Answer:
a)COP=5.01
b)
KW
c)COP=6.01
d)
Explanation:
Given
= -12°C,
=40°C
For refrigeration
We know that Carnot cycle is an ideal cycle that have all reversible process.
So COP of refrigeration is given as follows
,T in Kelvin.

a)COP=5.01
Given that refrigeration effect= 15 KW
We know that 
RE is the refrigeration effect
So
5.01=
b)
KW
For heat pump
So COP of heat pump is given as follows
,T in Kelvin.

c)COP=6.01
In heat pump
Heat rejection at high temperature=heat absorb at low temperature+work in put

Given that
KW
We know that 


d)
Yes, they have to make certain sacrifices ! for many many benefits.
Answer: the answer is a marine biologist
Explanation: the answer is marine biologist because Jylan likes to be in and out the water which means he would be perfect as marine biologist.
Answer:
1. B. False
2. B. False
3. A. True
4. B. False
5. A. True
6. A. True
7. A. True
Explanation:
1. B. False
The relation of Reynolds' number, Reₓ to boundary layer thickness δ at a point x is given by the relation

That is the boundary layer thickness is inversely proportional to the square root of the Reynolds' number so that if the Reynolds' number were to increase, the boundary layer thickness would decrease
Therefore, the correct option is B. False
2. B. False
From the relation

As the outer flow velocity increases, the boundary layer thickness diminishes
3. A. True
As the viscous force is increased the boundary layer thickness increases
4. B. False
Boundary layer thickness is inversely proportional to velocity
5. A. True
The boundary layer model developed by Ludwig Prandtl is a special case of the Navier-Stokes equation
6. A. True
Given a definite boundary layer thickness, the curve representing the boundary layer thickness is a streamline
7. A. True
The boundary layer approximation by Prandtl Euler bridges the gap between the Euler (slip boundary conditions) and Navier-Stokes (no slip boundary conditions) equations.
Answer:
9.248 <
< 9.253 mm
Explanation:
Given data:
standard deviation = 0.021 mm
sample mean = 9.251 mm
total sample = 409 m
confidence interval level = 95%
mean diameter of entire batch 
where
- sample mean = 9.251 mm
z = critical value
plugginf all value in the above relation to get the mean diameter

9.248 <
< 9.253 mm