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3 years ago

State the condition under which a satellite would move with escape velocity

1 answer:

7 0

When the launch velocity is a bit less than the escape velocity, the satellite with time will find itself back to earth and when the speed is far beyond the escape velocity, the satellite with time, be lost in space.

The velocity of escape from the less massive Moon is about 2.4 km per second at its surface. ... A planet (or satellite) cannot long retain an atmosphere if the planet's escape velocity is low enough to be near the average velocity of the gas molecules making up the atmosphere.

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Please help me please help

Anna [14]

Answer:

2.12x 10 -3

Explanation:

i dont really have explanation but like that's how it works

5 0

3 years ago

Cart A and cart B are traveling in the same direction on a straight track. Cart A is moving at 9.25 m/s and cart B is moving at

Liula [17]

Answer:

The speed of cart B is 11.21 m/s.

Explanation:

Given that,

Speed of cart A = 9.25 m/s

Speed of cart B = 7.15 m/s

Mass of cart A = 72.0 kg

Mass of cart B = 55.0 kg

Speed of card A after collision = 6.15 m/s

We need to calculate the speed of cart B

Using conservation of momentum

$m_{A}v_{A}+m_{B}v_{B}=m_{A}v_{A}+m_{B}v_{B}$

Put the value into the formula

$72.0\times9.25+55.0\times7.15=72.0\times6.15+55.0\times v_{B}$

$v_{B}=\dfrac{72.0\times9.25+55.0\times7.15-72.0\times6.15}{55.0}$

$v_{B}=11.21\ m/s$

Hence, The speed of cart B is 11.21 m/s.

7 0

3 years ago

How can people who live near rivers control flodding

dimaraw [331]

By building dams and levees . Hope this helps

8 0

3 years ago

Read 2 more answers

The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 2.0 rev/s in 10.0 s. At th

Arte-miy333 [17]

Answer:

22 revolutions

Explanation:

2 rev/s = 2*(2π rad/rev) = 12.57 rad/s

The angular acceleration when it starting

$\alpha_a = \frac{\Delta \omega}{\Delta t} = \frac{12.57}{10} = 1.257 rad/s^2$

The angular acceleration when it stopping:

$\alpha_o = \frac{\Delta \omega}{\Delta t} = \frac{-12.57}{12} = -1.05 rad/s^2$

The angular distance it covers when starting from rest:

$\omega^2 - 0^2 = 2\alpha_a\theta_a$

$\theta_a = \frac{\omega^2}{2\alpha_a} = \frac{12.57^2}{2*1.257} = 62.8 rad$

The angular distance it covers when coming to complete stop:

$0 - \omega^2 = 2\alpha_o\theta_o$

$\theta_o = \frac{-\omega^2}{2\alpha_o} = \frac{-12.57^2}{2*(-1.05)} = 75.4 rad$

So the total angular distance it covers within 22 s is 62.8 + 75.4 = 138.23 rad or 138.23 / (2π) = 22 revolutions

6 0

3 years ago

A magnetic dipole <img src="https://tex.z-dn.net/?f=%5Cvec%7Bm%7D" id="TexFormula1" title="\vec{m}" alt="\vec{m}" align="absmidd

ser-zykov [4K]

We are asked to know the reason why magnetic dipole can be written as U = -m *B. The reason for this was that the negative sign is just showing or indicating the direction of the force. If it if negative, it means that the work done is opposite to the force field.

7 0

3 years ago