In the graph of the force vs the time:
Force is the gradient of the momentum versus the time.
If we get the area under the curve in this graph:
it will be calculated as : force * time
which gives the change in momentum or the impulse.
Therefore, the area under the curve <span>represents the impulse of the force in a graph of force versus time</span>
Pitch of the sound increases as frequency increases.
choose B pitch
This is very good conceptual question and can clear your doubts regarding work-energy theorem.
Whenever force is perpendicular to the direction of the motion, work done by that force is zero.
According to work-energy theorem,
Work done by all the force = change in kinetic energy.
here, work done = 0.
Therefore,
0=change in kinetic energy
This means kinetic energy remains constant.
Hope this helps
Answer
given,
k = 250 N/m
q = 900 N/m³
(FSp)s=−kΔs−q(Δs)^3
work done = Force x displacement

limits are x = 0 to x = 0.15 m
work done

![W = [\dfrac{kx^2}{2}+\dfrac{qx^4}{4}+ C]_0^0.15](https://tex.z-dn.net/?f=W%20%3D%20%5B%5Cdfrac%7Bkx%5E2%7D%7B2%7D%2B%5Cdfrac%7Bqx%5E4%7D%7B4%7D%2B%20C%5D_0%5E0.15)

W = 3.375 + 0.1139
W = 3.3488 J
b) % cubic term =
% cubic term =
Answer:
75.6J
Explanation:
Hi!
To solve this problem we must use the first law of thermodynamics that states that the heat required to heat the air is the difference between the energy levels of the air when it enters and when it leaves the body,
Given the above we have the following equation.
Q=(m)(h2)-(m)(h1)
where
m=mass=1.3×10−3kg.
h2= entalpy at 37C
h1= entalpy at -20C
Q=m(h2-h1)
remember that the enthalpy differences for the air can approximate the specific heat multiplied by the temperature difference
Q=mCp(T2-T1)
Cp= specific heat of air = 1020 J/kg⋅K
Q=(1.3×10−3)(1020)(37-(-20))=75.6J