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denis-greek [22]
3 years ago
12

The pressure at the pump outlet is 2372 psi. If the pump has a mechanical efficiency of 0.93, what input horsepower is required

to drive the pump
Physics
1 answer:
11Alexandr11 [23.1K]3 years ago
8 0

Answer:

HP = 27.08 hp

Explanation:

The complete question has a theoretical flow rate of 18.2 GPM. So, to find the input horsepower, we will use the following formula:

HP = QP/1714(E)

where,

HP = Input Horse Power = ?

Q = Volume Flow Rate in Gallons Per Minute (GPM) = 18.2 GPM

P = Outlet Pressure in psi = 2372 psi

E = Mechanical Efficiency = 0.93

Therefore,

HP = (18.2 GPM)(2372 psi)/(1714)(0.93)

<u>HP = 27.08 hp</u>

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A woman is 1.6 m tall and has a mass of 50 kg. She moves past an observer with the direction of the motion parallel to her heigh
eduard

To solve this problem it is necessary to apply the concepts related to linear momentum, velocity and relative distance.

By definition we know that the relative velocity of an object with reference to the Light, is defined by

V_0 = \frac{V}{\sqrt{1-\frac{V^2}{c^2}}}

Where,

V = Speed from relative point

c = Speed of light

On the other hand we have that the linear momentum is defined as

P = mv

Replacing the relative velocity equation here we have to

P = \frac{mV}{\sqrt{1-\frac{V^2}{c^2}}}

P^2 = \frac{m^2V^2}{1-\frac{V^2}{c^2}}

P^2 = \frac{P^2V^2}{c^2}+m^2V^2

P^2 = V^2 (\frac{P^2}{c^2}+m^2)

V^2 = \frac{P^2}{\frac{P^2}{c^2}+m^2}

V^2 = \frac{(2.1*10^10)^2}{\frac{(2.1*10^10)^2}{(3.8*10^8)^2}+50^2}

V = 2.81784*10^8m/s

Therefore the height with respect the observer is

l = l_0*\sqrt{1-\frac{V^2}{c^2}}

l = 1.6*\sqrt{1-\frac{(2.81*10^8)^2}{(3*10^8)^2}}

l = 0.56m

Therefore the height which the observerd measure for her is 0.56m

8 0
3 years ago
A cell supplies current of 0.6A and 0.2A through 1ohms and 4.0ohms resistor respectively. Calculate the internal resistance of t
Vlad [161]
<h2>Answer:</h2>

0.5Ω

<h2>Explanation:</h2>

Since different currents are passing through the resistors, then the resistors are most probably connected in parallel. This also means that the same voltage will pass across them.

Using Ohm's law, the voltage across a resistor in a circuit is given by;

V = I(R + r)            -----------(i)

<em>For the 1ohm resistor, the voltage across it is given by;</em>

<em>Where;</em>

I = current passing through the 1 ohm resistor = 0.6A

R = resistance of the 1 ohm resistor = 1Ω

r = internal resistance of the cell = r

Substitute these values into equation (i) as follows;

V = 0.6(1 + r)                 -------------------(ii)

<em>For the 4.0ohm resistor, the voltage across it is given by;</em>

<em>Where;</em>

I = current passing through the 4.0 ohms resistor = 0.2A

R = resistance of the 4.0 ohms resistor = 4.0Ω

r = internal resistance of the cell = r

Substitute these values into equation (i) as follows;

V = 0.2(4.0 + r)                 -------------------(iii)

<em>Now solve equations (ii) and (iii) simultaneously;</em>

V = 0.6(1 + r)

V = 0.2(4.0 + r)

Substitute the value of V in equation (ii) into equation (iii). Therefore, we have;

0.6(1 + r) = 0.2(4.0 + r)

<em>Solve for r</em>

0.6 + 0.6r = 0.8 + 0.2r

0.6r - 0.2r = 0.8 - 0.6

0.4r = 0.2

r = \frac{0.2}{0.4}

r = 0.5

Therefore, the internal resistance of the cell is 0.5Ω

4 0
3 years ago
WILL MARK BRAINLIEST
d1i1m1o1n [39]

Answer:

aww that's simple! its B can i get brainliest? i really need some points rn <3

Explanation:

4 0
3 years ago
Read 2 more answers
Why do cars need a cooling system in conjunction with the engine
snow_lady [41]
So it doesnt get over heated ...
A cooling system<span> works by sending a liquid </span>coolant<span> through passages in the </span>engine<span>block and heads. As the </span>coolant<span> flows through these passages, it picks up heat from the </span>engine<span>. The heated fluid then makes its way through a rubber hose to the radiator in the front of the </span>car<span>.</span>
7 0
3 years ago
A stationary skateboarder I with a mass of 50 kg pushes a stationary skateboarder II with a mass of 75 kg. After the push the sk
Mice21 [21]

The final velocity of the skateboarder I after the collision is 3 m/s to the left.

<h3>Velocity of skateboarder I</h3>

The velocity of skateboarder I is determined from the principle of conservation of linear momentum.

m1u1 + m2u2 = m1v1 + m2v2

50(0) + 75(0) = 50v1 + 75(2)

0 = 50v1 + 150

v1 = (-150)/50

v1 = - 3

v1 = 3 m/s to the left

Thus, the final velocity of the skateboarder I after the collision is 3 m/s to the left.

Learn more about linear momentum here: brainly.com/question/7538238

#SPJ1

8 0
2 years ago
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