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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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You hold a bucket in one hand. In the bucket is a 500 g rock. You swing the bucket so the rock moves in a vertical circle 2.2 m
slavikrds [6]

Answer:v=3.28 m/s

Explanation:

Given

mass of rock m=500 gm

diameter of circle d=2.2 m

radius r=\frac{2.2}{2}=1.1 m

At highest Point

mg+N=\frac{mv^2}{r}

At highest Point N=0 because mass is just balanced by centripetal Force

thus mg=\frac{mv^2}{r}

v=\sqrt{gr}

v=\sqrt{9.8\times 1.1}

v=\sqrt{10.78}

v=3.28 m/s

6 0
3 years ago
An electron and a proton are held on an x axis, with the electron at x = + 1.000 m
mixas84 [53]

Answer:

  r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

Explanation:

For this exercise we must use conservation of energy

the electric potential energy is

          U = k \frac{q_1q_2}{r_{12}}

for the proton at x = -1 m

          U₁ =- k \frac{e^2 }{r+1}

for the electron at x = 1 m

          U₂ = k \frac{e^2 }{r-1}

starting point.

        Em₀ = K + U₁ + U₂

        Em₀ = \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1}

final point

         Em_f = k e^2 ( -\frac{1}{r_2 +1} + \frac{1}{r_2 -1})

   

energy is conserved

        Em₀ = Em_f

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e^2 (- \frac{1}{r_2 +1} + \frac{1}{r_2 -1})              

       

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e²(  \frac{2}{(r_2+1)(r_2-1)} )

we substitute the values

½ 9.1 10⁻³¹ 450 + 9 10⁹ (1.6 10⁻¹⁹)² [ - \frac{1}{20+1} + \frac{1}{20-1} ) = 9 109 (1.6 10-19) ²( \frac{2}{r_2^2 -1} )

          2.0475 10⁻²⁸ + 2.304 10⁻³⁷ (5.0125 10⁻³) = 4.608 10⁻³⁷ ( \frac{1}{r_2^2 -1} )

          2.0475 10⁻²⁸ + 1.1549 10⁻³⁹ = 4.608 10⁻³⁷     \frac{1}{r_2^2 -1}

          \frac{2.0475 \ 10^{-28} }{1.1549 \ 10^{-37} } = \frac{1}{r_2^2 -1}

          r₂² -1 = (4.443 10⁸)⁻¹

           

          r2 = \sqrt{1 + 2.25 10^{-9}}

          r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

4 0
3 years ago
4. The blades on a fan have a frequency of 15 Hz.
vichka [17]

Answer:

a) 4500 cycles b) 0.0667s c) 6.67s

Explanation:

a) 15 Hz= 15 cycles/ s

   5 mins= 300s

   15 cycles/s * 300s= 4500 cycles

b) Period= 1/ frequency

   Period= 1/ 15 cycles/s

   Period= 0.0667s

c) Period * number of revolutions= time

  0.0667 * 100= 6.67s

6 0
3 years ago
A mass of 0.5 kg hangs motionless from a vertical spring whose length is 1.10 m and whose unstretched length is 0.50 m. Next the
ser-zykov [4K]

Answer:

The maximum length during the motion is L_{max} = 1.45m

Explanation:

From the question we are told that

           The mass  is  m =0.5 kg

            The vertical spring  length is  L = 1.10m

            The unstretched  length is  L_{un} = 1.30m

          The initial speed is v_i = 1.3m/s

          The new length of the spring L_{new} =  1.30 m

The spring constant k is mathematically represented as

                           k = -\frac{F}{y}

Where F is the force applied  = m * g = 0.5 * 9.8=4.9N

           y is the difference in weight which is   =1.10-0.50=0.6m

The negative sign is because the displacement of the spring (i.e its extension occurs against the force F)

    Now  substituting values accordingly

                    k =  \frac{4.9}{0.6}

                       = 8.17 N/m

The  elastic potential energy is given as E_{PE} = \frac{1}{2} k D^2

  where D is this the is the displacement  

Since Energy is conserved the total elastic potential energy would be

             E_T = initial  \ elastic\ potential \ energy + kinetic \ energy

            E_T = \frac{1}{2} k D_{max}^2 =   \frac{1}{2} k D^2 + \frac{1}{2} mv^2

Substituting value accordingly

                \frac{1}{2} *8.17 *D_{max}^2 =\frac{1}{2} * 8.17*(1.30 - 0.50)^2 + \frac{1}{2} * 0.5 *1.30^2

                4.085 * D_{max}^2 = 3.69

                 D^2_{max} = 0.9033

                D_{max} = 0.950m

So to obtain total length we would add the unstretched length

 So we have

                  L_{max} = 0.950 + 0.5 = 1.45m

                               

               

               

                 

                     

5 0
2 years ago
Read 2 more answers
Please help I need this fast
STALIN [3.7K]

Answer:

0 m/sec

Explanation:

b/c they were at rest and initial means at rest ,at rest means 0 HOPE THIS HELPS

8 0
2 years ago
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