Answer:
76969.29 W
Explanation:
Applying,
P = F×v............. Equation 1
Where P = Power, F = force, v = velocity
But,
F = ma.......... Equation 2
Where m = mass, a = acceleration
Also,
a = (v-u)/t......... Equation 3
Given: u = 0 m/s ( from rest), v = 12.87 m/s, t = 3.47 s
Substitute these values into equation 3
a = (12.87-0)/3.47
a = 3.71 m/s²
Also Given: m = 1612 kg
Substitute into equation 2
F = 1612(3.71)
F = 5980.52 N.
Finally,
Substitute into equation 1
P = 5980.52×12.87
P = 76969.29 W
Answer:
The work done shall be 14715 Joules
Explanation:
The work done by a force 'F' in a displacement 'dy' is given by

At any position 'y' the weight shall be sum of weft of water and weight of string

Thus applying values we get

The force needed to stretch the steel wire by 1% is 25,140 N.
The given parameters include;
- diameter of the steel, d = 4 mm
- the radius of the wire, r = 2mm = 0.002 m
- original length of the wire, L₁
- final length of the wire, L₂ = 1.01 x L₁ (increase of 1% = 101%)
- extension of the wire e = L₂ - L₁ = 1.01L₁ - L₁ = 0.01L₁
- the Youngs modulus of steel, E = 200 Gpa
The area of the steel wire is calculated as follows;

The force needed to stretch the wire is calculated from Youngs modulus of elasticity given as;


Thus, the force needed to stretch the steel wire by 1% is 25,140 N.
Learn more here: brainly.com/question/21413915
Answer:
Explanation:
The stunt will likely sustain serious injury in case of concrete blocks because the average force acting on the person will be more because concrete blocks do not squeeze to provide more time for the force to act on the body instead it acts for a small amount of interval.

As impulse is constant so time requires to act force on the body is more as compared to concrete block and thus average force in mattress case is less.
Sound waves are known to be the one that's not considered as a type of electromagnetic energy. As for microwaves and x-rays, they tend to share the same frequencies that can be considered as electromagnetic, and sound waves have a different frequency than them.