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Papessa [141]
3 years ago
12

suppose that a particle at rest with mass m decays into a photon and another particle with mass m/2 , which recoilds in a direct

ion opposite to the photon. Using the conservation of energy and momentum find the momentum of the recoiling particle
Physics
1 answer:
Nady [450]3 years ago
5 0

Answer:

p_r=\frac{h}{\lambda}

Explanation:

To find the momentum of the recoiling particle you can use the momentum formula for a photon:

p=\frac{h}{\lambda}

before the decay the momentum is zero. Hence, after the decay the momentum of the photon plus the momentum of the recoiling particle must be zero:

p_a=p_b\\\\\frac{h}{\lambda}-\frac{m}{2}v=0\\\\p_r=\frac{m}{2}v=\frac{h}{\lambda}

where pr is the momentum of the recoiling particle.

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The two speakers at S1 and S2 are adjusted so that the observer at O hears an intensity of 6 W/m² when either S1 or S2 is sounde
Zanzabum

Answer:

The minimum frequency is 702.22 Hz

Explanation:

The two speakers are adjusted as attached in the figure. From the given data we know that

S_1 S_2=3m

S_1 O=4m

By Pythagoras theorem

                                 S_2O=\sqrt{(S_1S_2)^2+(S_1O)^2}\\S_2O=\sqrt{(3)^2+(4)^2}\\S_2O=\sqrt{9+16}\\S_2O=\sqrt{25}\\S_2O=5m

Now

The intensity at O when both speakers are on is given by

I=4I_1 cos^2(\pi \frac{\delta}{\lambda})

Here

  • I is the intensity at O when both speakers are on which is given as 6 W/m^2
  • I1 is the intensity of one speaker on which is 6  W/m^2
  • δ is the Path difference which is given as

                                           \delta=S_2O-S_1O\\\delta=5-4\\\delta=1 m

  • λ is wavelength which is given as

                                             \lambda=\frac{v}{f}

      Here

              v is the speed of sound which is 320 m/s.

              f is the frequency of the sound which is to be calculated.

                                  16=4\times 6 \times cos^2(\pi \frac{1 \times f}{320})\\16/24= cos^2(\pi \frac{1f}{320})\\0.667= cos^2(\pi \frac{f}{320})\\cos(\pi \frac{f}{320})=\pm0.8165\\\pi \frac{f}{320}=\frac{7 \pi}{36}+2k\pi \\ \frac{f}{320}=\frac{7 }{36}+2k \\\\ {f}=320 \times (\frac{7 }{36}+2k )

where k=0,1,2

for minimum frequency f_1, k=1

                                  {f}=320 \times (\frac{7 }{36}+2 \times 1 )\\\\{f}=320 \times (\frac{79 }{36} )\\\\ f=702.22 Hz

So the minimum frequency is 702.22 Hz

5 0
3 years ago
A mass is attached to a vertical spring, which then goes into oscillation. At the high point of the oscillation, the spring is i
andrew-mc [135]

Answer:

0.34 sec

Explanation:

Low point of spring ( length of stretched spring ) = 5.8 cm

midpoint of spring = 5.8 / 2 = 2.9 cm

Determine the oscillation period

at equilibrum condition

Kx = Mg

g= 9.8 m/s^2

x = 2.9 * 10^-2 m

k / m = 9.8 / ( 2.9 * 10^-2 ) =  337.93

note : w = \sqrt{k/m}   = \sqrt{337.93} = 18.38 rad/sec

Period of oscillation =  2\pi  / w

                                  = 0.34 sec

8 0
3 years ago
the threshold of hearing is defined as the minimum discernible intensity of the sound. it is approximately 10−12w/m2 . find the
vivado [14]

The distance is 97720.5 m

From the question, we have

P = 0.06 W × 2 = 0.12 W

d = ?

Sound intensity, I = P/4πd²

I = 10⁻¹² W/m²

10⁻¹² = 0.12/4πd²

d = 97720.5 m

The distance is 97720.5 m

Sound intensity :

The power carried by sound waves per unit area in the direction  perpendicular to that region is known as sound intensity or acoustic intensity. The watt per square meter (W/m2) is the SI unit of intensity, which also covers sound intensity. Sound intensity is a measure of how quickly energy moves across a given space. The unit area in the SI measurement system is 1 m2. So Watts per square meter are used to measure sound intensity. As there will be energy flow in certain directions but not in others, sound intensity also provides a measure of direction.

To learn more about Sound intensity visit: brainly.com/question/12899113

#SPJ4

3 0
1 year ago
An electron passes through a point 2.83 cm 2.83 cm from a long straight wire as it moves at 35.5 % 35.5% of the speed of light p
igor_vitrenko [27]

Answer:

The magnitude of electron acceleration is 2.34 \times 10^{15} \frac{m}{s^{2} }

Explanation:

Given:

Distance from the wire to the field point r = 2.83 \times 10^{-2} m

Speed of electron v = 35.5 \%c

Current I = 17.7 A

For finding the acceleration,

First find the magnetic field due to wire,

  B = \frac{\mu _{o}I }{2\pi r }

Where \mu_{o} = 4\pi   \times 10^{-7}

  B = \frac{4\pi \times 10^{-7}  \times 17.7 }{2\pi (2.83 \times 10^{-2} ) }

  B = 12.50 \times 10^{-5} T

The magnetic force exerted on the electron passing through straight wire,

  F = qvB  

  F = 1.6 \times 10^{-19} \times 0.355 \times 3 \times 10^{8} \times 12.50 \times 10^{-5}

  F = 21.3 \times 10^{-16} N

From the newton's second law

  F = ma

Where m = mass of electron = 9.1 \times 10^{-31} kg

So acceleration is given by,

   a = \frac{F}{m}

   a = \frac{21.3 \times 10^{-16} }{9.1 \times 10^{-31} }

   a = 2.34 \times 10^{15} \frac{m}{s^{2} }

Therefore, the magnitude of electron acceleration is 2.34 \times 10^{15} \frac{m}{s^{2} }

7 0
2 years ago
A merry-go-round of radius 2 m is rotating at one revolution every 5 s. A
galben [10]

Answer:

a) The angular speed of the child is approximately 1.257 rad/s

b) The angular speed of the teenager is approximately 1.257 rad/s

c) The tangential speed of the child is approximately 1.257 m/s

d) For the child, r = 2 m

The tangential speed of the teenager is approximately 2.513 m/s

Explanation:

The revolutions per minute, r.p.m. of the merry-go-round = 1 revolution/(5 s)

The radius of the merry-go-round = 2 m

The location of the child = 1 m from the axis

The location of the teenager = 2 m from the axis

1 revolution = 2·π radians

Therefore, we have;

The angular speed, ω = (Angle turned)/(Time elapsed) = (2·π radians)/(5 s)

∴ The angular speed of the merry-go-round, ω = 2·π/5 radians/second

a) The angular speed of the child = The angular speed of the merry-go-round = 2·π/5 radians/second ≈ 1.257 rad/s

b) The angular speed of the teenager = The angular speed of the merry-go-round = 2·π/5 radians/second ≈ 1.257 rad/s

c) The tangential speed, v = r × The angular speed, ω

Where;

r = The radius of rotation of the object

For the child, r = 1 m

The tangential speed of the child = 1 m × 2·π/5 radians/second = 2·π/5 m/s ≈ 1.257 m/s

d) For the child, r = 2 m

The tangential speed of the teenager = 2 m × 2·π/5 radians/second = 4·π/5 m/s ≈ 2.513 m/s

8 0
2 years ago
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