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2 years ago

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Jordan wants to know the difference between using a 60-W and 100-W lightbulb in her lamp. She calculates the energy it would tak

e to use each bulb for 30 s by using this equation: Power = Energy transferred (J)/time (s) What is the difference in the amount of energy transferred by the two bulbs during this time?

1 answer:

Andreas93 [3]2 years ago

4 0

Answer:

The difference in the amount of energy transferred by the two bulbs is 1200 J.

Explanation:

The energy transferred by the two lightbulbs can be calculated with the given equation:

$E = P*t$

Where P is the power and t is the time

For the 60 W lightbulb:

$E_{60} = P*t = 60 W*30 s = 1800 J$

For the 100 W lightbulb:

$E_{100} = P*t = 100 W*30 s = 3000 J$

Hence, the difference in the amount of energy transferred is:

$E_{t} = E_{100} - E_{60} = 3000 J - 1800 J = 1200 J$

Therefore, the difference in the amount of energy transferred by the two bulbs is 1200 J.

I hope it helps you!

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What is a the kinetic energy, in joules, of this baseball when it is thrown by a major-league pitcher at 96.0 mi/h?

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=0.5*(5.12 o *0.02835 kg/1 ounce)* (95 miles/h*1609.34m/1 miles* 1hr/3600s^)2
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By what factor with the kinetic energy change if the speed of the baseball is decreased to 55.0 mi/h?

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Answer:

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Explanation:

The relationship between rotational speed in radians per second and acceleration is ...

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We want the rotation rate in RPM, so we need the conversion ...

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2 years ago

A 1.2-m radius cylindrical region contains a uniform electric field along the cylinder axis. It is increasing uniformly with tim

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Answer:

The magnitude of rate of change of electric field is $49.95\ V/m{\cdot} s$ .

Explanation:

Given that,

Radius of the cylindrical region contains a uniform electric field along the cylinder axis, r = 1.2 m

Total displacement current through a cross section of the region, $I=2\times 10^{-9}\ A$

We need to find the rate of change of electric field. Its is given by the formula as follows :

$\dfrac{dE}{dt}=\dfrac{I}{A\epsilon_o}\\\\\dfrac{dE}{dt}=\dfrac{2\times 10^{-9}}{\pi (1.2)^2\times 8.85\times 10^{-12}}\\\\\dfrac{dE}{dt}=49.95\ V/m{\cdot} s$

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Calculate the final temperature of a mixture of 0.350 kg of ice initially at 218°C and 237 g of water initially at 100.0°C.

kramer

Answer:

115 ⁰C

Explanation:

<u>Step 1:</u> The heat needed to melt the solid at its melting point will come from the warmer water sample. This implies

$q_{1} +q_{2} =-q_{3}$ -----eqution 1

where,

$q_{1}$ is the heat absorbed by the solid at 0⁰C

$q_{2}$ is the heat absorbed by the liquid at 0⁰C

$q_{3}$ the heat lost by the warmer water sample

Important equations to be used in solving this problem

$q=m *c*\delta {T}$ , where -----equation 2

q is heat absorbed/lost

m is mass of the sample

c is specific heat of water, = 4.18 J/0⁰C

$\delta {T}$ is change in temperature

Again,

$q=n*\delta {_f_u_s}$ -------equation 3

where,

q is heat absorbed

n is the number of moles of water

tex]\delta {_f_u_s}[/tex] is the molar heat of fusion of water, = 6.01 kJ/mol

<u>Step 2:</u> calculate how many moles of water you have in the 100.0-g sample

$=237g *\frac{1 mole H_{2} O}{18g} = 13.167 moles of H_{2}O$

<u>Step 3: </u>calculate how much heat is needed to allow the sample to go from solid at 218⁰C to liquid at 0⁰C

$q_{1} = 13.167 moles *6.01\frac{KJ}{mole} = 79.13KJ$

This means that equation (1) becomes

79.13 KJ + $q_{2} = -q_{3}$

<u>Step 4:</u> calculate the final temperature of the water

$79.13KJ+M_{sample} *C*\delta {T_{sample}} =-M_{water} *C*\delta {T_{water}$

Substitute in the values; we will have,

$79.13KJ + 237*4.18\frac{J}{g^{o}C}*(T_{f}-218}) = -350*4.18\frac{J}{g^{o}C}*(T_{f}-100})$

79.13 kJ + 990.66J* $(T_{f}-218})$ = -1463J* $(T_{f}-100})$

Convert the joules to kilo-joules to get

79.13 kJ + 0.99066KJ* $(T_{f}-218})$ = -1.463KJ* $(T_{f}-100})$

$79.13 + 0.99066T_{f} -215.96388= -1.463T_{f}+146.3$

collect like terms,

2.45366 $T_{f}$ = 283.133

∴ $T_{f} =$ = 115.4 ⁰C

Approximately the final temperature of the mixture is 115 ⁰C

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2 years ago