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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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4 0
3 years ago
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A slinky forms it’s third harmonic standing wave when the input frequency is 24 Hz. What is the fundamental frequency of the sli
jarptica [38.1K]

Answer: The fundamental frequency of the slinky = 8Hz

An input frequency of 28 Hz will not create a standing wave

Explanation:

Let Fo = fundamental frequency

At third harmonic,

F = 3Fo

If F = 24Hz

24 = 3Fo

Fo = 24/3 = 8Hz

If an input frequency = 28 Hz at 3rd harmonic

Let find the fundamental frequency

28 = 3Fo

Fo = 28/3

Fo = 9.33333Hz

Since Fo isn't a whole number, it can't create a standing wave

6 0
3 years ago
Initially a car accelerates at 2 m/s2 for x seconds. The car then travels at a velocity of -6 m/s for x seconds. If the car disp
Luda [366]

Answer:

The time travel is

t=8 s

Explanation:

a= 2 \frac{m}{s^{2} } \\v=-6 \frac{m}{s} \\x=16m

x_{f}=x_{o}+v_{o}*t+\frac{1}{2} *a*t^{2}

x_{f}=0-6*t+\frac{1}{2} *2*t^{2}

t^{2}-6*t-16=0\\ using :\\\frac{-b+/-\sqrt{b^{2}-4*c*a } }{2} \\\frac{-(-6)+/-\sqrt{(-6)^{2}-4*(-16)*(1) } }{2}=\frac{3}{2} +/- \frac{10}{2} \\t_{1} = 2s \\t_{2} = 8s

Check

t_{2}=8s

x_{f}=x_{o}+v_{o}*t+\frac{1}{2} *a*t^{2}

x_{f}=0-6*+\frac{1}{2} *2*8^{2}

x_{f}=-48+64\\x_{f}=16

5 0
3 years ago
At 6: 00 am, a motorbike set off from town A to town B at a speed of 40km/h. At the same time, a car set off from town B to town
Keith_Richards [23]

Answer:

One would need to know how far apart the towns are:

T = SA / 40      time it takes for first cyclist to travel S1

T = SB / 60       time it takes for cyclist B to travel distance S2

SA + SB = S     the distance between the towns

SB = 60 / 40 SA = 1.5 SA

SA + 1.5 SA = S

S = 2.5 SA where cyclist travels distance SA

The time will depend on the separation of the towns.

4 0
2 years ago
Objects with masses of 165 kg and a 465 kg are separated by 0.340 m. (a) Find the net gravitational force exerted by these objec
Dmitrij [34]

Answer:

(a) F_{net} = 4.19 x 10^{-5} N, and its direction is towards m_{2}.

(b) It must be placed inside a hollow shell.

Explanation:

Let, m_{1} = 165 kg, m_{2} = 465 kg, m_{3} = 60 kg, and the distance between m_{1} and m_{2} is 0.340 m.

(a) Since m_{3} is placed midway between m_{1} and m_{2}, then its distance to both masses is 0.170 m.

From the Newton's law of universal gravitation,

F = \frac{Gm_{1}m_{2}  }{r^{2} }

Where all variables have their usual meaning.

Then,

a. F_{net} = F_{23} - F_{13}

F_{13} = \frac{6.67*10^{-11}*165*60 }{(0.17)^{2} }

     = 2.25 x 10^{-5} N

F_{23} = \frac{6.67*10^{-11}*465*60 }{(0.17)^{2} }

     = 6.44 x 10^{-5} N

∴ F_{net} =  = 6.44 x 10^{-5} - 2.25 x 10^{-5}

              = 4.19 x 10^{-5} N

The net force exerted by the two masses on the 60 kg object is 4.19 x 10^{-5}  N.

(ii) /F_{net}/ = /F_{23}/ - /F_{13}/

              = 6.44 x 10^{-5} - 2.25 x 10^{-5}

              = 4.19 x 10^{-5} N

(iii) The direction of the net force is to the right i.e towards m_{2}.

(b) For the net force experienced by the 60 kg object to be zero, it must be placed inside a hollow shell.

7 0
3 years ago
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