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sweet [91]
3 years ago
13

A satellite at a particular point along an elliptical orbit has a gravitational potential energy of 5100 MJ with respect to Eart

h's surface and a kinetic energy of 4200 MJ . Later in its orbit the satellite's potential energy is 5700 MJ . Part A Use conservation of energy to find its kinetic energy at that point.
Physics
1 answer:
serious [3.7K]3 years ago
5 0

To solve this problem we will apply the theorem given in the conservation of energy, by which we have that it is conserved and that in terms of potential and kinetic energy, in their initial moment they must be equal to the final potential and kinetic energy. This is,

E_{initial} = E_{final}

PE_{initial}+KE_{initial} = PE_{final}+KE_{final}

Replacing the 5100MJ for satellite as initial potential energy, 4200MJ for initial kinetic energy and 5700MJ for final potential energy we have that

KE_{final} = (PE_{initial}+KE_{initial} )-PE_{final}

KE_{final} = (5100+4200)-5700

KE_{final} = 3600MJ

Therefore the final kinetic energy is 3600MJ

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The pressure in an automobile tire depends on the temperature of the air in the tire. When the air temperature is 25°C, the pres
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Answer:0.0704 kg

Explanation:

Given

initial Absolute pressure(P_1)=210+101.325=311.325

T_1=25^{\circ}\approx 298 K

V=0.025 m^3

T_2=50^{\circ}\approx 323 K

as the volume remains constant therefore

\frac{P_1}{T_1}=\frac{P_2}{T_2}

\frac{311.325}{298}=\frac{P_2}{323}

P_2=337.44 KPa

therefore Gauge pressure is 337.44-101.325=236.117 KPa

Initial mass m_1=\frac{P_1V}{RT_1}=\frac{311.325\times 0.025}{0.0287\times 298}

m_1=0.91 kg

Final mass m_2=\frac{P_2V}{RT_2}=\frac{311.325\times 0.025}{0.0287\times 323}

m_2=0.839

Therefore m_1-m_2=0.91-0.839=0.0704 kg of air needs to be removed to get initial pressure back

4 0
3 years ago
The three main types of nuclear radiation are alpha, beta, and gamma
valkas [14]

Answer:

Gamma beta alpha

Explanation:

6 0
2 years ago
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The attraction of liquid particles for a solid surface is due to ____.
White raven [17]
This attraction occurs from adhesion, also known as adsorption <span />
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An object is hanging by a string from the ceiling of an elevator. The elevator is moving upward with a constant speed. What is t
xeze [42]

Answer:

The magnitude of the tension in he string is equal to the magnitude of the weight of the object.

Explanation:

According to the Newton's 1st law, An object will remain at rest or in uniform motion in a straight line unless acted upon by an unbalanced force.

In here, the elevator is moving with a constant speed. So the object must have the equal constant speed. Which means, it has a uniform motion. According to Newton's 1st law, the total unbalanced force on the object must be zero . As we know, there are only two forces are on the object and they are,

The tension in string(T)   ,   The weight of the object(W) .

                                  ∴ F    =   0

                                 T  -  W  =   0

So to balanced those forces, the magnitude of the tension in the string must be equal to the magnitude of the weight of  the object.  

       

8 0
2 years ago
a shot putter releases the shot some distance above the level ground with a velocity of 12.0 m/s, 51.0 ∘above the horizontal. th
Savatey [412]

Answer:

15.7 m

Explanation:

The range (horizontal distance) of the projectile is determined only by its horizontal motion.

The horizontal motion is a motion with constant speed, which is equal to the initial horizontal velocity of the object:

v_x = v cos \theta

where

v = 12.0 m/s is the initial velocity

\theta=51.0^{\circ} is the angle between the direction of v and the horizontal

Substituting,

v_x = (12.0 m/s)(cos 51.0^{\circ} )=7.55 m/s

We know that the projectile hits the ground in a time of

t = 2.08 s

so the horizontal distance covered is

d = v_x t = (7.55 m/s)(2.08 s)=15.7 m

8 0
2 years ago
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