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storchak [24]
3 years ago
7

Choose the correct statement concerning units of power or energy. a. A kilojoule (kJ) is a unit of power. b. A gigawatt (GW) is

a unit of energy. c. A watt (W) is a unit of energy. d. A kilowatt x hour per year (kWh/yr) is a unit of energy. e. A kilowatt (kW) is a unit of power.
Physics
1 answer:
sergiy2304 [10]3 years ago
3 0

Answer:

A kilowatt (kW) is a unit of power.

Explanation:

The power of an object is given by :

P=\dfrac{E}{t}

Here,

E is the energy required

t is time

The SI unit of power is Watts and the SI unit of energy is Joule. the commercial unit of energy is kilowatt per hour.    

Option (1) :  A kilojoule (kJ) is a unit of power is incorrect.

Option (2) : A gigawatt (GW) is a unit of energy is incorrect.

Option (3) : A watt (W) is a unit of energy is incorrect.

Option (4) : A kilowatt x hour per year (kWh/yr) is a unit of energy is incorrect.

Option (4) : A kilowatt (kW) is a unit of power is correct.

Hence, the correct option is (d).

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A 220 g mass is on a frictionless horizontal surface at the end of a spring that has force constant of 7.0
Talja [164]

The concept of conservation of energy and harmonic motion allows to find the result for the power where the kinetic and potential energy are equal is:

        x = 0.135 cm

Given parameters

  • The mass m = 220 g = 0.220 kg
  • The spring cosntnate3 k = 7.0 N / m
  • Initial displacement A = 5.2 cm = 5.2 10-2 m

To find

  • The position where the kinetic and potential energy are equal

 

A simple harmonic movement is a movement where the restoring force is proportional to the displacement, the result of this movement is described by the expression.

          x = A cos wt + fi

          w² = \frac{k}{m}

Where x is the displacement from the equilibrium position, A the initial amplitude of the system, w the angular velocity t the time, fi a phase constant determined by the initial conditions, k the spring constant and m the mass.

The speed is defined by the variation of the position with respect to time.

       v = \frac{dx}{dt}

let's evaluate

       v = - A w sin (wt + Ф)

Since the body releases for a time t = 0 the velocity is zero, therefore the expression remains.

       0 = - A w sin Ф

For the equality to be correct, the sine function must be zero, this implies that the phase constant is zero

        x = A cos wt

Let's find the point where the kinetic and potential energy are equal.

        K = U

        ½ m v² = m g x

       

we substitute

        ½ A² w² sin² wt = g A cos wt

        sin² wt = \frac{2g}{A}  cos wt

let's calculate

      w = \sqrt{\frac{7}{0.220} }  

      w = 5.64 rad / s

      sin² 5.64t = 2 9.8 / 0.052 cos 5.64t

      sin² 5.64t = 376.92 cos 5.64 t

      1 - cos² 5.64t = 376.92 cos 5.64t

      cos² 5.64t -376.92 cos564t -1 = 0

we make the change of variable

       x = cos 5.64t

      x²- 376.92 x - 1 = 0

      x = 0.026

      cos 5.64t = 0.026

   

Let's find the displacement for this time

       x = 5.2 10-2 0.026

       x = 1.35 10-3 m

In conclusion Using the concepts of conservation of energy and harmonic motion we can find the result for the could where the kientic and potential enegies are equal is:

        x = 0.135 cm

Learn more here: brainly.com/question/15707891

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A train is speeding down a railroad track at the speed of 50 miles per hour. From whose reference point is the train not moving?
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A: a person sitting on a train

Hence person could have a meal and not get food all over them.

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How does electricity turn into light in a common lightbulb?
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Answer:

C or D

Explanation:

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If a power utility were able to replace an existing 500 kV transmission line with one operating at 1 MV, it would change the amo
mestny [16]

Answer:

It would change the amount of heat produced in the transmission line to four times the previous value.

Explanation:

Given;

initial voltage in the transmission line, V₁ = 500 kV = 500,000 V

Final voltage in the transmission line, V₂ = 1 MV = 1,000,000

The power lost in the transmission line due to heat is given by;

P = \frac{V^2}{R}

Power lost in the first wire;

P_1 = \frac{V_1^2}{R}

R = \frac{V_1^2}{P_1}

Power lost in the second wire

P_2 = \frac{V_2^2}{R}\\\\ R = \frac{V_2^2}{P_2}

Keeping the resistance constant, we will have the following equation;

\frac{V_2^2}{P_2} = \frac{V_1^2}{P_1} \\\\P_2 = \frac{V_2^2P_1}{V_1^2}\\\\

P_2 = \frac{(1,000,000)^2P_1}{(500,000)^2}\\\\P_2 =4P_1

Therefore, it would change the amount of heat produced in the transmission line to four times the previous value.

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