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Anvisha [2.4K]
3 years ago
12

A project is located in an area with a demand-response program and on a site that has enough room for a wind-turbine to allow fo

r on-site renewable energy. If the project implements both of these strategies, which of the following will occur?
A. The project's energy costs will decrease B. The project's renewable energy production will decrease C. The project's energy demands will decrease D. the project minimum energy performance will decrease
Physics
1 answer:
Ksju [112]3 years ago
5 0

Answer:

A. The project's energy costs will decrease

Explanation:

Since the project is located in an area with a demand-response program and on a site that has enough room for a wind-turbine to allow for on-site renewable energy.

Hence, the project's energy costs will decrease very well because it's implementing both of these strategies;

- Area with demand-response program.

- On-site renewable energy.

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Which of the following might be the material list for an experiment?
Alisiya [41]

Answer:

C

Explanation:

8 0
3 years ago
Identify the type of reaction shown in each of the following chemical equations
tensa zangetsu [6.8K]

i need to know the chemical eqaitions so......

6 0
3 years ago
Hello help me please :)
hjlf

Answer: 75 N

Because we're finding the force, we need to use F = ma. M (Mass) = 25 and A (Acceleration) = 3. Multiply 25(3) to get 75. So, you're answer should be 75 N.

7 0
2 years ago
A Carnot engine operates between two heat reservoirs at temperatures THTH and TCTC. An inventor proposes to increase the efficie
Marat540 [252]

Answer:

e_12=1-Tc/Th

This is same as the original Carnot engine.  

Explanation:

For original Carnot engine, its efficiency is given by

e = 1-Tc/Th

For the composite engine, its efficiency is given by

e_12=(W_1+W_2)/Q_H1

where Q_H1 is the heat input to the first engine, W_1 s the work done by the first engine and W_2 is the work done by the second engine.  

But the work done can be written as  

W= Q_H + Q_C with Q_H as the heat input and Q_C as the heat emitted to the cold reservoir. So.  

e_12=(Q_H1+Q_C1+Q_H2+Q_C2)/Q_H1

But Q_H2 = -Q_C1 so the second and third terms in the numerator cancel  

each other.

 e_12=1+Q_C2/Q_H1

but, Q_C2/Q_H2= -T_C/T'

⇒ Q_C2 = -Q_H2(T_C/T')

               = Q_C1(T_C/T')

(T1 is the intermediate temperature)  

But, Q_C1 = -Q_H1(T'/T_H)

so, Q_C2 =  -Q_H1(T'/T_H)(T_C/T') = Q_H1(T_C/T_H) So the efficiency of the composite engine is given by  

e_12=1-Tc/Th

This is same as the original Carnot engine.  

7 0
3 years ago
An object is thrown upwards with a speed of 14 m/s. How long does it take to reach a height of 5.0 m above the projection point
Mamont248 [21]

Answer:

2.43 s

Explanation:

Using newton's equation of motion.

T = (v-u)/g

Where T = time taken for the object to return to the point of projection , u = initial velocity, v = final velocity, g = acceleration due to gravity.

Given: v =-14 m/s, u = 14 m/s, g = -9.8 m/s²

T = (-14-14)/-9.81

T = 2.85 s

Note: We look for the object's speed at 5.0 m.

using

v² = u²+2gs.................................... Equation 1

Where v = final velocity, u = initial velocity, g = acceleration due to gravity, s = distance.

Given:  u = 14 m/s, g = -9.81 m/s², s = 5.0 m

Substitute into equation 1

v² = 14²+(-9.81×5×2)

v² = 196-98.1

v = √97.9

v = 9.89

We look for the time taken for the velocity to decrease from 14 m/s to 9.89 m/s.

using

v = u+gt

t =(v-u)/g........................... Equation 2

Where t = time taken for the object to decrease it velocity from 14 m/s to 9.89 m/s

Given: v = 9.89 m/s, u =14 m/s g = -9.81 m/s²

t = (14-9.89)/-9.81

t = -4.11/-9.81

t = 0.42 s

Thus,

Time taken to reach 5.0 m above projection point = T-t

=2.85-0.42

2.43 s

4 0
3 years ago
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