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loris [4]
3 years ago
10

6. How much power does it take to lift 30.0 N 10.0 m high in 5.00 s?

Physics
1 answer:
Zina [86]3 years ago
8 0

Answer:

60 W

Explanation:

The work done in lifting the object is given by:

W=Fd

where

F = 30.0 N is the weight of the object

d = 10.0 m is the displacement covered by the object

Using the formula,

W=(30.0 N)(10.0 m)=300 J

Now we can calculate the power, which is given by

P=\frac{W}{t}

where

W = 300 J is the work done

t = 5.00 s is the time interval

Substituting,

P=\frac{300 J}{5 s}=60 W

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An electric heater draws a steady 15.0A on a 120-V
morpeh [17]

Answer:

1. 1800 W

2. $ 17.3

Explanation:

From the question given above, the following data were obtained:

Current (I) = 15 A

Voltage (V) = 120 V

Time (t) = 20 h per day

Duration = 31 days

Cost = 15.5 cents per kWh

1. Determination of the power.

Current (I) = 15 A

Voltage (V) = 120 V

Power (P) =?

P = IV

P = 15 × 120

P = 1800 W

Thus, 1800 W of power is required.

2. Determination of the cost per month (31 days).

We'll begin by converting 1800 W to KW.

1000 W = 1 KW

Therefore,

1800 W = 1800 W × 1 KW / 1000 W

1800 W = 1.8 KW

Next, we shall determine the energy consumption for 31 days. This can be obtained as follow:

Power (P) = 1.8 KW

Time (t) = 2 h per day

Time (t) for 31 days = 2 × 31 = 62 h

Energy (E) =?

E = Pt

E = 1.8 × 62

E = 111.6 KWh

Finally, we shall determine the cost of consumption. This can be obtained as follow:

1 KWh = 15.5 cents

Therefore,

111.6 KWh = 111.6 KWh × 15.5 cents / 1 KWh

111.6 KWh = 1729.8 cents

Converting 1729.8 cents to dollar, we have:

100 cents = $ 1

Therefore,

1729.8 cents = 1729.8 cents × $ 1 / 100 cents

1729.8 cents = $ 17.3

Thus, it will cost $ 17.3 per month to run the electric heater.

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A rock thrown with a horizontal velocity of 20m/s from a cliff that is 125m above level ground. If air resistance is negligible,
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Answer: c

Explanation: 125/20 =6.25

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