The work done by a rotating object can be calculated by the formula Work = Torque * angle.
This is analog to the work done by the linear motion where torque is analog to force and angle is analog to distance. This is Work = Force * distance.
An example will help you. Say that you want to calculate the work made by an engine that rotates a propeller with a torque of 1000 Newton*meter over 50 revolution.
The formula is Work = torque * angle.
Torque = 1000 N*m
Angle = [50 revolutions] * [2π radians/revolution] = 100π radians
=> Work = [1000 N*m] * [100π radians] = 100000π Joules ≈ 314159 Joules of work.
Answer:
The answer is not able to be solved, because we dont know what objects are in it, and how heavy they are. More information please!
Explanation:
Answer:Fg = mg however newtons second law states that the net force acting on an object is equal to it's mass times it's acceleration so what allows us to say that Fg = mg because certainly not for every single situation the net force is going to equal to the force of gravity please explain... what allows us to say Fg = mg
Source https://www.physicsforums.com/threads/fg-mg-questioned.336776/
Explanation:
Answer:
a) 6076 m
b) 43.33 m/s
c) 68 m/s
Explanation:
(a) If the airplane rounds half the circle in 156s, its displacement is the circle diameter in 156s, or twice the circle's radius
s = 2r = 2* 3.38km = 6.76 km or 6760 m
(b) The average velocity would be displacement over unit of time
v = s/t = 6760 / 156 = 43.33 m/s
(c) The length of the chord it's swept in 156s is half of the circle perimeter
c = πr = π3.38 = 10.62 km or 10620 m
The airplane average speed is its chord length over a unit of time
c / t = c / 156 = 68 m/s
Answer:
m = 14*26 = 364
Explanation:
overall magnification is given as m

mo magnification of objective lens
me magnification of EYE lens
where mo is given as

and me as

d is distant of distinct vision = 25.0 cm for normal eye
fe = focal length of eye piece
focal length of objective lense is 0.140 cm
we know that








m = 14*26 = 364