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drek231 [11]
3 years ago
14

190 kg of water is to be raised by a water pump to a height of 25 meters from the bottom of a well in 60 seconds. What should be

the minimum power rating of the pump in kilowatts and in horsepower?
Physics
1 answer:
ladessa [460]3 years ago
6 0

Answer:

776.6 w

1.04 hp

Explanation:

given:

Mass, m = 190kg

height change, h = 25m

time elapsed, t = 60 s

acceleration due to gravity, g = 9.81 m/s²

Potential energy required raising 190 kg of water to a height of 25m

= mgh

= 190 x 9.81 x 25

= 46,597.5 J

Power required in 60 s

= Energy required ÷ time elapsed

= 46,597.5 ÷ 60

= 776.6 Watts  (Use conversion 1 W = 0.00134102 hp)

= 776.6 w x 0.00134102 hp/w

= 1.04 hp

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Is the energy of a wave is calculated by squaring the frequency of the wave?
Misha Larkins [42]
Nope.
Energy is directly proportional to frequency. and when you calculate energy, you multiply frequency with a constant number called "Planck's Constant"

E = hf

Hope this helps!
5 0
3 years ago
a wave in a rope is traveling at 6 m/s and at a frequency of 2 Hz. what is the wavelength of the wave producced
Tems11 [23]

Answer:

3

Explanation:

4 0
3 years ago
What happens to the electric potential energy of a negatively charged ion as it moves through the water from the negative probe
Nataly [62]

Answer:

Decreases.

Explanation:

Electric potential energy is the potential energy which is associated with the configuration of points charge in a system and it is the result of conservative coulomb force.

When the negatively charge ion is at the position of the negative probe than its potential energy is positive when it is move towards the positive probe it's potential energy becomes negative due to the negative ion.

Therefore, potential energy is decreases when negative charge ion moves through the water from negative probe to positive probe.

5 0
4 years ago
A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on
SIZIF [17.4K]

(a) 73.5 N

The velocity of the crate is constant: this means that the acceleration is zero (a=0), so according to Newton's second law

\sum F = ma

the resultant of the forces must be zero: \sum F = 0 (1)

The motion is along the horizontal direction, so we are only interested in the forces acting along this direction. There are two of them:

F, the push applied by the worker

F_f=-\mu mg, the force of friction, with \mu=0.25 being the coefficient of friction, m=30.0 kg being the mass of the crate, and g=9.8 m/s^2. The negative sign is due to the fact that the friction acts in the opposite direction to the motion. Eq.(1) then becomes

F-\mu mg=0\\F=\mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N

So, this is the force that the worker must apply.

(b) 330.8 J

The work done by the pushing force of the worker on the crate is given by:

W=Fd cos \theta

where

F = 73.5 N is the force

d = 4.5 m is the displacement

\theta=0^{\circ} is the angle between the direction of the force and the displacement (0 degrees, since they are in same direction)

Substituting, we have

W=(73.5 N)(4.5 m)(cos 0^{\circ})=330.8 J

(c) -330.8 J

To calculate the work done by friction, we apply the same formula:

W=F_f d cos \theta

where

F_f = \mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N is the magnitude of the force of friction

d = 4.5 m is the displacement

\theta=180^{\circ} is the angle between the direction of the force of friction and the displacement (it is 180 degrees since the two are into opposite directions)

Substituting, we find

W=(73.5 N)(4.5 m)(cos 180^{\circ})=-330.8 J

So, the work done by friction is negative.

(d) 0 J

As before, the work done by any force on the crate is

W=F_f d cos \theta

We notice that both gravity and normal force are perpendicular to the displacement: therefore, \theta=90^{circ}, and so

cos \theta=0

which means that the work done by both forces is zero.

(e) 0 J

The total work done on the crate is the sum of the work done by the four forces acting on it, so:

W=W_{push} + W_{friction}+W_{gravity}+W_{normal}=330.8J-330.8J+0+0=0

And this is in accordance with the work-energy theorem, which states that the variation of kinetic energy of the crate is equal to the work done on it: since the crate is moving at constant velocity, its variation of kinetic energy is zero, as well as the work done on it.

5 0
3 years ago
5. 3 women push a stalled car. Each woman pushes with a 400N force. What is the mass of the car if the car accelerates at 0.85 m
iris [78.8K]

Answer:

<h2>470.59 kg</h2>

Explanation:

The the mass of the car can be found by using the formula

m =  \frac{f}{a}  \\

f is the force

a is the acceleration

From the question we have

m =  \frac{400}{0.85}  \\  = 470.58823...

We have the final answer as

<h3>470.59 kg</h3>

Hope this helps you

7 0
3 years ago
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