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blagie [28]
3 years ago
13

The complex power of a load is 10-10j VA. What component should be added in parallel with the load so that the new load has a un

ity power factor
Engineering
1 answer:
Korolek [52]3 years ago
7 0
Need points bdjdjdhdhd
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Take another picture i cant see nun
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2 years ago
In an aircraft jet engine at takeoff, the combustion products expand adiabatically in the exhaust nozzle. At entrance to the noz
tensa zangetsu [6.8K]

At entrance to the nozzle, the pressure is 0.180 MPa and the temperature is 1200 K. The kinetic energy of the gas entering the nozzle is very much smaller than ... The specific heat of the exhaust gas varies with temperature approximately as follows: ... Problem 4P: In an aircraft jet engine at takeoff, the combustion product.

5 0
2 years ago
A steel bar 100 mm long and having a square cross section 20 mm x 20 mm is pulled in
Ierofanga [76]

Answer:

222.5 Gpa

Explanation:

From definition of engineering stress, \sigma=\frac {F}{A}

where F is applied force and A is original area

Also, engineering strain, \epsilon=\frac {\triangle l}{l} where l is original area and \triangle l is elongation

We also know that Hooke's law states that E=\frac {\sigma}{\epsilon}=\frac {\frac {F}{A}}{\frac {\triangle l}{l}}=\frac {Fl}{A\triangle l}

Since A=20 mm* 20 mm= 0.02 m*0.02 m

F= 89000 N

l= 100 mm= 0.1 m

\triangle l= 0.1 mm= 0.1\times 10^{-3} m

By substitution we obtain

E=\frac {89000\times 0.1}{0.02^{2}\times 0.1\times 10^{-3}}=2.225\times 10^{11}= 225.5 Gpa

5 0
3 years ago
John has just graduated from State University. He owes $35,000 in college loans, but he does not have a job yet. The college loa
IRISSAK [1]

Answer:

The correct response is "821.88". A further explanation is given below.

Explanation:

According to the question,

The largest amount unresolved after five years would have been:

= 35000\times (\frac{F}{P}, 4 \ percent,5 )

= 35000\times 1.216 7

= 42584.50

Now,

time (t) will be:

= 5\times 12

= 60 \ monthly \ payments

So, monthly payment will be:

= 42582.85\times (\frac{A}{P}, 0.5 \ percent,60 )

= 42584.50\times 0.0193

= 821.88

6 0
3 years ago
Compared to 15 mph on a dry road, about how much longer will it take for
Marysya12 [62]

Answer:

8 to 10 times

Explanation:

For dry road

u= 15 mph        ( 1 mph = 0.44 m/s)

u= 6.7 m/s

Let take coefficient of friction( μ) of dry road is 0.7

So the de acceleration a = μ g

a= 0.7 x 10  m/s ²                         ( g=10 m/s ²)

a= 7 m/s ²

We know that

v= u - a t

Final speed ,v=0

0 = 6.7 - 7 x t

t= 0.95 s

For snow road

μ = 0.4

de acceleration a = μ g

a = 0.4 x 10 = 4 m/s ²

u= 30 mph= 13.41 m/s

v= u - a t

Final speed ,v=0

0 = 30 - 4 x t'

t'=7.5 s

t'=7.8 t

We can say that it will take 8 to 10 times more time as compare to dry road for stopping the vehicle.

8 to 10 times

7 0
3 years ago
Read 2 more answers
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