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NISA [10]
3 years ago
10

A particle zips by us with a Lorentz factor (γ) of 1.12. Then another particle zips by us moving at twice the speed of the first

particle.
a) What is the Lorentz factor (γ) of the second particle?

b) If the particles were moving with a speed much less than c, the magnitude of the
momentum of the second particle would be twice that of the first. However, what is the ratio of the magnitudes of momentum for these relativistic particles?
Physics
2 answers:
Firdavs [7]3 years ago
6 0
B yes this is right when I’m test on that
mixer [17]3 years ago
3 0

Answer:

Explanation:

Part A

Lorentz factor = 1/(sqrt(1 - (v/c)^2 ))

1.12 = 1/(sqrt(1 - v^2/c^2))                         square both sides.

1.2544 = 1 / (1 - v^2/c^2)                          Multiply both sides by 1 - v^2/c^2

1.2544 * (1 - v^2/c^2) = 1                         Remove the brackets

1.2544  - 1.2544 v^2 /c^2 = 1                 Multiply c^2 through the entire equation

1.2544*c^2 - 1.2544v^2 = c^2                Subtract 1.2544 c^2 from both sides

-1.2544v^2 = - 0.2544 c^2                     Divide by - 1.2544

v^2 = 0.2544 c^2 /1.2544

v^2 = 0.2028 c

v =  0.4534 * c

v = 1.360 * 10^8

2*v = 2.272 * 10*8

Lorentz Factor = LF = 1/square root( (1 - 2.272* 10^8/3*10^3))

LF = 2.345

Part B

Momentum is mv

I still get two even though the Lorentz factors are different.

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A helium atom (mass 4.0 u) moving at 598 m/s to the right collides with an oxygen molecule (mass 32 u) moving in the same direct
labwork [276]

Answer:

Speed of the helium after collision = 246 m/s

Explanation:

Given that

Mass of helium ,m₁ = 4 u

u₁=598 m/s

Mass of oxygen ,m₂ = 32 u

u₂  = 401 m/s

v₂ =445 m/s

Given that initially both are moving in the same direction and lets take they are moving in the right direction.

Speed of the helium after collision = v₁

There is no any external force on the masses that is why the linear momentum will be conserve.

Initial linear momentum = Final linear momentum

P = m v

m₁u₁+m₂u₂ = m₁v₁+m₂v₂

598 x 4 + 32 x 401 = 4 x v₁+ 32 x 445

v₁ = 246 m/s

Speed of the helium after collision = 246 m/s

6 0
3 years ago
A force F=0.12N is aplied on spring and spring elongates by 3cm . specific constant of spring ​
PilotLPTM [1.2K]

The spring constant is 4 N/m

Explanation:

When a spring is stretched/compressed by the application of a force, the relationship between the magnitude of the force applied and the elongation of the spring is given by Hooke's law:

F=kx

where

F is the magnitude of the spring applied

k is the spring constant

x is the elongation of the spring, relative to its equilibrium position

For the spring in this problem, we have:

F = 0.12 N (force applied)

x = 3 cm = 0.03 m (elongation of the spring)

Therefore, we can solve the formula for k to find the spring constant:

k=\frac{F}{x}=\frac{0.12}{0.03}=4 N/m

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4 0
3 years ago
How many atoms are in a sample of 2.39 moles of neon (Ne) atoms?
Reil [10]
<span>Avogadro's number represents the number of units in one mole of any substance. This has the value of 6.022 x 10^23 units / mole. This number can be used to convert the number of atoms or molecules into number of moles.

 </span>2.39 moles Ne ( 6.022 x 10^23 atoms / mole ) = 1.44 × 10^24 atoms Ne
4 0
3 years ago
A uniformly charged conducting sphere of 1.1 m diameter has a surface charge density of 6.2 µC/m2. (a) Find the net charge on th
ira [324]

Answer:

(a) q = 2.357 x 10⁻⁵ C

(b) Φ = 2.66 x 10⁶ N.m²/C

Explanation:

Given;

diameter of the sphere, d = 1.1 m

radius of the sphere, r = 1.1 / 2 = 0.55 m

surface charge density, σ = 6.2 µC/m²

(a)  Net charge on the sphere

q = 4πr²σ

where;

4πr² is surface area of the sphere

q is the net charge on the sphere

σ is the surface charge density

q = 4π(0.55)²(6.2 x 10⁻⁶)

q = 2.357 x 10⁻⁵ C

(b) the total electric flux leaving the surface of the sphere

Φ = q / ε

where;

Φ is the total electric flux leaving the surface of the sphere

ε is the permittivity of free space

Φ = (2.357 x 10⁻⁵) / (8.85 x 10⁻¹²)

Φ = 2.66 x 10⁶ N.m²/C

8 0
3 years ago
A closely wound search coil has an area of 3.21 cm2, 120 turns, and a resistance of 58.7 O. It is connected to a charge-measurin
Alexxx [7]

Answer:

The magnetic field in the System is 0.095T

Explanation:

To solve the exercise it is necessary to use the concepts related to Faraday's Law, magnetic flux and ohm's law.

By Faraday's law we know that

\epsilon = \frac{NBA}{t}

Where,

\epsilon  =electromotive force

N = Number of loops

B = Magnetic field

A = Area

t= Time

For Ohm's law we now that,

V = IR

Where,

I = Current

R = Resistance

V = Voltage (Same that the electromotive force at this case)

In this system we have that the resistance in series of coil and charge measuring device is given by,

R = R_c + R_d

And that the current can be expressed as function of charge and time, then

I = \frac{q}{t}

Equation Faraday's law and Ohm's law we have,

V = \epsilon

IR = \frac{NBA}{t}

(\frac{q}{t})(R_c+R_d) = \frac{NBA}{t}

Re-arrange for Magnetic Field B, we have

B = \frac{q(R_c+R_d)}{NA}

Our values are given as,

R_c = 58.7\Omega

R_d = 45.5\Omega

N = 120

q = 3.53*10^{-5}C

A = 3.21cm^2 = 3.21*10^{-4}m^2

Replacing,

B = \frac{(3.53*10^{-5})(58.7+45.5)}{120*3.21*10^{-4}}

B = 0.095T

Therefore the magnetic field in the System is 0.095T

3 0
3 years ago
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