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Mandarinka [93]

2 years ago

5

What is the energy of a photon whose frequency is 5.0x10^14 hz

1 answer:

igomit [66]2 years ago

5 0

Answer:

Energy of a photon is $3.313\times 10^{-19}\textrm{ J}$

Explanation:

Given:

Frequency of the photon is, $f=5.0\times 10^{14}\textrm{ Hz}$ .

Energy of a photon of frequency $f$ is given as:

$E=hf$ , where, $h$ is called Planck's constant.

$h = 6.626\times 10^{-34}\textrm{ J.s}$

So, energy of a photon is:

$E= 6.626\times 10^{-34}\times 5.0\times 10^{14}=3.313\times 10^{-19}\textrm{ J}$

Therefore, the energy of a photon whose frequency is $5.0\times 10^{14}$ Hz is $3.313\times 10^{-19}\textrm{ J}$ .

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In Hooke’s law, what does the x represent?

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Answer: C

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In another words The force F is equal to the constant K times the disparagement.

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If the spring moves in opposite direction then,

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3 years ago

What is the net torque on the square plate, with sides 0.2 m, from each of the three forces? F1=18 N, F2=26 N, and F3=14 N. Use

marshall27 [118]

Answer:

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Explanation:

Given that,

Side = 0.2 m

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Force $F_{2}=26\ N$

Force $F_{3}=14\ N$

We need to calculate the torque due to force F₁

Using formula of torque

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$\tau_{1}=-F_{1}\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{1}=-18\times\dfrac{0.2}{2}$

$\tau_{1}=-1.8\ N-m$

We need to calculate the torque due to force F₂

Using formula of torque

$\tau_{2}=F_{2}d_{2}$

$\tau_{2}=F_{2}\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{2}=26\times\dfrac{0.2}{2}$

$\tau_{2}=2.6\ N-m$

We need to calculate the torque due to force F₃

Using formula of torque

$\tau_{3}=F_{3}d_{3}$

$\tau_{3}=(F_{3}\sin45+F_{3}\cos45)\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{3}=0.1(14\sin45+14\cos45)$

$\tau_{3}=1.92\ N-m$

We need to calculate the net torque on the square plate

$\tau=\tau_{1}+\tau_{2}+\tau_{3}$

$\tau=-1.8+2.6+1.92$

$\tau=2.72\ N-m$

Hence, The net torque on the square plate is 2.72 N-m.

3 0

3 years ago

In normal light conditions, how well do plants grow when there are 10 plants in the

Sati [7]

Answer

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7 0

3 years ago

An electric field of magnitude 2.35 V/m is oriented at an angle of 25.0° with respect to the positive z-direction. Determine the

zzz [600]

Answer:

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Explanation:

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Using formula of the magnetic flux

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Hence, The magnitude of the electric flux is $3.53\ N-m^2/C$

8 0

2 years ago