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3 years ago

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A photon in a laboratory experiment has an energy of 5 eV. What is the frequency of this photon? (using the idea of the electron

volt)

1 answer:

andreyandreev [35.5K]3 years ago

7 0

Answer and working shown on photo

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AnnZ [28]

Answer:

friction help to slow motion in other word it oppose motion, but in a frictionless environment object would move with difficult stopping point.

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3 years ago

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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

Photon Lighting Company determines that the supply and demand functions for its most popular lamp are as follows: S(p) = 400 - 4

torisob [31]

Answer: The correct answer is (C).

Explanation:

Supply function:

$S(p)=400-4p+0.00002p^4$

Demand function:

$D(p)=2800-0.0012p^3$

S(p)=D(p), price for which supply is equal to demand:

$S(p)=D(p)=400-4p+0.00002p^4=2800-0.0012p^3$

After solving this above equation graphically, we will get the values of p.

1) p =96.236

2) p=( -118.258)

We will reject the negative value of p.

So, the value of p that price for which the supply equals the demand is $ 96.236 $\approx$ $96.24. Hence, the correct answer is option (C).

6 0

3 years ago

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A light woman and a heavy man jump from an airplane at the same time and open their same-size parachutes at the same time. which

Airida [17]

The heavy man will fall faster, because he has more mass and heavy things fall faster than light things

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3 years ago

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A water main pipe of diameter 10 cm enters a house 2 m below ground. A smaller diameter pipe carries water to a faucet 5 m above

Lemur [1.5K]

Explanation:

Given that,

Diameter = 10 cm

Distance = 2 m

Speed $v_{1}= 2\ m/s$

Speed $v_{2}=7\ m/s$

Pressure in main pipe $P_{1}=2\times10^{5}\ Pa$

(I). We need to calculate the diameter

Using equation of continuity

$Av_{1}=Av_{2}$

$\pi(\dfrac{d_{1}}{2})^2\times v_{1}=\pi(\dfrac{d_{2}}{2})^2\times v_{2}$

$(\dfrac{10}{2})^2\times2=(\dfrac{d_{2}}{2})^2\times7$

$d_{2}=\sqrt{\dfrac{25\times2\times4}{7}}$

$d_{2}=5.345\ cm$

(II). We need to calculate the pressure the gauge pressure

Using Bernoulli equation

$P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho gh_{1}=P_{2}+\dfrac{1}{2}\rho v_{2}^2+\rho g h_{2}$

$P_{2}=P_{1}+\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)-\rho g(h_{1}-h_{2})$

$P_{2}=2\times10^{5}+\dfrac{1}{2}\times1000(4-49)-1000\times 9.8\times(5)$

$P_{2}=1.28500\times10^{5}\ Pa$

(III). If it is possible to carry water to a faucet 17 m above ground,

Using Bernoulli equation

$P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho gh=P_{3}+\dfrac{1}{2}\rho v_{3}^2+\rho g h_{3}$

$P_{3}=P_{1}+\dfrac{1}{2}\rho v_{1}^2-\rho g(h_{1}-h_{3})$

Here, $h_{3}=0$

Put the value in the equation

$P_{3}=2\times10^{5}+\dfrac{1}{2}\times1000\times4-1000\times 9.8\times17$

$P_{3}=3.5400\times10^{5}\ Pa$

Hence, This is required solution.

7 0

3 years ago