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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 thin spherical shell has a radius of 0.70 m. An applied torque of 860 N m gives the shell an angular acceleration of 4.70 rad/
Artyom0805 [142]

Answer:

I=182.97\ kg-m^2

Explanation:

Given that,

Radius of a spherical shell, r = 0.7 m

Torque acting on the shell, \tau=860\ N

Angular acceleration of the shell, \alpha =4.7\ m/s^2

We need to find the rotational inertia of the shell about the axis of rotation. The relation between the torque and the angular acceleration is given by :

\tau=I\alpha

I is the rotational inertia of the shell

I=\dfrac{\tau}{\alpha }\\\\I=\dfrac{860}{4.7}\\\\I=182.97\ kg-m^2

So, the rotational inertia of the shell is 182.97\ kg-m^2.

7 0
3 years ago
What is the definition of cyclone???
notsponge [240]
A Cyclone is a hurricane.

6 0
3 years ago
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What is the kinetic energy in Joules of a 1,500kg car traveling at 75mph?What is the kinetic energy in Joules of a 1,500kg car t
Sergio039 [100]

Answer: 2812500 joules

Explanation:

Mass of car = 1500kg

Velocity of car = 75mph

Kinetic energy = ?

Recall that kinetic energy is the energy possessed by a moving object, and it depends on its mass M and velocity, V

Thus, Kinetic energy = 1/2 x mv^2

= 1/2 x 1000kg x (75mph)^2

= 0.5 x 1000kg x (75mph)^2

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Thus, the car travels with a kinetic energy of 2812500 joules

5 0
3 years ago
10. How far does a transverse pulse travel in 1.23 ms on a string with a density of 5.47 × 10−3 kg/m under tension of 47.8 ?????
KATRIN_1 [288]

Answer: Tension = 47.8N, Δx = 11.5×10^{-6} m.

              Tension = 95.6N, Δx = 15.4×10^{-5} m

Explanation: A speed of wave on a string under a tension force can be calculated as:

|v| = \sqrt{\frac{F_{T}}{\mu} }

F_{T} is tension force (N)

μ is linear density (kg/m)

Determining velocity:

|v| = \sqrt{\frac{47.8}{5.47.10^{-3}} }

|v| = \sqrt{0.00874 }

|v| = 0.0935 m/s

The displacement a pulse traveled in 1.23ms:

\Delta x = |v|.t

\Delta x = 9.35.10^{-2}*1.23.10^{-3}

Δx = 11.5×10^{-6}

With tension of 47.8N, a pulse will travel Δx = 11.5×10^{-6}  m.

Doubling Tension:

|v| = \sqrt{\frac{2*47.8}{5.47.10^{-3}} }

|v| = \sqrt{2.0.00874 }

|v| = \sqrt{0.01568}

|v| = 0.1252 m/s

Displacement for same time:

\Delta x = |v|.t

\Delta x = 12.52.10^{-2}*1.23.10^{-3}

\Delta x = 15.4×10^{-5}

With doubled tension, it travels \Delta x = 15.4×10^{-5} m

4 0
3 years ago
Can someone help me with this science question
Anettt [7]
Mechanical layers of the earth:

-Lithosphere
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-Outer Core
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Chemical layers of the earth:

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Hope this helps

Have a great weekend! :)
6 0
3 years ago
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