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dusya [7]
3 years ago
12

How much current must be applied across a 60 Ω light bulb filament in order for it to consume 55 W of power? Unserious answers w

ill be deleted, and reported. (Please show some work)
Physics
1 answer:
sergij07 [2.7K]3 years ago
4 0

Answer: current I = 0.96 Ampere

Explanation:

Given that the

Resistance R = 60 Ω 

Power = 55 W

Power is the product of current and voltage. That is

P = IV ...... (1)

But voltage V = IR. From ohms law.

Substitutes V in equation (1) power is now

P = I^2R

Substitute the above parameters into the formula to get current I

55 = 60 × I^2

Make I^2 the subject of formula

I^2 = 55/60

I^2 = 0.92

I = sqr(0.92)

I = 0.957 A

Therefore, 0.96 A current must be applied.

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Calculate the amount of heat transferred when 710 grams of water warms from an initial temperature of 4.0 ºC to a final temperat
tigry1 [53]

Answer:

Q = 62383.44 Joules

Explanation:

Given that,

Mass of water, m = 710 gm

Initial temperature of water, T_i=4^{\circ} C

Final temperature of water, T_f=25^{\circ} C

The specific heat capacity of liquid water is, c=4.184\ J/g\ ^oC

Heat transferred is given by :

Q=mc(T_f-T_i)

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3 0
3 years ago
A boat has a mass of 7660 kg. Its engines generate a drive force of 4080 N due west, while the wind exerts a force of 680 N due
makvit [3.9K]

Answer:

0.29 m/s due west.

Explanation:

According to newton's second law,

Net force acting on an object = mass×acceleration

From the question,

F+F₁+F₂ = ma................ Equation 1

Where F = The force generated from the engine, F₁ = Force exerted by the wind, F₂ = Force exerted due to the water, m = mass of the boat, a = acceleration of the boat.

Given: F = 4080 N , F₁ = -680 N(east), F₂ = -1160 N(east). m = 7660 kg

substitute into equation 1

4080-680-1160 = 7660(a)

2240 = 7660a

Therefore,

a = 2440/7660

a = 0.29 m/s due west.

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Explanation:

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3 years ago
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