Answer:
Explanation:
Power (p)= 40 watt
time (t)= 3 minutes = 180 seconds
Energy (E)= ?
we know
P = E/t
40 = E / 180
E = 7200 Joules
Therefore it does 7200 joules of work.
hope it helps :)
Answer:
30m/s^2
Explanation:
Acceleration=Final Velocity-Initial Velocity/Time
10m/s^2= Final Velocity-0m/s/3
30m/s^2= Final Velocity
Final Velocity=30m/s^2
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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Answer:
Explained below
Explanation:
1) The human arm: This is a type of simple machine called "Lever". In this type of machine, the elbow acts as the fulcrum, the palm serves as the load because that's where we place the load we want to carry. While the inner part of the arm which is the inner part of the elbow represents the effort because that is the joint we mover when making use of our arms.
2) Pulleys: An example of this in the human body is the knee cap where the direction of an applied force is changed. Thus means as it is in motion, it alters the direction for which the quadriceps tendon pulls on the tibia.
3) wheel and axle: An example of this in the human body is the lateral rotation of the shoulder joint medial. The humerus which is the bone between the shoulder and elbow will act as the axle while the rotator will be the will because when it is rotated a little bit, the humerus will move along with it.
Answer:

Explanation:
As we know that the length of the conductor is given as

now if it is converted into a square then we have


now the are of the loop will be

now the magnetic flux is defined as

here we know
B = 1.0 T


