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2 years ago

Part C

1 answer:

3 0

Answer: The battery in circuit 3 would run out more quickly when the switches are closed because the two bulbs are brighter, which means the battery is providing more energy to that circuit.

Explanation: Answer from plato

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What is the focal length of a lens

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Answer:

The focal length of a lens is refers to the distance from the center of the lens to the principal foci.

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When driving across Death Valley in the summertime, it is recommended that you release some air from your tires before making th

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An engine absorbs 1749 J from a hot reservoir and expels 539 J to a cold reservoir in each cycle. a. What is the engine’s effici

Masja [62]

Answer:

a)η = 69.18 %

b)W= 1210 J

c)P=3967.21 W

Explanation:

Given that

Q₁ = 1749 J

Q₂ = 539 J

From first law of thermodynamics

Q₁ = Q₂ +W

W=Work out put

Q₂=Heat rejected to the cold reservoir

Q₁ =heat absorb by hot reservoir

W= Q₁- Q₂

W= 1210 J

The efficiency given as

$\eta=\dfrac{W}{Q_1}$

$\eta=\dfrac{1210}{1749}$

$\eta=0.6918$

η = 69.18 %

We know that rate of work done is known as power

$P=\dfrac{W}{t}$

$P=\dfrac{1210}{0.305}\ W$

P=3967.21 W

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3 years ago

In an electric field, 0.90 joule of work is required to bring 0.45 coulomb of charge from point a to point

jarptica [38.1K]

The difference in electric potential energy between the two points is
$\Delta U = q \Delta V$
where q is the magnitude of the charge and $\Delta V$ is the electric potential difference.

But for energy conservation, the difference in electric potential energy $\Delta U$ between the two points is equal to the work done to move the charge between A and B:
$W=\Delta U$
so we have
$W=q \Delta V$

and by substituting the numbers of the problem, we find the value of $\Delta V$ :
$\Delta V = \frac{W}{q}= \frac{0.90 J}{0.45 C}=2 V$

3 0

3 years ago

Suppose the ski patrol lowers a rescue sled carrying an injured skier, with a combined mass of 97.5 kg, down a 60.0-degree slope

Kitty [74]

a. 1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b.21,835 J work, in joules, is done by the rope on the sled this distance.

c. 23,170 J the work, in joules done by the gravitational force on the sled d. The net work done on the sled, in joules is 43,670 J.

<h3>What is friction work?</h3>

The work done by friction is the force of friction times the distance traveled times the cosine of the angle between the friction force and displacement

a. How much work is done by friction as the sled moves 28m along the hill?

ans. We use the formula:

friction work = -µ.mg.dcosθ

= -0.100 * 97.5 kg * 9.8 m/s² * 28 m * cos 60

= -1337.3 J

-1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b. How much work is done by the rope on the sled in this distance?

We use the formula:

Rope work = -m.g.d(sinθ - µcosθ)

rope work = - 97.5 kg * 9.8 m/s² * 28 m (sin 60 – 0.100 * cos 60)

= 26,754 (0.816)

= 21,835 J

21,835 J work, in joules, is done by the rope on the sled this distance.

c. What is the work done by the gravitational force on the sled?

By using the formula:

Gravity work = mgdsinθ

= 97.5 kg * 9.8 m/s² * 28 m * sin 60

= 23,170 J

23,170 J the work, in joules done by the gravitational force on the sled .

D. What is the total work done?

By adding all the values

work done = -1337.3 + 21,835 + 23,170

= 43,670 J

The net work done on the sled, in joules is 43,670 J.

Learn more about friction work here:

brainly.com/question/14619763

#SPJ1

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2 years ago