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1 year ago

If you have a mass of 30 kilograms and you are resting on top of a hill that is 20 meters high how much energy would you have.

Physics

1 answer:

myrzilka [38]1 year ago

4 0

Answer:

we can have 2 values one is with g taken as 10 m/s² and the second with g taken as 9.8m/s² where g is the acceleration due to gravity

when g is 10,

30kg×10m/s²×20m. using m×g×h

6000 J

when g is 9.8

30×9.8×20

4080 J

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2 years ago

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2 years ago

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3 years ago

Read 2 more answers

Circuit A in a house has a voltage of 208 V and is limited by a 40.0-A circuit breaker. Circuit B is at 120.0 V and has a 20.0-A

Setler79 [48]

Given Information:

Voltage of circuit A = Va = 208 Volts

Current of circuit A = Ia = 40 Amps

Voltage of circuit B = Vb = 120 Volts

Current of circuit B = Ib = 20 Amps

Required Information:

Ratio of power = Pa/Pb = ?

Answer:

Ratio of power = Pa/Pb = 52/15

Explanation:

Power can be calculated using Ohm's law

P = VI

Where V is the voltage and I is the current flowing in the circuit.

The power delivered by circuit A is

Pa = Va*Ia

Pa = 208*40

Pa = 8320 Watts

The power delivered by circuit B is

Pb = Vb*Ib

Pb = 120*20

Pb = 2400 Watts

Therefore, the ratio of the maximum power delivered by circuit A to that delivered by circuit B is

Pa/Pb = 8320/2400

Pa/Pb = 52/15

4 0

3 years ago

Along the line connecting the two charges, at what distance from the charge q1 is the total electric field from the two charges

Nostrana [21]

Answer:

$r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})$

Explanation:

Here two charges are placed at distance "d" apart

now the net value of electric field at some position between two charges will be ZERO

so we will have

electric field due to charge 1 = electric field due to charge 2

$E_1 = E_2$

Let the position where net field is zero will lie at distance "r" from q1

$\frac{kq_1}{r^2} = \frac{kq_2}{(d-r)^2}$

now we will have

$\frac{(d - r)^2}{r^2} = \frac{q_2}{q_1}$

now square root both sides

$\frac{d}{r} - 1 = \sqrt{\frac{q_2}{q_1}}$

now we have

$\frac{d}{r} = \sqrt{\frac{q_2}{q_1}} + 1$

so we have

$r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})$

8 0

3 years ago