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Fed [463]
2 years ago
11

Calculate the power transfer when a pump moves 50kg of water through a vertical height of 8meters in 5seconds (take g =10meters

per second)​
Physics
1 answer:
Vika [28.1K]2 years ago
4 0

Hi there!

Recall the following relationships:

W = \Delta U = mg\Delta h\\\\P = \frac{W}{t}

W = Work (J)

U = Potential Energy (J)

m = mass (kg)

g = acceleration due to gravity (9.8 m/s²)

h = height (m)

t = time (s)

Begin by calculating the work:

W = 50(10)(8) = 4000 J

Now, divide by the time to solve for power:

P = \frac{4000}{5} = \boxed{800W}

**W is the unit for power (Watts).  Be careful not to get the two confused.

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When released , what is the kinetic energy of the 1c charge of the preceding problem if it flies past its starting position?
Alekssandra [29.7K]
A) When a charge is moved in an electric field the work done (W) is calculated as charge*(change in potential). We can write W = q*V or V = W/q = 10/1 = 10V . This voltage is a difference in electric potential between 2 points within the field. If the charge is positive, and positive work is done upon it, then the final position is more positive than the original one. 

<span>b) If a charge (Q) is released from rest and falls through a potential difference V, then its gain in energy (KE if no other force acts on the charged body) is q*V = 10J. This is the same as the work done in moving the charge to its new position in part (a), and is an example of the conservation of energy.</span>
5 0
3 years ago
A body travels 10 meters during the first 5 seconds of its travel, and it travels a total of 30 meters over the first 10 seconds
Nesterboy [21]

Answer:

c. 4 meters/second

Explanation:

The formula to calculate average speed is:

s = \frac{x_{2} - x_{1}  }{ t_{2} - t_{1}  }

7 0
3 years ago
A rectangular loop of area A is placed in a region where the magnetic field is perpendicular to the plane of the loop. The magni
mina [271]

Answer:

Induced emf, \epsilon=-A\dfrac{B_{max}e^{-t/\tau}}{\tau}

Explanation:

The varying magnetic field with time t is given by according to equation as :

B=B_{max}e^{-t/\tau}

Where

B_{max}\ and\ t are constant

Let \epsilon is the emf induced in the loop as a function of time. We know that the rate of change of magnetic flux is equal to the induced emf as:

\epsilon=-\dfrac{d\phi}{dt}

\epsilon=-\dfrac{d(BA)}{dt}

\epsilon=-A\dfrac{d(B)}{dt}

\epsilon=-A\dfrac{d(B_{max}e^{-t/\tau})}{dt}

\epsilon=A\dfrac{B_{max}e^{-t/\tau}}{\tau}

So, the induced emf in the loop as a function of time is A\dfrac{B_{max}e^{-t/\tau}}{\tau}. Hence, this is the required solution.

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