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bixtya [17]
3 years ago
13

An electric furnace runs 13 hours a day to heat a house during January (31 days). The heating element has a resistance of 7.2 an

d carries a current of 16.7 A. The cost of electricity is $0.10/kWh. Find the cost of running the furnace during January.
Physics
2 answers:
liq [111]3 years ago
4 0

Answer:

cost of running the furnace during January is $5619.62

Explanation:

given data

runs a day = 13 hours

January days = 31 days

resistance = 7.2 ohm

current = 16.7 A

cost of electricity = $0.10/kWh

to find out

cost of running the furnace during January

solution

first we get her power consumed by furnace that is

Power consumed = \frac{I^2}{R}  ........1

put here value we get

Power consumed = \frac{16.7^2}{7.2}

Power consumed = 38.7347 W

and

Power consumed by furnace in one hour is

Power consumed by furnace in one hour is = Power consumed × 3600

Power consumed by furnace in one hour is = 38.7347 × 3600  

Power consumed by furnace in one hour is 139.445kWh

and

Power consumed by furnace in the month of January is

Power consumed by furnace in the month of January = 139.445kWh × 13 hours × 31 days

Power consumed by furnace in the month of January = 56196.335 kWh

so

cost of running the furnace during January is = $0.10/kWh × 56196.335 kWh

cost of running the furnace during January is $5619.62

Vesna [10]3 years ago
4 0

Answer:

$ 80.9

Explanation:

Resistance, R = 7.2 ohm

current, i = 16.7 A

Cost = $ 0.10 / kWh

time = 13 hours per day , 1 month

Energy = i²Rt

E = 16.7 x 16.7 x 7.2 x 13 x 31 Wh

E = 809.28 kWh

Cost of 1 kWh = $ 0.10

Cost of 809.28 = $ 0.10 x 809.28 = $ 80.9

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A family is skating at an ice rink. The 58.2 kg mother is holding the
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Answer:

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Explanation:

use this formula :

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then you fill it in :

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a_{y} = \frac{100N}{93.7kg}

a_{y} = 1.0672 m/s^{2}

then you multiply that with the daughters weight :

T_{2} x= m_{2} a_{y}

T_{2} x = 35.5kg (1.0672 m/s^{2})

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\\ \rm\Rrightarrow \mu =0.75

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Karla Ayala pulls a sled on an icy road (dangerous!). Because of Karla's pull, the tension force is 151 N, and the rope makes a
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Answer:

W = 1418.9 J = 1.418 KJ

Explanation:

In order to find the work done by the pull force applied by Karla, we need to can use the formula of work done. This formula tells us that work done on a body is the product of the distance covered by the object with the component of force applied in the direction of that displacement:

W = F.d

W = Fd Cosθ

where,

W = Work Done = ?

F = Force = 151 N

d = distance covered = 10 m

θ = Angle with horizontal = 20°

Therefore,

W = (151 N)(10 m) Cos 20°

<u>W = 1418.9 J = 1.418 KJ</u>

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Answer:

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\frac{21600}{1000}= 21.6 km

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