Answer:


Explanation:
<u>Net Force And Acceleration
</u>
The Newton's second law relates the net force applied on an object of mass m and the acceleration it aquires by

The net force is the vector sum of all forces. In this problem, we are not given the magnitude of each force, only their angles. For the sake of solving the problem and giving a good guide on how to proceed with similar problems, we'll assume both forces have equal magnitudes of F=40 N
The components of the first force are


The components of the second force are


The net force is


The magnitude of the net force is


The acceleration has a magnitude of



The direction of the acceleration is the same as the net force:


Answer:

Explanation:
Given that
d= 1.5 in ( 1 in = 0.0254 m)
d= 0.0381 m
P= 75 hp ( 1 hp = 745.7 W)
P= 55927.5 W
N= 1800 rpm
We know that power P is given as

T=Torque
N=Speed

T=296.85 N.m
The maximum shear stress is given as



We know that 1 MPa =0.145 ksi

Answer:
I hope it is no too late
Explanation:
hmmm,
In a gas, for example, the molecules are traveling in random directions at a variety of speeds - some are fast and some are slow. ... If more energy is put into the system, the average speed of the molecules will increase and more thermal energy or heat will be produced.
Answer:
α = 3×10^-5 K^-1
Explanation:
let ΔL be the change in length of the bar of metal, ΔT be the change in temperature, L be the original length of the metal bar and let α be the coefficient of linear expansion.
then, the coefficient of linear expansion is given by:
α = ΔL/(ΔT×L)
= (0.3×10^-3)/(100)(100×10^-3)
= 3×10^-5 K^-1
Therefore, the coefficient of linear expansion is 3×10^-5 K^-1