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.
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Explanation:
a) I = V / R
1.70 = 115 / R
R = 115 / 1.70
R = 67.647
R = 67.65 ohms
Therefore, equivalent resistance is 67.65 ohms
b) Equivalent resistance of circuit from above sum is 67.65 ohms
Given resistance of each bulb is 1.50 ohms
Number bulbs = Equivalent resistance / Resistance of each bulb
= 67.65 / 1.50
= 45
Answer:
197.5072.
Explanation:
According to the Coulomb's law, the magnitude of the electrostatic force of interaction between two charges 
and 
which are separated by the distance 
is given by
<em>where,</em> k is the Coulomb's constant.
For the case, when,
Then, using Coulomb's law,
For the case, when,
Then, using Coulomb's law, the new electric force between the charges is given by,
Answer:
Coefficient of static friction = 0.547
Explanation:
The maximum resistive force offered by the body(coin)against the applied force(turntable) to continue its state of motion is called Coefficient Of Static Friction
When given
The angular velocity which is 51.1rpm
We are going to calculate velocity of coin
= r*w(angular velocity) = 0.108*(51.1*2π/60) = 0.58
Coefficient=F/N
F= coefficient* N = coefficient*m*g
F= mv^2/r which is centripetal force
Therefore, mv^2/r = coefficient*m*g
coefficient = mv^2/mgr = v^2/rg
Where g,gravity = 9.81
Substitute into the equation coefficient = 0.58^2/9.81*0.108
= 0.547
<h2>
Answer:</h2>
Answer to this question is (A)
<h2>
Explanation</h2>
A ball bouncing on the floor is not the example of simple harmonic motion. SHM is the special kind of to and fro motion in which a particle oscillate about its mean position in a straight line. The acceleration of the particle is always directed towards its mean position and is directly proportional to its displacement from its mean position.
In case of a ball bouncing on the ground, the motion of the ball is not SHM, as neither it’s a to and fro motion nor the acceleration is proportional to its displacement from its mean position.
