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swat32
4 years ago
15

A (11.4 A) ohm device is connected to a 115 V outlet. If the cost of the electrical energy is $0.16/kWh, determine the cost of r

unning this device for (36.0 B) hours. Round your answer in dollars ($) with 2 decimal places.
Physics
1 answer:
Scrat [10]4 years ago
4 0

Answer:

$6.68

Explanation:

The current running through the device would be its outlet voltage divided by the resistance:

I = U/R = 115 / 11.4 = 10.09 A

The power generated by this device is the product of its current and voltage

P = IU = 10.09 * 115 = 1160 W or 1.16 kW

If running this device for 36 hours then the total energy it would consume is

E = Pt = 1.16 * 36 = 41.76 kWh

Therefore the total cost of electrical energy is

C = Ec = 41.76*0.16 = $6.68

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A force of 14 lb is required to hold a spring stretched 4 in. beyond its natural length. How much work W is done in stretching i
enyata [817]

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W = 19.8 J

Explanation:

14 lb force is required to stretch the spring by 4 inch distance

So we have

F = 14 lbf

F = 6.35 \times 9.8 N

F = 62.3 N

stretch in the spring is given as

x = 4 in = 0.1016 m

now we will have

F = kx

62.3 = k(0.1016)

k = 613.125 N/m

Now we need to find the work to stretch it by x = 10 in = 0.254 m

so we have

W = \frac{1}{2}kx^2

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6 0
3 years ago
Three charges, q1 = +2.06 x 10-9 C, q2 = -3.27 x 10-9 C, and q3 = +1.05 x 10-9 C, are located on the x-axis at x1 = 0, x2 = 10.0
lisov135 [29]

Answer:

The resultant force on charge 3 is Fr= -2,11665 * 10^(-7)

Explanation:

Step 1: First place the three charges along a horizontal axis. The first positive charge will be at point x=0, the second negative charge at point x=10 and the third positive charge at point x=20. Everything is indicated in the attached graph.

Step 2: I must calculate the magnitude of the forces acting on the third charge.

F13: Force exerted by charge 1 on charge 3.

F23: Force exerted by charge 2 on charge 3.

K: Constant of Coulomb's law.

d13: distance from charge 1 to charge 3.

d23: distance from charge 2 to charge 3

Fr: Resulting force.

q1=+2.06 x 10-9 C

q2= -3.27 x 10-9 C

q3= +1.05 x 10-9 C

K=9-10^9 N-m^2/C^2

d13= 0,20 m

d23= 0,10 m

F13= K * (q1 * q3)/(d13)^2

F13=9,7335*10^(-8) N

F23=K * (q2 * q3)/(d23)^2

F23= -3,09 * 10^(-7)

Step 3: We calculate the resultant force on charge 3.

Fr=F13+F23= -2,11665 * 10^(-7)

3 0
3 years ago
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