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Anika [276]
2 years ago
8

A photon of red light (wavelength = 710 nm) and a ping-pong ball (mass = 2. 40 × 10^-3 kg) have the same momentum. At what speed

is the ball moving?
Physics
1 answer:
Paha777 [63]2 years ago
3 0

The speed of the ball moving is

v = 3.94 \times 10 {}^{ - 25}m/s

what is momentum?

The momentum p of a classical object of mass m and velocity v is given by pclassical =mv.

For photons with wavelength λ,this equation does not hold.Instead, the momentum of the Photon is given by p Photon = h/λ

where,h is the planck's constant.

The momentum of the red Photon is

given:

h = 6.626 \times 10 {}^{ - 34}kgm {}^{2}/s(plancks \: constant)

λ = 700 \times 10 {}^{ - 9} m(photons \: wavelength)

p \: photons =  \frac{6.626 \times 10 {}^{ - 34}kgm {}  ^{2}/s }{700 \times 10 {}^{ - 9}m  }

= 9.47 \times 10 {}^{ - 28}kgm /s

since,the Photon and the ping-pong ball have the same momentum,we have

pball = pphotons =  \frac{6.626 \times 10 {}^{ - 34}kgm/s  }{700 \times 10 {}^{ - 9} }

pball = mv,m = 2.40 \times 10 {}^{ - 3}kg

v = 3.94 \times 10 {}^{ - 25} m/s

Therefore, if the red photon and the ping-pong ball have the same momentum, the ping-pong ball must have a speed of approximately

v = 3.94 \times 10 {}^{ - 25}m/s

learn more about momentum of photon from here: brainly.com/question/28197406

#SPJ4

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1) The unit of power is
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Answer:

1) the unit of power is A.Watt

2)100J=50W

100=50s

divide both sides by 50s

S=2s(A)

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What mechanism of energy is transferred by mass motion of fluid from one region of space to another?​
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3 years ago
3. A 2kg wooden block whose initial speed is 3 m/s slides on a smooth floor for 2 meters before it comes to a
serious [3.7K]

Answer:

Calculating Coefficient of friction is 0.229.

Force is 4.5 N that keep the block moving at a constant speed.

Explanation:

We know that speed expression is as \mathrm{V}^{2}=\mathrm{V}_{\mathrm{i}}^{2}+2 . \mathrm{a} . \Delta \mathrm{s}.

Where, {V}_{i} is initial speed, V is final speed, ∆s displacement and a acceleration.

Given that,

{V}_{i} =3 m/s, V = 0 m/s, and  ∆s = 2 m

Substitute the values in the above formula,

0=3^{2}-2 \times 2 \times a

0 = 9 - 4a

4a = 9

a=2.25 \mathrm{m} / \mathrm{s}^{2}

a=2.25 \mathrm{m} / \mathrm{s}^{2} is the acceleration.

Calculating Coefficient of friction:

\mathrm{F}=\mathrm{m} \times \mathrm{a}

\mathrm{F}=\mu \times \mathrm{m} \times \mathrm{g}

Compare the above equation

\mu \times m \times g=m \times a

Cancel "m" common term in both L.H.S and R.H.S

\text { Equation becomes, } \mu \times g=a

\text { Coefficient of friction } \mu=\frac{a}{g}

\mathrm{g} \text { on earth surface }=9.8 \mathrm{m} / \mathrm{s}^{2}

\mu=\frac{2.25}{9.8}

\mu=0.229

Hence coefficient of friction is 0.229.

calculating force:

\text { We know that } \mathrm{F}=\mathrm{m} \times \mathrm{a}

\mathrm{F}=2 \times 2.25 \quad(\mathrm{m}=2 \mathrm{kg} \text { given })

F = 4.5 N

Therefore, the force would be <u>4.5 N</u> to keep the block moving at a constant speed across the floor.

7 0
2 years ago
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