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frutty [35]
2 years ago
9

Escape velocity of an object from the surface of a planet depends upon:

Physics
1 answer:
andrey2020 [161]2 years ago
7 0

Answer:

Escape velocity: Measuring the gravitational strength of an object

The escape velocity is the exact amount of energy you would need to escape the gravitational clutches of an object with mass. Since all objects have mass, they all have a measureable gravitational strength. A good way to think about escape velocity is to think about a deep well (physicists like to think of this as an energy well). If you are at the bottom of the well and want to get out (to escape), you need enough energy to climb out. The deeper the well, the more energy you will have to expend in order to climb to

the top. If you have only enough energy to get half way out, you will eventually fall back to the bottom. The escape velocity is a way of measuring the exact amount of energy needed to reach the lip of the well -- and have no energy left over for walking away.

When a ball is thrown up into the air from the surface of the Earth, it does not have enough energy to escape. So it falls back down. How might we enable the ball to escape? Throw it harder, give it more energy. How hard must we throw it? Just hard enough to get over the top, over the edge of the well.

We can find this energy directly by saying that the kinetic energy of the thrown ball must exactly equal the 'potential energy' of the well. From basic physics we know that the potential energy for an object at a height above a surface is:

Epotential= GMm/R

where

G = Newton's universal constant of gravity = 6.67 x 10-11 N-m2/kg3

M = the mass of the 'attracting object' [the planet] [in units of kg]

m = the mass of the object trying to escape [e.g., me or a ball or a rocket or a molecule] [in kg]

R = the distance between the centers of objects M and m [in units of m]

note: provided we do everything in the same units, we don't have to worry about units

while the kinetic energy we know from above:

Ekinetic=0.5 m v2

where

m = mass of the moving object [in kg]

v = the velocity of object m [in m/sec]

If we set these two energies equal to each other, and solve for v, we find the exact velocity needed to escape from the energy well:

0.5 m v2= GMm/R

v= (2GM/R)0.5

and since this velocity is exactly what is needed to 'escape,' it is called the escape velocity:

vescape= (2GM/R)0.5

Explanation:

that's my all i know

correct me if I'm wrong❤️

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on cloudless nighs, heat escapes from our planet and goes into empty space A.conduction B.convection C.radiation
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A ball is thrown into the air with 100 J of kinetic energy, which is transformed to gravitational potential energy
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Answer:

Air resistance

Explanation:

Despite the law of conservation of energy stating that energy can neither be created nor destroyed but can only be transformed from one state to another, some energy is usually lost in the process of transformation and its majorly attributed to frictional loss. Friction opposes normal movement hence in air, air resistance tends to reduce the original energy compared to the initial. That is why the final energy in this case is slightly less than the original energy.

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In Ancient Greece, athletes competing in the long jump used handheld weights called halteres to lengthen their jumps. You are a
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The halter add the distance to the jump in meters is 0.55 m.

<h3>What is projectile?</h3>

When an object is thrown at an angle from the horizontal direction, the object is said to be in projectile motion. The object which follows the projectile motion is called the projectile.

The magnitude of velocity u =10.3 m/s, angle of jumping θ = 22.8 degrees.

Components of velocity in x and y direction are

Vx = 10.3 cos 22.8 = 9.5 m/s

Vy = 10.3 sin 22.8 = 4 m/s

Maximum Range of athlete achieved using halter is given by

R = u²sin2θ /g

where, u = initial velocity, θ is the angle of projection and g is the gravitational acceleration.

Substituting the values, we get

R = (10.3)² sin(2 x 22.8 °) / 2 x 9.81

R = 7.75m

At the peak of jump you throw two 5.5 kg masses horizontally behind you such that their velocity is zero in the ground's reference frame.

The momentum is conserved in this situation,

(M+2m)Vxo =MVx'

Vx' = (M+2m)/M x Vxo'

Change in x component of velocity ΔVx = Vx' -Vxo

Vxo = 2m/M x Vx

Vxo = 2 x 5.5 /78 x 9.5

Vxo = 1.34 s

Maximum height gained when final velocity is zero

Vy = 0 = Vyo -gt

time t = Vyo/g = 4/9.8 = 0.41s'

Increase in range by using of halters is

ΔR = ΔVx' x t

ΔR = 1.34 x 0.41

ΔR =0.55m

Thus, the halter add the distance to the jump in meters is 0.55 m.

Learn more about projectile.

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A 15.0 kg fish swimming at 1.10 m/s suddenly gobbles up a 4.50 kg fish that is initially stationary. Ignore any drag effects of
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Answer

given.

Mass of big fish = 15 Kg

speed of big fish = 1.10 m/s

mass of the small fish = 4.50 Kg

speed of the fish after eating small fish =?

a) using conservation of momentum

m₁v₁ + m₂v₂ = (m₁+m₂) V

15 x 1.10 + 4.50 x 0 = (15 + 4.5)V

16.5 = 19.5 V

V = 0.846 m/s

b) Kinetic energy before collision

KE_1 = \dfrac{1}{2}m_1v_1^2 + \dfrac{1}{2}m_2v_2^2

KE_1 = \dfrac{1}{2}\times 15 \times 1.1^2 + \dfrac{1}{2}m_2\times 0^2

KE₁ = 9.075 J

Kinetic energy after collision

KE_2= \dfrac{1}{2}(15+4.5)\times 0.846^2

KE₂ = 6.98 J

Change in KE = 6.98 - 9.075 = -2.096 J

hence,

mechanical energy was dissipated during this meal = -2.096 J

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