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Anastaziya [24]

3 years ago

8

A car goes around in a circular curve on a horizontal road at constant speed. what is the direction of the friction force on the

car due to the road?

1 answer:

velikii [3]3 years ago

3 0

Outside. but I may be wrong

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Why are noise considerations important in optical fiber communications? 3. Describe the principle of "population inversion".

pochemuha

Answer and Explanation:

the electronic devices always have some noises present in the signal

there are some important considerations in optical fiber communications these are.

the noise which is contributed by transmitter are electronic random noise, low frequency noise
noise which is contributed by laser are relative intensity noise, mode partition noise, conversion of phase noise to amplitude noise.
noise contributed by photo detector are quantum shot noise, shot noise from dark current, avalanche multiplication noise.

PRINCIPLE OF POPULATION INVERSION :

The principle of population inversion is defined as for production of high percentage of simulated emission for a laser beam the number of atoms in higher state should be greater than lower energy state

7 0

3 years ago

Math Exploration

Airida [17]

Question: What is the frequency of a wave that has a wave speed of 120 m/s and a wavelength of 0.40 m?

Answer: The equation that relates frequency of a wave to a waves speed and wavelength is Speed of Wave= Frequency X Wavelength. Since you are given speed and wavelength, you plug those two known numbers into the equation, 120= Frequency X 0.40. You then divide 120 by .4 to get your frequency of 300.

Explanation: this might help for

5 0

3 years ago

Issac newton how did he shape the new worldview?

gulaghasi [49]

Accordingly Newton's findings, astronomy and physics have industrialized hugely over the period. Scientists now recognize that every object in the world has a force that draws each other and the power of the force hinge on the mass of the object. Also, Newton's Laws of Motion offer individuals a better understanding of what is likely concerning movement. This is very helpful, particularly in mechanics and space travel. Generally, Newton had a huge and permanent impact on science.

3 0

2 years ago

When the current in a toroidal solenoid is changing at a rate of 0.0240 A/s , the magnitude of the induced emf is 12.4 mV . When

Gemiola [76]

Answer:

The number of turns in the solenoid is 230.

Explanation:

Given that,

Rate of change of current, $\dfrac{dI}{dt}=0.0240\ A/s$

Induced emf, $\epsilon=12.4\ mV=12.4\times 10^{-3}\ V$

Current, I = 1.5 A

Magnetic flux, $\phi=0.00338\ Wb$

The induced emf through the solenoid is given by :

$\epsilon=L\dfrac{dI}{dt}$

or

$L=\dfrac{\epsilon}{(di/dt)}$ ........(1)

The self inductance of the solenoid is given by :

$L=\dfrac{N\phi}{I}$ .........(2)

From equation (1) and (2) we get :

$\dfrac{\epsilon}{(di/dt)}=\dfrac{N\phi}{I}$

N is the number of turns in the solenoid

$N=\dfrac{\epsilon I}{\phi (dI/dt)}$

$N=\dfrac{12.4\times 10^{-3}\times 1.5}{0.00338 \times 0.024}$

N = 229.28 turns

or

N = 230 turns

So, the number of turns in the solenoid is 230. Hence, this is the required solution.

3 0

2 years ago

How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe

Rus_ich [418]

Explanation:

(a) For an isothermal process, work done is represented as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

Putting the given values into the above formula as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

= $- 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})$

= $-12280.82 \times ln (0.09)$

= $-12280.82 \times -2.41$

= 29596.78 J

or, = 29.596 kJ (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

W = $\frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}$

= $\frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}$

= $\frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}$

= 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c) We know that for an isothermal process,

$P_{1}V_{1} = P_{2}V_{2}$

or, $P_{2} = \frac{P_{1}V_{1}}{V_{2}}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})$

= 11 atm

Hence, the required pressure is 11 atm.

(d) For adiabatic process,

$P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}$

or, $P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}$

= 28.7 atm

Therefore, required pressure is 28.7 atm.

6 0

3 years ago