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PilotLPTM [1.2K]
3 years ago
11

A pursuit spacecraft from the planet Tatooine is attempting to catch up with a Trade Federation cruiser. As measured by an obser

ver on Tatooine, the cruiser is traveling away from the planet at 0.600c. The pursuit ship is traveling at 0.800c relative to Tatooine, in the same direction as the cruiser. What is the speed of the cruiser relative to the pursuit ship?

Physics
2 answers:
Lady bird [3.3K]3 years ago
4 0

Answer:

0.385c

Explanation:

The speed of the cruiser in relative to the pursuit ship as they are moving in the same direction is 0.385c.

Kindly go through the attached file for a rundown of how the answer is gotten.

madam [21]3 years ago
3 0

Answer:

the speed of the cruiser relative to the pursuit ship is 0.3846c

Explanation:

the solution is in the attached Word file

Download docx
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It would exert the same back right?
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3 years ago
Two equally charged tiny spheres of mass 1.0 g are placed 2.0 cm apart. When released, they begin to accelerate away from each o
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Answer:

The magnitude of the charge on each sphere is 0.135 μC

Explanation:

Given that,

Mass = 1.0

Distance = 2.0 cm

Acceleration = 414 m/s²

We need to calculate the magnitude of charge

Using newton's second law

F= ma

a=\dfrac{F}{m}

Put the value of F

a=\dfrac{kq^2}{mr^2}

Put the value into the formula

414=\dfrac{9\times10^{9}\times q^2}{1.0\times10^{-3}\times(2.0\times10^{-2})^2}

q^2=\dfrac{414\times1.0\times10^{-3}\times(2.0\times10^{-2})^2}{9\times10^{9}}

q^2=1.84\times10^{-14}

q=0.135\times10^{-6}\ C

q=0.135\ \mu C

Hence, The magnitude of the charge on each sphere is 0.135μC.

7 0
3 years ago
Read 2 more answers
When the area of block increases for constant force, the
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Answer: B

adding force will add accesion

7 0
2 years ago
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Estimate the final temperature of a mole of gas at 200.0 atm and 19.0°C as it is forced through a porous plug to a final pressur
Yanka [14]

Answer : The final temperature of gas is 266.12 K

Explanation :

According to the Joule-Thomson experiment, it states that when a gas is expanded adiabatically from higher pressure region to lower pressure region, the change in temperature with respect to change in pressure at constant enthalpy is known as Joule-Thomson coefficient.

The formula will be:

\mu_{J,T}=(\frac{dT}{dP})_H

or,

\mu_{J,T}=(\frac{dT}{dP})_H\approx \frac{\Delta T}{\Delta P}

As per question the formula will be:

\mu_{J,T}=\frac{T_2-T_1}{P_2-P_1}   .........(1)

where,

\mu_{J,T} = Joule-Thomson coefficient of the gas = 0.13K/atm

T_1 = initial temperature = 19.0^oC=273+19.0=292.0K

T_2 = final temperature = ?

P_1 = initial pressure = 200.0 atm

P_2 = final pressure = 0.95 atm

Now put all the given values in the above equation 1, we get:

0.13K/atm=\frac{T_2-292.0K}{(0.95-200.0)atm}

T_2=266.12K

Therefore, the final temperature of gas is 266.12 K

5 0
3 years ago
a child's toy consists of a piece of plastic attached to a spring. the spring is compressed against the floor a distance of 2.0
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Answer:

1.7N

Explanation:

Force = kx

Where x = spring compression and

K = spring constant

K =85N/m

x = 2.0cm / 100

= 0.02m

Force = 85 x 0.02

= 1.7N

5 0
3 years ago
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