Answer:
if there is no friction in a simple machine, work output and work input are found equal in that machine
Explanation:
Answer:The net force acting on the car is 3×103 Newtons
Explanation:
Answer:
728 N
Explanation:
= length of the wire = 0.680 m
= mass of the steel wire = 0.0046 kg
= Fundamental frequency = 261.6 Hz
= tension force in the steel wire
Fundamental frequency in wire is given as
Answer:
Terminal speed, v = 6901.07 m/s
Explanation:
It is given that,
Mass of the horizontal bar, m = 30 g = 0.03 kg
Length of the bar, l = 13 cm = 0.13 m
Magnetic field, 
Resistance, R = 1.2 ohms
We need to find the terminal speed oat which the bar falls. When terminal speed is reached,
Force of gravity = magnetic force
..................(1)
i is the current flowing
l is the length of the rod
Due to the motion in rods, an emf is induced in the coil which is given by :
, v is the speed of the bar
Equation (1) becomes,
v = 6901.07 m/s
So, the terminal speed at which the bar falls is 6901.07 m/s. Hence, this is the required solution.
The work done when a spring is stretched from 0 to 40cm is 4J.
What is work done?
Work done is the magnitude of force multiplied by displacement of an object. It is also the amount of energy transferred to an object when work is done on that.
The work done on the spring to stretch to 40cm is,
F = kx
where F is force, k is force constant.
k = F / x = 10 N / 20 * 10^-2 m = 50 N/m
W = 0.5 * k * (x)^2
where W = work done, k = force constant.
W = 0.5 x 50 x (40 x 10^-2)^2 = 4 J.
Therefore, the work done on the spring when it is stretched to 40cm is 4J.
To learn more about work done click on the given link brainly.com/question/25573309
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