Which statement is true about a planet’s orbital motion?
1 answer:
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,
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.
Hence, the orbital motion results when the object’s forward motion is balanced by a second object’s gravitational pull.
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Option ( c ) is correct.
Using the formula for power
P= power
i= current= 0.4 A
R= resistance= 30 ohm
so power= (0.4)² (30)
Power=4.8 W
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 +
200 = 0 +
200 =
200 / 4.9 =
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 +
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 (
)
- a = acceleration due to gravity = 9.8 m/s^{2}
substituting all required values we have
s = 0 + ()
s = 0 + ()
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 +
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 (
- a = acceleration due to gravity = 9.8 m/s^{2}
substituting all required values we have
s = 0 + ()
s = 0 + ()
s = 22.22 cm
