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IceJOKER [234]
3 years ago
8

Show that pair production is not possible without a neighbouring particle that takes part of the momentum.

Physics
1 answer:
egoroff_w [7]3 years ago
3 0

Answer:

By conservation of momentum

Explanation:

By conservation of energy, the photon needs at least the same amount of energy as the energy rest mass of the electron and positron together, which is around 1,02 MeV. But, by conservation of momentum, the electron-positron do not have momentum, while the photon had some. This means that, in order to conserve momentum, we need another neighboring particle, normally an atomic nucleus,  to receive the photon's momentum.

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The question is asking to calculate the tension that the string has to adjust the string so that when vibrating in its second overtone, it produces sound of wavelength of 0.761m, base on my calculation, the calculation must be done by the formula of <span>v=λf</span><span>., I hope this would help </span>
3 0
3 years ago
Five properties of magnet​
Olenka [21]

Answer:

1. The magnet is magnetic and can attract iron articles.

2. The magnet has magnetic poles. Each magnet has two kinds of poles: N pole and S pole. They are in pairs.

3. Temporary magnet and permanent magnet: when the ferromagnetic material is magnetized, it is easy to lose the magnetic property, which is called temporary magnet (for example: iron); when the ferromagnetic material is magnetized, it is not easy to lose the magnetic property, which is called permanent magnet (for example: steel).

4. When two magnets are close to each other, the same poles will repel and push away from each other, and the different poles will attract and stick to each other. Therefore: the same pole repels each other, the different pole attracts each other.

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3 0
3 years ago
An oil gusher shoots crude oil 25.0 m into the air through a pipe with a 0.100-m diameter. Neglecting air resistance but not the
Mariana [72]

Answer:3764.282 KPa

Explanation:

Given gusher shoots oil at h=25 m

i.e. the velocity of jet is

v=\sqrt{2gh}[/tex]

v=22.147 m/s

Now the pressure loss in pipe is given by hagen poiseuille equation

\Delta P=\frac{32L\mu v}{D^2}

\Delta P=\frac{32\times 50\times 22.147\times 1}{10^{-2}}

\Delta P=3543.557 KPa

For  25 m head in terms of Pressure

\Delta P_2=\rho \times g\times h=220.725 KPa

Total Pressure=\Delta P+\Delta P_2=3543.557+220.725=3764.282 KPa

4 0
3 years ago
To tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.50•10^6 N, one at an angle 19.0° west of north, a
irina1246 [14]

Answer:

Work done by a tug boat, W = 1.735 x 10⁸ J

Explanation:

Given,

The of each tugboat, F = 1.5 x 10⁶ N

The angle of each tugboat forms with the resultant force, θ = 19°

The displacement of the supertanker, s = 710 m

The individual tugboat will be responsible for the displacement, d = 710/2

                                                                                                               = 355 m

The displacement component in each tugboat direction = 355 · sin θ meter

Therefore, the work done by each tugboat is

                                           W = F x S    joules

Substituting the values in the above equation

                                            W = 1.5 x 10⁶  x  355 · sin θ

                                                = 1.735 x 10⁸ J

Hence, the work done by each tugboat is, W = 1.735 x 10⁸ J                        

6 0
3 years ago
Work is being done in which of these situations? all motions are at a constant velocity
fredd [130]
Where the force is not perpendicular to the path of motion 

are you missing the the situations ?

7 0
3 years ago
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