Answer:
According to newton's second law of motion F=ma so
Explanation:
Data:-M=25kg ,a=7m/s² F=? ,solution:-F=ma ,F=25×7=175kgm/s² or N
Answer:
The Required pressure for this situation is P= 735000Pa
Explanation:
In Determining the required pressure in this situation we use two equations
First one is
F = mg = (ρhA)g
And Second one is
P =
= (ρhAg)/A Where P is pressure
We get
P = ρhg
since g = 9.8 m/s and h is given that is 75m and ρ = 
so
P = (9.8 m/s)(
)(75) we get
P = 735000 Pa
Answer:

Explanation:
A) In order to solve the table it is necessary to consult tables A11-E and A10E for refrigerant R134-a
In this way we obtain that:

In this way,


In this way the entropy change is,

B) Whenever entropy yields a positive result, the process can be carried out adiabatically.
Answer:

Explanation:
According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.
Being
and
the masses of pucks a and b respectively, the initial momentum of the system is

Since b is initially at rest

After the collision and being
and
the respective velocities, the total momentum is

Both momentums are equal, thus
Solving for 


The initial kinetic energy can be found as (provided puck b is at rest)


The final kinetic energy is


The change of kinetic energy is

Answer: minimum theoretical value of T = 750k
Explanation:
Assuming the the cycle is reversible and is ideal then
Wnet/Qh = Nmin .... equa 1
Equation 1 can be rewritten as
(Th -TL)/ Th ...equation 2
Th= temp of hot reservoir
TL= temp of low reservoir= 300
Wnet = power generated=10kw
He = energy transfer=10kj per cycle
Qhe = power transfer = (100/60)*10Kj = 16.67kw
Sub into equat 1
Nmin = 10/16.67 = 0.6
Sub Nmin into equation 2
Th = -TL/(Nmin -1) = -300k/(0.6 - 1)
Th =750k