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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.Calculate the average speed (in km/h) of Charlie, who runs to the store 4 kilometers away in 30 minutes. b.Calculate the dista
sveta [45]

as per the question charlie runs to the store which is  4 km away

hence the total distance covered [S] is 4 km

he takes 30 minutes to reach the store.

hence the total time taken [t] = 30 minutes=0.5 hour

We have to calculate the average speed.

the average speed[v]= =\frac{S}{t}

                                   =\frac{4 km}{0.5 hour}

                                    =8 km/hour

then we have to calculate the total distance traveled by charlie in 1 hour.

the distance covered S= V_{avg} *t

                                      =8 km/hour ×1 hour

                                      =8 km

Hence the average speed of charlie is 8 km/hour and he covers a distance of 8 km in 1 hour.

7 0
3 years ago
Which kind of road surface is easier to see when driving at night, a pebbled uneven surface or a mirror smooth surface?
pshichka [43]
A pebbled, uneven road would be easier to see at night because it minimizes the reflection of light from car’s light coming in the opposite direction. It is difficult to see when driving on the rainy day because the roadway reflects light from cars coming in the opposite <span>directions.</span>
6 0
3 years ago
A rod of length L and electrical resistance R moves through a constant uniform magnetic field ; both the magnetic field and the
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6 0
3 years ago
A 0.877 mol sample of N2(g) initially at 298 K and 1.00 atm is held at constant volume while enough heat is applied to raise the
MissTica

Answer : The value of q\text{ and }\Delta U is 286.2 J and 286.2 J respectively.

Explanation : Given,

Moles of sample = 0.877 mol

Change in temperature = 15.7 K

First we have to calculate the heat absorbed by the system.

Formula used :

q=n\times c_v\times \Delta T

where,

q = heat absorbed by the system = ?

n = moles of sample = 0.877 mol

\Delta T = Change in temperature = 15.7 K

c_v = heat capacity at constant volume of N_2 (diatomic molecule) = \frac{5}{2}R

R = gas constant = 8.314 J/mol.K

Now put all the given value in the above formula, we get:

q=0.877mol\times \frac{5}{2}\times 8.314J/mol.K\times 15.7K

q=286.2J

Now we have to calculate the change in internal energy of the system.

\Delta U=q+w

As we know that, work done is zero at constant volume. So,

\Delta U=q=286.2J

Therefore, the value of q\text{ and }\Delta U is 286.2 J and 286.2 J respectively.

8 0
4 years ago
Read 2 more answers
To calibrate your calorimeter cup, you first put 45 mL of cold water in the cup, and measure its temperature to be 24.7 °C. You
drek231 [11]

Answer : The heat change of the cold water in Joules is, 1.6\times 10^3J

Explanation :

First we have to calculate the mass of cold water.

As we know that the density of water is 1 g/mL. The volume of cold water is 45 mL.

Density=\frac{Mass}{Volume}

Mass=Density\times Volume=1g/mL\times 45mL=45g

Now we have to calculate the heat change of cold water.

Formula used :

Q=m\times c\times (T_2-T_1)

where,

Q = heat change of cold water = ?

m = mass of cold water = 45 g

c = specific heat of water = 4.184J/g^oC

T_1 = initial temperature of cold water = 24.7^oC

T_2 = final temperature  = 33.4^oC

Now put all the given value in the above formula, we get:

Q=45g\times 4.184J/g^oC\times (33.4-24.7)^oC

Q=1638.036J=1.6\times 10^3J

Therefore, the heat change of cold water is 1.6\times 10^3J

4 0
3 years ago
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