Ask question

Ask question

All categories

2 years ago

7

10. As a spaceship is moving toward Earth, an Earthling measures its length to be 325 m, while the captain on board radios that

her spaceship's length is 1150 m. (c = 3.00 × 108 m/s) (a) How fast is the rocket moving relative to Earth? (b) What is the TOTAL energy of a 75.0-kg crewman as measured by (i) the captain in the rocket and (ii) the Earthling?

1 answer:

Oliga [24]2 years ago

3 0

Answer:

Explanation:

10. As a spaceship is moving toward Earth, an Earthling measures its length to be 325 m, while the captain on board radios that her spaceship's length is 1150 m. (c = 3.00 × 108 m/s) (a) How fast is the rocket moving relative to Earth? (b) What is the TOTAL energy of a 75.0-kg crewman as measured by (i) the captain in the rocket and (ii) the Earthling?

You might be interested in

A bowling ball of mass 5.8 kg moves in a straight line at 4.34 m/s How fast must a Ping-Pong ball of mass 2.214 g move in a stra

lilavasa [31]

Answer: 11369.46 m/s

Explanation:

We have the following data:

$m_{1}=5.8 kg$ is the mass of the bowling ball

$V_{1}=4.34 m/s$ is the velocity of the bowling ball

$m_{2}=2.214 g \frac{1 kg}{1000 g}=0.002214 kg$ is the mass of the ping-pong ball

$V_{2}$ is the velocity of the ping-pong ball

Now, the momentum $p_{1}$ of the bowling ball is:

$p_{1}=m_{1}V_{1}$ (1)

$p_{1}=(5.8 kg)(4.34 m/s)$

$p_{1}=25.172 kg m/s$ (2)

And the momentum $p_{2}$ of the ping-pong ball is:

$p_{2}=m_{2}V_{2}$ (3)

If the momentum of the bowling ball is equal to the momentum of the ping-pong ball:

$p_{1}=p_{2}$ (4)

$m_{1}V_{1}=m_{2}V_{2}$ (5)

Isolating $V_{2}$ :

$V_{2}=\frac{m_{1}V_{1}}{m_{2}}$ (6)

$V_{2}=\frac{25.172 kg m/s}{0.002214 kg}$ (7)

Finally:

$V_{2}=11369.46 m/s$

6 0

3 years ago

If the mass of the spacecraft were doubled, how much would the force of gravity would change?

vodka [1.7K]

F=mg
if m doubled, F would double as well

5 0

3 years ago

2. How long must a 400 W electrical engine work in order to produce 300 kJ of work?

yanalaym [24]

Answer:

Explanation:

400 W = 400 J/s

300000 J / 400 J/s = 750 s or 12.5 minutes

7 0

3 years ago

What displacement in cm would occur with a 75 N/m spring if you placed a 300 N weight on the spring?

ollegr [7]

Surface tension=75N/m

Weight=300N

$\\ \bull\tt\longmapsto Surface\:Tension=\dfrac{Weight}{Displacement}$

$\\ \bull\tt\longmapsto Displacement=\dfrac{Weight}{Surface\:Tension}$

$\\ \bull\tt\longmapsto Displacement=\dfrac{300}{75}$

$\\ \bull\tt\longmapsto Displacement=4m$

$\\ \bull\tt\longmapsto Displacement=400cm$

6 0

3 years ago

At noon on a clear day, sunlight reaches the earth\'s surface at Madison, Wisconsin, with an average intensity of approximately

Dmitrij [34]

Intensity of sunlight at given position is defined as power received per unit area

so here we can say

$I = 2 kJ/s*m^2$

area on which photons are received is given as

$A = 4.80 cm^2 = 4.80 * 10^-4 m^2$

now we can find the power received due to sunlight

$P = I*A$

$P = 2* 10^3 * 4.80 * 10^-4$

$P = 0.96 Watt$

now we can say this power is due to photons that strikes on surface of earth

so here we can say

$P = N\frac{hc}{\lambda}$

given here that

$\lambda = 510 nm$

$0.96 = N\frac{6.6 * 10^{-34}* 3 * 10^8}{510*10^{-9}}$

$0.96 = N * 3.88 * 10^{-19}$

$N = \frac{0.96}{3.88*10^{-19}}$

$N = 2.47 * 10^{18}$

so it will strike 2.47 * 10^18 photons on given area per second

3 0

3 years ago