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dolphi86 [110]
3 years ago
11

What is escape velocity​

Physics
2 answers:
Bogdan [553]3 years ago
3 0
The minimum speed needed a non-propelled object to escape from the gravitational influence of a massive body, that is to achieve an infinite distance from it. I think
Eddi Din [679]3 years ago
3 0

<em>Answer:</em>

<em>E</em><em>scape velocity is defined as the velocity which will take a body to go the infinite distance away above the surface of the earth when projected upwards.</em>

<em>Hope</em><em> </em><em>this</em><em> </em><em>helps</em><em>.</em><em>.</em><em>.</em>

<em>Good</em><em> </em><em>luck</em><em> on</em><em> your</em><em> assignment</em><em>.</em><em>.</em><em>.</em><em>.</em>

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anyanavicka [17]
The Answer: Parallel
6 0
3 years ago
Read 2 more answers
Please help me
RSB [31]
The potentially harmful effects of mining can be reduced by A. Flushing mines with water.
5 0
3 years ago
After being struck by a bowling ball, a 1.3 kg bowling pin sliding to the right at 5.0 m/s collides head-on with another 1.3 kg
GuDViN [60]

Answer:

a) 4.2m/s

b) 5.0m/s

Explanation:

This problem is solved using the principle of conservation of linear momentum which states that in a closed system of colliding bodies, the sum of the total momenta before collision is equal to the sum of the total momenta after collision.

The problem is also an illustration of elastic collision where there is no loss in kinetic energy.

Equation (1) is a mathematical representation of the the principle of conservation of linear momentum for two colliding bodies of masses m_1 and m_2 whose respective velocities before collision are u_1 and u_2;

m_1u_1+m_2u_2=m_1v_1+m_2v_2..............(1)

where v_1 and v_2 are their respective velocities after collision.

Given;

m_1=1.3kg\\u_1=5m/s\\m_2=1.3kg\\u_2=0m/s

Note that u_2=0 because the second mass m_2 was at rest before the collision.

Also, since the two masses are equal, we can say that m_1=m_2=m so that equation (1) is reduced as follows;

mu_1+mu_2=mv_1+mv_2\\\\m(u_1+u_2)=m(v_1+v_2)..............(2)

m cancels out of both sides of equation (2), and we obtain the following;

u_1+u_2=v_1+v_2.............(3)

a) When v_1=0.8m/s, we obtain the following by equation(3)

5+0=0.8+v_2\\hence\\v_2=5-0.8\\v_2=4.2m/s

b) As m_1 stops moving v_1=0, therefore,

5+0=0+v_2\\v_2=5m/s

5 0
3 years ago
What happens when gasoline is used to power a vehicle?
mojhsa [17]
<span>Energy is neither lost nor gained as it transforms from chemical, to heat, to mechanical energy.</span>
5 0
3 years ago
A 1400-kg car traveling east at 25m/s collides with a 1800-kg car traveling at a speed of 20m/s in a direction that makes angle
julia-pushkina [17]
M1*V1 + M2*V2 = M1*V + M2*V.
1400*25 + 1800*20[180+40]=1400*V+1800*V.
Divide both sides by 100:
14*25 + 18*20[220o] = 14V + 18V.
350 + 360[220o] = 32V.
350 - 276-231i = 32V.
74 - 231i = 32V.
242.6[-72.2o] = 32V.
V = 7.6m/s[-72.2o]=7.6m/s[72o] S. of E.
7 0
3 years ago
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