Option D is correct. The component of college life known as general education deals with the knowledge, abilities, attitudes, and values that define educated people.
It respects the linkages between different bodies of knowledge and is not constrained by disciplines. Graduates with a NOVA degree will have the ability to write effectively and engage in civic participation, critical thinking, professional preparation, numeric literacy, and scientific literacy. As a result, the College has chosen the following list of general education electives in accordance with the general education standards of the Virginia Community College System. In order to choose the best course for their curriculum and/or transferability to another college, it is strongly advised that students speak with their academic advisor or counselor.
Learn more about system here-
brainly.com/question/19548032
#SPJ4
Answer:
(a) Effectiveness of the regenerator= 0.433
(b) The rate of heat removal=21.38 kW
Explanation:
The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.
Answer:
The answer is "0.147 nm and 99.63 mol %"
Explanation:
In point (a):




find:
d(062)=?
formula:



In point (b):


formula:

that's why the composition value equal to 99.63 %
Answer:
31.1 slug, 453.4 Kg
Explanation:
given,
mass of the beam is 1000 lb
to convert mass of beam into slugs and kilograms.
1 lb is equal to 0.0311 slug
1000 lb = 1000 × 0.0311
= 31.1 slug
now, conversion of lb into kg
1 lb is equal to 0.4534 kg
so,
1000 lb = 1000 × 0.4534
= 453.4 Kg
hence, 1000 lb of beam in slugs is equal to 31.1 slugs and in kilo gram is 453.4 Kg.
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)