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3 years ago

8

The speed of light changes as it passes from one medium to the next. In which image does the speed of light change the most?

2 answers:

blagie [28]3 years ago

7 0

I believe it changes the most in water

Ber [7]3 years ago

6 0

Answer:

IDK

Explanation:

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How is the q of an rlc parallel resonant circuit calculated?

notka56 [123]

Answer:

It is calculated by dividing Resistance, R, by Inductive reactance, XL.

Explanation:

Q is called the Q factor of a resonance circuit. In a parallel resonance circuit, it is calculated by finding the ratio of the power stored in the circuit to the power distributed in the circuit. It is a way of measuring the quality of a circuit or how effective the circuit is.

Q factor is the inverse in the resonance series circuit.

Q factor of a resonance parallel circuit,

<h3>Q = R/XL</h3>

R = Resistance

XL = Inductive reactance

3 0

3 years ago

A real heat engine operates between temperatures TcTcT_c and ThThT_h. During a certain time, an amount QcQcQ_c of heat is releas

Nookie1986 [14]

The maximum amount of work performed is

$W_{max}=\frac{T_H-T_C}{T_C}Q_C$

Explanation:

The efficiency of a real heat engine is given by the equation:

$\eta = 1-\frac{T_C}{T_H}$ (1)

where

$T_C$ is the temperature of the cold reservoir

$T_H$ is the temperature of the hot reservoir

However, the efficiency of a real heat engine can be also written as:

$\eta = \frac{W_{max}}{Q_H}$

where

$W_{max}$ is the maximum work done

$Q_H$ is the heat absorbed from the hot reservoir

$Q_H$ can be written as

$Q_H=W_{max}+Q_C$

where

$Q_C$ is the heat released to the cold reservoir

So the previous equation can be also written as

$\eta=\frac{W_{max}}{W_{max}+Q_C}$ (2)

By combining eq.(1) and (2) we get

$1-\frac{T_C}{T_H}=\frac{W_{max}}{W_{max}+Q}$

And re-arranging the equation and solving for $W_{max}$ , we find

$W_{max}=\frac{T_H-T_C}{T_C}Q_C$

Learn more about work and heat:

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#LearnwithBrainly

8 0

3 years ago

If this decay has half-life of 2 years, how many years would it take for 10.8 g Protactinium-231 to remain given an initial mass

sergij07 [2.7K]

Answer:

Time = 6 years

Explanation:

First, we will calculate the no. of half life periods required to reduce the mass of Protactinium to the given value:

$m' = \frac{m}{2^{n} } \\\\2^n = \frac{m}{m'}$

where,

n = no. of half-life periods = ?

m = initial mass = 86.3 g

m' = remaining mass = 10.8 g

Therefore,

$2^n = \frac{86.3\ g}{10.8\ g}\\\\2^n = 8\\2^n = 2^3$

Since the bases are the same. Therefore equating powers:

n = 3

Now we calculate the time:

$Time = (n)(Half-Life)\\Time =(3)(2\ years)$

<u>Time = 6 years</u>

3 0

3 years ago

Calculate the magnitude of the electric field at one corner of a square 2.42 m on a side if the other three corners are occupied

Anettt [7]

Answer:

loloeuhsh

Explanation:

shwvwgnwajejjeisus

5 0

3 years ago

________ is the si unit of angular momentum

navik [9.2K]

Answer:

kg m2/s

Explanation:

I think it is :)

4 0

3 years ago