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Juli2301 [7.4K]
3 years ago
10

An electron traveling with a speed v enters a uniform magnetic field directed perpendicular to its path. The electron travels fo

r a time t0 along a half-circle of radius R before leaving the magnetic field traveling opposite the direction it initially entered the field. Which of the following quantities would change if the electron had entered the field with a speed 2v? (There may be more than one correct answer.)
A. The time the electron is in the magnetic field
B. The magnitude of the net force acting on the electron inside the field
C. The magnitude of the electron's acceleration inside the field
D. The radius of the circular path the electron travels
Physics
1 answer:
pogonyaev3 years ago
4 0

Answer:

C. The magnitude of the electron's acceleration inside the field

D. The radius of the circular path the electron travels

Explanation:

The radius of the electron's motion in a uniform magnetic field is given by

R = \frac{MV}{qB}

where;

m is the mass of the electron

q is the charge of the electron

B is the magnitude of the magnetic field

V is speed of the electron

R is the radius of the electron's

Thus, the radius of the of the electron's motion will change since it depends on speed of the electron.

The magnitude of the electron's acceleration inside the field  is given by;

a_c = \frac{V^2}{R}

where;

a_c is centripetal acceleration of electron

Thus, the magnitude of the electron's acceleration inside the field will change since it depends on the electron speed.

The time the electron is in the magnetic field is given by;

T = \frac{2\pi M}{qB}

The time of electron motion will not change

The magnitude of the net force acting on the electron inside the field will not change;

qVB = \frac{MV^2}{R} \\\\qVB - \frac{MV^2}{R}  = 0

Therefore, the correct options are "C" and "D"

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The force of gravity increases as the product of their individual masses' increases

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2 years ago
The density of gasoline is 730 kg/m3 at 0°C. Its average coefficient of volume expansion is 9.60 10-4(°C)−1. Assume 1.00 gal of
kipiarov [429]

Answer: 0.4911 kg

Explanation:

We have the following data:

\rho_{0\°C}= 730 kg/m^{3} is the density of gasoline at 0\°C

\beta=9.60(10)^{-4} \°C^{-1} is the average coefficient of volume expansion

We need to find the extra kilograms of gasoline.

So, firstly we need to transform the volume of gasoline from gallons to m^{3}:

V=8.50 gal \frac{0.00380 m^{3}}{1 gal}=0.0323 m^{3} (1)

Knowing density is given by: \rho=\frac{m}{V}, we can find the mass m_{1} of 8.50 gallons:

m_{1}=\rho_{0\°C}V

m_{1}=(730 kg/m^{3})(0.0323 m^{3})=23.579 kg (2)

Now, we have to calculate the factor f by which the volume of gasoline is increased with the temperature, which is given by:

f=(1+\beta(T_{f}-T_{o})) (3)

Where T_{o}=0\°C is the initial temperature and T_{f}=21.7\°C is the final temperature.

f=(1+9.60(10)^{-4} \°C^{-1}(21.7\°C-0\°C)) (4)

f=1.020832 (5)

With this, we can calculate the density of gasoline at 21.7\°C:

\rho_{21.7\°C}=730 kg/m^{3} f=(730 kg/m^{3})(1.020832)

\rho_{21.7\°C}=745.207 kg/m^{3} (6)

Now we can calculate the mass of gasoline at this temperature:

m_{2}=\rho_{21.7\°C}V (7)

m_{2}=(745.207 kg/m^{3})(0.0323 m^{3}) (8)

m_{2}=24.070 kg (9)

And finally calculate the mass difference \Delta m:

\Delta m=m_{2}-m_{1}=24.070 kg-23.579 kg (10)

\Delta m=0.4911 kg (11) This is the extra mass of gasoline

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if the velocity of a car is halved the fc required to keeo it in a path of constant radius is multiplied/divided by?
damaskus [11]
I believe this is it
The centripetal force is given by
F = mv^2 / r
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So the centripetal force is divided by 4.
7 0
3 years ago
Read 2 more answers
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