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Harman [31]
3 years ago
10

Is the gap store an example of commercial construction? Yes or no

Engineering
1 answer:
user100 [1]3 years ago
5 0

Answer:

no

Explanation:

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If a 9V battery produces a current of 3 A through a load, what is the resistance of the load
Elden [556K]

3 ohms hope this helps :D ❤

7 0
2 years ago
Read 2 more answers
A geothermal heat pump absorbs 15 KJ/s of heat from the Earth 15 m below a house. This heat pump uses a 7.45 kJ/s compressor.
Anna007 [38]

Answer:

COP of the heat pump is 3.013

OP of the cycle is  1.124

Explanation:

W = Q₂ - Q₁

Given

a)

Q₂ = Q₁ + W

     = 15 + 7.45

     = 22.45 kw

COP = Q₂ / W = 22.45 / 7.45 = 3.013

b)

Q₂ = 15 x 1.055 = 15.825 kw

therefore,

Q₁ = Q₂ - W

Q₁ = 15.825 - 7.45 = 8.375

∴ COP = Q₁ / W = 8.375 / 7.45 = 1.124

4 0
3 years ago
Read 2 more answers
The capacity of a battery is 1800 mAh and its OCV is 3.9 V. a) Two batteries are placed in series. What is the combined battery
Lynna [10]

Answer:

capacity  = 0.555 mAh

capacity  = 3600 mAh

Explanation:

given data

battery = 1800 mAh

OCV = 3.9 V

solution

we get here capacity when it is in series

so here Q = 2C  

capacity  = 2 × ampere × second   ...............1

put here value and we get

and 1 Ah = 3600 C

capacity  = \frac{2}{3600}

capacity  = 0.555 mAh

and

when it is in parallel than capacity will be

capacity = Q1 +Q2   ...............2

capacity  = 1800 + 1800

capacity  = 3600 mAh

3 0
3 years ago
A car accelerates from rest with an acceleration of 5 m/s^2. The acceleration decreases linearly with time to zero in 15 s, afte
Tpy6a [65]

Answer: At time 18.33 seconds it will have moved 500 meters.

Explanation:

Since the acceleration of the car is a linear function of time it can be written as a function of time as

a(t)=5(1-\frac{t}{15})

a=\frac{d^{2}x}{dt^{2}}\\\\\therefore \frac{d^{2}x}{dt^{2}}=5(1-\frac{t}{15})

Integrating both sides we get

\int \frac{d^{2}x}{dt^{2}}dt=\int 5(1-\frac{t}{15})dt\\\\\frac{dx}{dt}=v=5t-\frac{5t^{2}}{30}+c

Now since car starts from rest thus at time t = 0 ; v=0 thus c=0

again integrating with respect to time we get

\int \frac{dx}{dt}dt=\int (5t-\frac{5t^{2}}{30})dt\\\\x(t)=\frac{5t^{2}}{2}-\frac{5t^{3}}{90}+D

Now let us assume that car starts from origin thus D=0

thus in the first 15 seconds it covers a distance of

x(15)=2.5\times 15^{2}-\farc{15^{3}}{18}=375m

Thus the remaining 125 meters will be covered with a constant speed of

v(15)=5\times 15-\frac{15^{2}}{6}=37.5m/s

in time equalling t_{2}=\frac{125}{37.5}=3.33seconds

Thus the total time it requires equals 15+3.33 seconds

t=18.33 seconds

3 0
2 years ago
What is need for using fins?
antiseptic1488 [7]

Answer: It is a term of heat transfer process in which fins are surface that are the extension of the object to work for the heat exchangers to increase the heat exchanging rate.

 Explanation: Fins are considered to help the heat exchanger surface to lead the process of heat transfer by increasing the are of the surface which is exposed to the surroundings. Fins work really well with materials having high thermal conductivity and will  be more effective. They are preferred because they increase the rate of exchange of heat by increment in the convection.

7 0
3 years ago
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