A boy swings a ball on a string at constant speed in a circle that has a circumference equal to 6 m. What is the work done on th
e ball by the 10 N tension force in the string during one revolution of the ball
1 answer:
Answer:
0 J
Explanation:
From the diagram below; we would notice that the Force (F) = Tension (T)
Also the angle θ adjacent to the perpendicular line = 90 °
The Workdone W = F. d
W = Fd cos θ
W = Fd cos 90°
W = Fd (0)
W = 0 J
Hence the force is perpendicular to the direction of displacement and the net work done in a circular motion in one complete revolution is = 0
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The orbital period of the satellite[T] is given as .
The radius of the satellite is given [R] .
we are asked here to calculate the tangential speed of the satellite.
Before going to get the solution first we have understand the tangential speed.
The tangential speed of a satellite is given as the speed required to keep the satellite along the orbit. If satellite speed is less than tangential speed,there is the chance of it falling down towards earth. If it is more,then it will deviate from it orbit and can't stick to the orbit further.In a simple way the tangential speed is the linear speed of an object in a circular path.
Now we have to calculate the tangential speed [V].
Mathematically the tangential speed [V] written as -
V
where T is the time period of the satellite and R is the radius of the satellite.
There is also another way through which we can get the solution as explained below-
We know that the tangential speed of a satellite V
where G is the gravitational constant and M is the mas of central object.
But we know that
⇒ where g is the acceleration due to gravity of that central object.
Hence
⇒
By knowing the value of g due to that central object we can also calculate its tangential speed.