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kenny6666 [7]
3 years ago
6

When heat is converted into another form of energy, the total amount of energy is constant. Which law best illustrates this stat

ement?
the first law of thermodynamics
the second law of thermodynamics
the third law of thermodynamics
none of the above
Physics
2 answers:
Lorico [155]3 years ago
6 0
The correct option is (A) <span>the first law of thermodynamics

Explanation:
The first law of thermodynamics takes the idea of law of conservation of energy and modify it for thermodynamics systems. It is the total internal energy of the systems equals to the amount of heat added "to" the system and the workdone "by" the system given as:
</span>ΔU = Q - W
<span>Q = heat added TO the system
W = work BY the system
</span>ΔU = Total internal energy
<span>
But the total internal energy is CONSERVED; it means that energy cannot be created or destroyed; it can only be transformed from one form to another. Hence the correct option is (A).</span>
zalisa [80]3 years ago
5 0

Answer:

it is A

Explanation:

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The distance from one crest to the next crest in a set of waves is called the wavelength. The distance from the crest of one wave to the equilibrium point is called frequency.
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3 years ago
Read 2 more answers
A long, straight wire is surrounded by a hollow metal cylinder whose axis coincides with that of the wire. The wire has a charge
Alexxx [7]

Explanation:

It is given that, a long, straight wire is surrounded by a hollow metal cylinder whose axis coincides with that of the wire.

The charge per unit length of the wire is \lambda and the net charge per unit length is 2 \lambda.

We know that there exist zero electric field inside the metal cylinder.

(a) Using Gauss's law to find the charge per unit length on the inner and outer surfaces of the cylinder. Let \lambda_i\ and\ \lambda_o are the charge per unit length on the inner and outer surfaces of the cylinder.

For inner surface,

\phi=\dfrac{q_{enclosed}}{\epsilon_o}

E.A=\dfrac{q_{enclosed}}{\epsilon_o}

0=\dfrac{\lambda_i+\lambda}{\epsilon_o}

\lambda_i=-\lambda  

For outer surface,

\lambda_i+\lambda_o=2\lambda

-\lambda+\lambda_o=2\lambda

\lambda_o=3\lambda

(b) Let E is the electric field outside the cylinder, a distance r from the axis. It is given by :

E_o=\dfrac{\lambda_o}{2\pi \epsilon_o r}

E_o=\dfrac{3\lambda}{2\pi \epsilon_o r}

Hence, this is the required solution.

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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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Answer:

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6 0
2 years ago
Which of the following is an example of the Doppler effect? A water bug on the surface of a pond is producing small ripples in t
noname [10]

Answer:

A police car with its siren on is driving towards you, and you perceive the pitch of the siren to increase.

Explanation:

In Physics, Doppler effect can be defined as the change in frequency of a wave with respect to an observer in motion and moving relative to the source of the wave.

Simply stated, Doppler effect is the change in wave frequency as a result of the relative motion existing between a wave source and its observer.

The term "Doppler effect" was named after an Austrian mathematician and physicist known as Christian Johann Doppler while studying the starlight in relation to the movement of stars.

<em>The phenomenon of Doppler effects is generally applicable to both sound and light. </em>

An example of the Doppler effect is a police car with its siren on is driving towards you, and you perceive the pitch of the siren to increase. This is so because when a sound object moves towards you, its sound waves frequency increases, thereby causing a higher pitch. However, if the sound object is moving away from the observer, it's sound waves frequency decreases and thus resulting in a lower pitch.

<em>Other fields were the Doppler effects are applied are; astronomy, flow management, vibration measurement, radars, satellite communications etc. </em>

3 0
3 years ago
A 0.0450-kg golf ball initially at rest is given a speed of 25.2 m/s when a club strikes. part a part complete if the club and b
Ksenya-84 [330]
We are given information:
m = 0.0450 kg
Δv = 25.2 m/s
Δt = 1.95 ms = 0.00195s

To find force we use formula:
F = m * a

a is acceleration. To find it we use formula:
a = Δv / Δt 
a = 25.2 / 0.00195
a = 12923.1 m/s^2

Now we can find force:
F = 0.0450 * 12923.1
F = 581.5 N 


To check the effect of the ball's weight on this movement we need to calculate it and then compare it to this force.
W = m * g
W = 0.0450 * 9.81
W = 0.44145 N 

We can see that weight is much smaller than the applied force so it's influence in negligible.
3 0
3 years ago
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