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Dmitrij [34]
3 years ago
7

What is escape velocity?

Physics
1 answer:
olga_2 [115]3 years ago
6 0
<span>It is the lowest velocity which a body must have in order to escape the gravitational attraction of a particular planet or other object.

Every planet has their own corresponding escape velocities. Example - Earth has escape velocity of 11.2 Km/s. It means, if you want to leave the Earth's gravitational field then it's the lowest speed which you need to acquire otherwise you wouldn't do that!

Hope this helps!</span>
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Kruka [31]

Answer:

Immunization, or immunisation, is the process by which an individual's immune system becomes fortified against an infectious agent.

Explanation:

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4 0
3 years ago
The mechanical advantage of a simple machine is 3 and its velocity ratio is 4 find its efficiency ​
LiRa [457]

Answer:

75%

Explanation:

Efficiency=MA/VR*100

=3/4*100

=0.75*100

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6 0
2 years ago
What is the mass of a 600 gram object on the moon
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5 0
4 years ago
Read 2 more answers
someone made a copy of an analog wave in a digital wave compare the copies with the original waves and explain which type of wav
lilavasa [31]

Answer: Analog transfer information quicker, example electromagnetic waves which are used in our modern day phones.

Explanation:  Digital signals are a more reliable form of transmitting information because an error in the amplitude or frequency value would have to be very large in order to cause a jump to a different value. Signals are composed of infinite possible values. Signals are composed of only two possible values: 0 or 1.

6 0
3 years ago
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LOTS OF POINTSA rocket of mass 40 000 kg takes off and flies to a height of 2.5 km as its engines produce 500 000 N of thrust.
Svetradugi [14.3K]

Answer:

i) E = 269 [MJ]    ii)v = 116 [m/s]

Explanation:

This is a problem that encompasses the work and principle of energy conservation.

In this way, we establish the equation for the principle of conservation and energy.

i)

E_{k1}+W_{1-2}=E_{k2}\\where:\\E_{k1}= kinetic energy at moment 1\\W_{1-2}= work between moments 1 and 2.\\E_{k2}= kinetic energy at moment 2.

W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]

At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.

Er = 269*10^6[J]

ii ) With the energy calculated at the previous point, we can calculate the speed developed.

E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]

8 0
3 years ago
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