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Alex
3 years ago
15

Which statement is true about a planet’s orbital motion?

Physics
1 answer:
lana66690 [7]3 years ago
3 0

Answer:

Orbital motion results when the object’s forward motion is balanced by a second object’s gravitational pull.

Explanation:

The gravitational force is responsible for the orbital motion of the planet, satellite, artificial satellite, and other heavenly bodies in outer space.

When an object is applied with a velocity that is equal to the velocity of the orbit at that location, the body continues to move forward. And, this motion is balanced by the gravitational pull of the second object.

The orbiting body experience a centripetal force that is equal to the gravitational force of the second object towards the body.

The velocity of the orbit is given by the relation,

                                    V = \sqrt{\frac{GM}{R + h} }

Where

                   V - velocity of the orbit at a height h from the surface

                    R - Radius of the second object

                    G - Gravitational constant

                    h - height from the surface

The body will be in orbital motion when its kinetic motion is balanced by gravitational force.

                         1/2 mV^{2} = GMm/R

Hence, the orbital motion results when the object’s forward motion is balanced by a second object’s gravitational pull.

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A race car is accelerating at 5m/s and then speeds up to a final velocity of 12 m/s if the race car driver finished this let of
noname [10]

Answer:

the acceleration of the car is 1.167 m/s²

Explanation:

Given;

initial velocity of the race car, u = 5 m/s

final velocity of the race car, v = 12 m/s

time to finish the race, t = 6 s

The acceleration of the car is calculated as;

a = (v - u) / t

a = (12 - 5) / (6)

a = 1.167 m/s²

Therefore, the acceleration of the car is 1.167 m/s²

3 0
3 years ago
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How much voltage is required to run 0.025 A of current through a 36 Ω resistor?
aalyn [17]

Yes D is definitely the answer

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<em>V</em><em>=</em><em>0</em><em>.</em><em>0</em><em>2</em><em>5</em><em>×</em><em>3</em><em>6</em>

<em>V</em><em>=</em><em>0</em><em>.</em><em>9</em><em> </em><em>sa</em><em>me</em><em> </em><em>as</em><em> </em><em>0</em><em>.</em><em>9</em><em>0</em>

7 0
3 years ago
How much kinetic energy does a 4kg cat have while running at 9 m/s
ludmilkaskok [199]

Answer:

How much kinetic energy does a 4 Kg cat have while running at 9 m/s?

its 5 J of kinetic energy.

Explanation:

3 0
2 years ago
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A circuit has a voltage drop of 12.0 V across a 30.0 resistor that carries a current of 0.400 A. What is the power used by the r
Aloiza [94]

Option ( c ) is correct.

Using the formula for power

P= i^2 R

P= power

i= current= 0.4 A

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so power= (0.4)² (30)

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7 0
3 years ago
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Water drips from the nozzle of a shower onto the floor 200 cm below. The drops fall at regular (equal) intervals of time, the fi
Delicious77 [7]

Answer:

(A) 88.92 cm

(B) 22.22 cm

Explanation:

distance (s) = 200 cm = 0.2 m

initial velocity (v) = 0 m/s

acceleration due to gravity (g) = 9.8 m/s^{2}

lets first find the time (T) it takes for the first drop to strike the floor

from  s = ut + 0.5at^{2}

         200 = 0 + 0.5 x 9.8 x T^{2}

         200 = 4.9 x T^{2}

         200 / 4.9 = T^{2}

         T = 6.4

(A) When the first drop strikes the floor, how far below the nozzle is the second drop.

we can find how far the second drop was when the first drop hits the ground from the formula s = ut + 0.5at^{2}

where

  • s = distance
  • u = initial velocity = 0
  • t = time, since the drops fall at regular (equal) intervals of time, the first drop striking the floor at the instant the fourth drop begins to fall there wold be 3 time intervals and this can be seen illustrated in the attached diagram. therefore the time of the second drop = 2/3 of the time it takes the first drop to strike the ground (\frac{2}{3}.T)
  • a = acceleration due to gravity = 9.8 m/s^{2}

substituting all required values we have

s = 0 + (0.5 x 9.8 x (\frac{2}{3}T)^{2})

s = 0 + (0.5 x 9.8 x (\frac{2}{3} x 6.4)^{2})

s = 88.92 cm

(B) When the first drop strikes the floor, how far below the nozzle is the third drop.

we can find how far the third drop was when the first drop hits the ground from the formula s = ut + 0.5at^{2}

where

  • s = distance
  • u = initial velocity = 0
  • t = time, since the drops fall at regular (equal) intervals of time, the first drop striking the floor at the instant the fourth drop begins to fall there wold be 3 time intervals and this can be seen illustrated in the attached diagram. therefore the time of the third drop = 1/3 of the time it takes the first drop to strike the ground (\frac{2}{3}.T)
  • a = acceleration due to gravity = 9.8 m/s^{2}

substituting all required values we have

s = 0 + (0.5 x 9.8 x (\frac{1}{3}T)^{2})

s = 0 + (0.5 x 9.8 x (\frac{1}{3} x 6.4)^{2})

s = 22.22 cm

6 0
3 years ago
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