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klio [65]
3 years ago
9

.A 100.0-W lightbulb is 22 percent efficient. This means that 22 percent of the electrical energy is transformed to radiant ener

gy.
a. How many joules does the lightbulb transform into radiant energy each minute it is in operation?

b. How many joules of thermal energy does the lightbulb output each minute?
Physics
1 answer:
algol [13]3 years ago
6 0

Answer:

Explanation:

100 W bulb is using energy of 100 J in one second.

22 percent of the electrical energy is transformed to radiant energy.

a )

So , electrical energy is transformed to radiant energy per second

= 100 x .22 = 22 J

energy transformed in one minute = 22 x 60  J

= 1320 J

b )

electrical energy is transformed to heat  energy per second

= 100 x .78 = 78 J

energy transformed in one minute = 78 x 60  J

= 4680  J

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A 50-cm-long spring is suspended from the ceiling. A 230g mass is connected to the end and held at rest with the spring unstretc
alukav5142 [94]

Answer:

k = 25.07 N/m

Amplitude = 9 cm

f = 1.66 Hz

Explanation:

Given:

- The original length of the spring L_o = 50 cm

- The mass hanged m = 230 g

- The amount of stretch given 2x = 18 cm @lowest point.

Find:

a. What is the spring constant? (K=)

b. What is the amplitude of the oscillation?

c. What is the frequency of the oscillation?

Solution:

- Make a FBD of the hanging mass, There are two external forces acting on it that is the force of gravity due to its weight and the springs restoring force when its stretched to its lowest point. After hanging the mass on the spring a new equilibrium position is achieved which also causes the spring to stretch. We can apply the Equilibrium conditions at this point in vertical direction as:

                                      k*x - m*g = 0

                                      k = m*g / x

Where, x is the extension of the spring or mean stretch. which 0.5*amplitude (Lowest point). x = 9 cm

                                      k = 0.23*9.81 / 0.09

                                      k = 25.07 N/m

Answer: For part a we have the stiffness of the spring k = 25.07 N/m

- The amplitude of the oscillating motion is the half the amount of total stretch or the amount the spring extends above or below the mean position.

                                       Amplitude = x = 9 cm

- The frequency of any oscillatory motion which can be modeled by SHM can be expressed as:

                                       f = 1 / 2*p*  sqrt ( k / m )

- Plug the values in:                                        

                                       f = 1 / 2*pi* sqrt (25.07 / 0.23 )

                                       f = 1.66 Hz

Answer: For part c the frequency of oscillation is f = 1.66 Hz

           

8 0
3 years ago
A typical nuclear power plant converts 0.01% of the mass of its fuel into energy through nuclear reactions. how much fuel would
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The fuel will be for all the city is 2,000 kilograms
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Earth has a magnetic dipole moment of 8.0 × 1022 J/T. (a) What current would have to be produced in a single turn of wire extend
Bogdan [553]

Answer:

a

The current that would be produced is I = 6.26 *10 ^8 A

b

Yes this arrangement can be used to cancel out earths magnetic field at points well above the Earth's surface.This is because this current loop acts as a magnetic dipole for point above the earth surface  

c

No this arrangement can not be used to cancel out earths magnetic field at points on the Earth's surface .this because on the earth surface it shifts from its behavior as a magnetic dipole

Explanation:

    From the question we are told that

             The magnetic moment of earth is M = 8.0*10 ^{22} J/T

               The radius of earth generally has a value of R = 6378 *10^3 m

Magnetic moment is mathematically given as

                    M = IA

A is the area of the of the earth(assumption that the earth is circular ) and this evaluated as  

                     A = \pi R^2

Now making I the subject in the above formula

                  I = \frac{M}{A}

                     = \frac{M}{\pi R^2}

                     = \frac{8.0^10^{22}}{\pi (6378 *10^{3})^2}

                     = 6.26 *10^8 A

                   

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While doing an experiment you measure the length of an object to be 32.5cm. The measuring device that you are using measures to
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Answer and Explanation:

By measuring in millimeter we can decrease the associated error with the measurement because when we measure in smaller unit the measurement is more precise  and accurate rather than when we measure in larger unit so when we measure 325 millimeter instead of 32.5 cm then there is chance of less error produced.

3 0
4 years ago
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when force stays the same and the mass increases, the ACCELERATION DECREASES.

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