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GalinKa [24]
3 years ago
13

How much would it cost to operate five incandescent light bulbs for the months of April and May, collectively, given that each b

ulb is one hundred and fifty Watts, electricity for home use is 0.3AED per kilowatt-hour, and that the light s will be on for three hours per day during this two month period.
please i need an answer fast or my teacher will kill me
Physics
1 answer:
const2013 [10]3 years ago
3 0

Answer:

Total Cost is 41.18 AED

Explanation:

Per day number of hours light bulbs were on=3hours

5 bulbs power consumption = 150Wx5 = 750W

Electricity usage per day = 750 x 3 = 2250Wh = 2.25kWh

April has 30 days and May has 31 days. Therefore totals days = 61days

Total power consumption = 2.25x 61 = 137.25kWh

Total Cost = 137.25x 0.30 = 41.175 AED

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How much force is required to cause an object with a mass of 850 kg to accelerate at a rate of 2 meters per second squared (m/s^
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How much force is required to cause an object with a mass of 850 kg to accelerate at a rate of 2 meters per second squared (m/s^2)?

Explanation:

<em>1700N </em>

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A huge tank of glycerine with a density of 1.260 g/cm3 is vertically stationed on a platform which is 15 m above the ground. The
EleoNora [17]

Answer:

The tank is losing 4.976*10^{-4}  m^3/s

v_g = 19.81 \ m/s

Explanation:

According to the Bernoulli’s equation:

P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 +  \frac{1}{2}  \rho v_2^2 + \rho gh_2

We are being informed that both the tank and the hole is being exposed to air :

∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume  v_1 ≅ 0 ;

then v_2 can be determined as:\sqrt{[2g (h_1- h_2)]

h₁ = 5 + 15 = 20 m;

h₂ = 15 m

v_2 = \sqrt{[2*9.81*(20 - 15)]

v_2 = \sqrt{[2*9.81*(5)]

v_2= 9.9 \ m/s  as it leaves the hole at the base.

radius r = d/2  = 4/2 = 2.0 mm

(a) From the law of continuity; its equation can be expressed as:

J = A_1v_2

J = πr²v_2    

J =\pi *(2*10^{-3})^{2}*9.9

J =1.244*10^{-4}  m^3/s

b)

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : v_g = \sqrt{392.31}

v_g = 19.81 \ m/s

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