

   = 129.68 J

Quantity of heat required to melt the bullet:

Q = mH_f

Q = 0.00353 × 24500

   = 86.49 J

TOTAL energy required to melt the bullet = 129.68 J + 86.49 J

                                                                      = 216.17 J

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Answer:

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Answer:

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Density of block = mass of block / volume of block

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Answer:B.

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