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Reil [10]
3 years ago
7

The first artificial satellite to orbit the Earth was Sputnik I, launched October 4, 1957. The mass of Sputnik I was 83.5 kg, an

d its distances from the center of the Earth at apogee and perigee were 7330 km and 6610 km, respectively. Find the difference in gravitational potential energy for Sputnik I as it moved from apogee to perigee.
Physics
1 answer:
9966 [12]3 years ago
4 0

Answer:

-4.941*10^8J.

Explanation:

To solve this exercise it is necessary to take into account the concepts related to gravitational potential energy, as well as the concept of perigee and apogee of a celestial body.

By conservation of energy we know that,

\Delta U = \Delta_{perogee}-\Delta_{Apogee}

Where,

U= \frac{-GmM_e}{r}

Replacing

\Delta U = \frac{-GmM_e}{r_p}- \frac{-GmM_e}{r_a}

\Delta U = GmM_e (\frac{1}{r_A}-\frac{1}{r_p})

Our values are given by,

m = 85.5Kg

M_e = 5.97*10^{24}Kg

r_A = 7330Km

r_p = 6610Km

G = 6.67*10^{-11}Nm^2/Kg^2

Replacing at the equation,

\Delta U = (6.67*10^{-11})(85.5)(5.97*10^{24}) (\frac{1}{7330}-\frac{1}{6610})

\Delta U = -4.941*10^8J

Therefore the Energy necessary for Sputnik I as it moved from apogee to perigee was -4.941*10^8J.

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A positive point charge Q is fixed on a very large horizontal frictionless tabletop. A second positive point charge q is release
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Answer:

(A) As it moves farther and farther from Q, its speed will keep increasing.

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When a positive charge Q is fixed on a horizontal frictionless tabletop and a second charge q is released near to it then according to the Coulombs law the force acting on it decreases with the square of the distance between them.

Mathematically:

F=\frac{1}{4\pi.\epsilon_0} \times \frac{Q.q}{r^2}

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\epsilon_0= permittivity of free space

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7 0
3 years ago
A positive charge is fixed at point (−4,−3) and a negative charge − is fixed at point (−4,0). Determine the net electric force ⃗
Masteriza [31]

The net electric force acting on a positive test charge at the origin is determined as ¹/₉(kq₁q₂).

<h3>Net electric force on the charges</h3>

The net electric force on the charges is calculated as follows;

F = kq₁q₂/r²

where;

  • k is coulomb's constant
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<h3>Distance between the charges</h3>

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F_{net} = \frac{kq_1q_2}{3^2} \\\\F_{net} = \frac{1}{9} (kq_1q_2) , \  N

Thus, the net electric force acting on a positive test charge at the origin is determined as ¹/₉(kq₁q₂).

Learn more about electric force here: brainly.com/question/17692887

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3 0
2 years ago
A car accelerates uniformly from rest at a speed of 1.67 ft s^2 over a distance of 5 yards.What is the acceleration of a car?
Nina [5.8K]

Answer:

a= 17.69 m/s^2

Explanation:

Step one:

given data

A car accelerates uniformly from rest to 23 m/s

u= 0m/s

v= 23m/s

distance= 30m

Step two:

We know that

acceleration= velocity/time

also,

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23= 30/t

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t= 1.30 seconds

hence

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5 0
2 years ago
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