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
#SPJ4
Answer:
Really fast, usually would bounce up and down after it falls
Explanation:
Answer:
The angle the wire make with respect to the magnetic field is 30°
Explanation:
It is given that,
Length of straight wire, L = 0.6 m
Current carrying by the wire, I = 2 A
Magnetic field, B = 0.3 T
Force experienced by the wire, F = 0.18 N
Let θ be the angle the wire make with respect to the magnetic field. Magnetic force is given by :



So, the angle the wire make with respect to the magnetic field is 30°. Hence, this is the required solution.
Answer:
a. 0.01 C
b. dissipated to outside environment
Explanation:
Let the specific heat of copper be 0.3846 kJ/kg-K or 384.6 J/kg-C
(a)The original kinetic energy of the block is:

As 85% of this kinetic energy is converted to block internal heat energy, with specific heat we can calculate the rise in temperature:




(b) the remaining 15% energy would probably be dissipated to outside environment as heat energy
Answer:

Explanation:
Given data:
Height, 
Time, 
Let the initial velocity of the object be <em>u</em>.
From the kinematic equation,



(rounding to tenth place)