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3 years ago

What is the power of 4ohm resistor connected to a 12v battery?

Physics

2 answers:

NemiM [27]3 years ago

8 0

Answer:

36watts

Explanation:

Data obtained from the question include:

R (Resistor) = 4 Ohms

V (Voltage) = 12V

P (power) =?

P = IV

Recall: I = V/R

P = V^2 /R

P = (12)^2 /4

P = 144/4

P = 36watts

Therefore, power is 36watts

DochEvi [55]3 years ago

8 0

Answer:

= 36 watts

Explanation:

Power = I V

From ohms law

V = I R

divide both sides by R

V/R = I

I = V/R

therefore if, P = IV, put I = V/R into it

P= V/R × V

P = V² / R

Given, V = 12v

R = 4 ohms

Therefore

P= 12² / 4

P = 144/4

P = 36 watts

Power = 36 watts

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The 60.0 kg skier shown below is skiing down a 35.0 degree incline where the magnitude of the friction force is 38.5N

Sunny_sXe [5.5K]

Answer:

a) 4.98m/s²

b) 481.66N

Explanation:

a) Using the Newtons second law of motion

$\sum F_x = ma_x\\F_m - F_f = ma_x\\Wsin \theta - F_f = ma_x\\mgsin \theta - F_f = ma_x\\$

m is the mass of the object

g is the acceleration due to gravity

Fm is the moving force acting along the plane

Ff is the frictional force opposing the moving froce

a is the acceleration of the skier

Given

m = 60kg

g = 9.8m/s²

$\theta$ = 35°

Ff = 38.5N

Required

acceleration of the skier a

Substituting into the formula;

$60(9.8)sin 35^0 - 38.5 = 60a\\588sin35^0 - 38.5 = 60a\\337.26 - 38.5 = 60a\\298.76 = 60a\\a = 298.76/60\\a = 4.98m/s^2\\$

Hence the acceleration of the skier is 4.98m/s²

b) The normal force on the skier is expressed as;

N = Wcosθ

N = mgcosθ

N = 60(9.8)cos 35°

N = 588cos 35°

N = 481.66N

Hence the normal force on the skier is 481.66N

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3 years ago

Light enters water from air at an angle of 25° with the normal, Θ1. If water has an index of refraction of 1.33, determine Θ2.

vladimir1956 [14]

By Snell's law:

η = sini / sinr. i = 25, η = 1.33

1.33 = sin25° / sinr

sinr = sin25° / 1.33 = 0.4226/1.33 = 0.3177 Use a calculator.

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Option A.

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3 years ago

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3 years ago

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The half-wave rectifier circuit of vs(t) = 170 sin(377t) V and a load resistance R = 15Ω. Determine: a. The average load current

irina1246 [14]

Answer:

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b rms load current = 8.02A

c true power =962.64 W

d apparent power =962.64 W

e. power factor cosθ =1

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Resistors are electronic components that consume energy

the power in a resistor is given by P =IVcosθ ; in a resistor cosθ =1

P =IV

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and is obtained by multiplying the rms voltage with the rms current.

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120.03×8.02

P= 962.64 W ; true power

apparent power = Vrms × Irms

=120.03×8.02

= 962.64W ; apparent power

power factor cosθ = true power/ apparent power

cosθ = 962.64/962.64

cosθ = 1

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