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Vladimir [108]
3 years ago
15

A student's room has a TV (250 W) ,heater (1150 W) and lamp (200 W) Electricity costs 10 fills per KWh. Calculate how much it wo

uld cost to run these three appliances for a total of 2 hour 30 minutes
Physics
1 answer:
Yuki888 [10]3 years ago
6 0

Answer:

Cost = 40 fills

Explanation:

First, we will calculate the total power consumed by all appliances:

Power = P_{TV}+P_{heater}+P_{lamp}\\Power = 250\ W+1150\ W+200\ W\\Power = 1600\ W = 1.6\ KW

Now, we will calculate the energy required:

Energy = (Power)(Time)\\Energy = (1.6\ KW)(2.5 h)\\Energy = 4 KWh\\

Now, for the cost we use the following formula:

Cost = (Energy)(Unit\ Cost)\\Cost = (4\ KWh)(10\ fills/KWh)\\

<u>Cost = 40 fills</u>

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A positive charge +q1 is located to the left of a negative charge -q2. On a line passing through the two charges, there are two
AfilCa [17]

Answer:

please the answer below

Explanation:

(a) If we assume that our origin of coordinates is at the position of charge q1, we have that the potential in both points is

V_1=k\frac{q_1}{r-1.0}-k\frac{q_2}{1.0}=0\\\\V_2=k\frac{q_1}{r+5.2}-k\frac{q_2}{5.2}=0\\\\

k=8.89*10^9

For both cases we have

k\frac{q_1}{r-1.0}=k\frac{q_2}{1.0}\\\\q_1(1.0)=q_2(r-1.0)\\\\r=\frac{q_1+q_2}{q_2}\\\\k\frac{q_1}{r+5.2}=k\frac{q_2}{5.2}\\\\q_1(5.2)=q_2(r+5.2)\\\\r=\frac{5.2q_1-5.2q_2}{q_2}

(b) by replacing this values of r in the expression for V we obtain

k\frac{q_1}{\frac{5.2(q_1-q_2)}{q_2}+5.2}=k\frac{q_2}{5.2}\\\\\frac{q_1}{q_2}=\frac{(q_1-q_2)}{q_2}-1.0=\frac{q_1-q_2-q_2}{q_2}=\frac{q_1-2q_2}{q_2}

hope this helps!!

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One major difference between plant and animal cells is that plant cells are the only ones with A) cellulose. B) lysosomes. C) nu
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The correct answer is A .
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A rock falls 15 m. is this vertical or horizontal motion? what is the displacement?
Irina-Kira [14]
Displacement will be 15 too because
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A child (approximately 4 years old) takes her metal “slinky Toy” (a flexible coiled metal spring) and does various tests to dete
anzhelika [568]

Answer:

248

Explanation:

L = Inductance of the slinky = 130 μH = 130 x 10⁻⁶ H

l = length of the slinky = 3 m

N = number of turns in the slinky

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Area of slinky is given as

A = πr²

A = (3.14) (0.04)²

A = 0.005024 m²

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L = \frac{\mu _{o}N^{2}A}{l}

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