Generators convert mechanical energy to electrical energy by using ________

What is the frequency for a beam of electrons orbiting in a field of 4.62 x 10^-3 T? Let the mass of an electrons m = 9.31 x 10^

melisa1 [442]

Frequency: $1.27\cdot 10^8 Hz$

Explanation:

The force experienced by an electron in a magnetic field is

$F=qvB$

where

$q=1.6\cdot 10^{-19}C$ is the electron charge

v is the speed of the electron

B is the strength of the magnetic field

Since the force is perpendicular to the direction of motion of the electron, the force acts as centripetal force, so we can write:

$qvB=m\frac{v^2}{r}$

where

r is the radius of the orbit

$m=9.31\cdot 10^{-31}kg$ is the mass of the electron

Re-arranging the equation,

$\frac{v}{r}=\frac{qB}{m}$ (1)

We also know that in a circular motion, the speed is equal to the ratio between circumference of the orbit and orbital period (T):

$v=\frac{2\pi r}{T}$

Substituting into (1),

$\frac{2\pi}{T}=\frac{qB}{m}$

We also know that 1/T is equal to the frequency f, so

$f=\frac{qB}{2\pi m}$

In this problem,

$B=4.62\cdot 10^{-3}T$

Therefore, the frequency of the electrons is

$f=\frac{(1.6\cdot 10^{-19})(4.62\cdot 10^{-3})}{2\pi(9.31\cdot 10^{-31})}=1.27\cdot 10^8 Hz$

4 0

3 years ago

A 110 kg ice hockey player skates at 3.0 m/s toward a railing at the edge of the ice and then stops himself by grasping the rail

velikii [3]

Answer:

Δ KE = -495 J

Explanation:

given,

mass of the ice hockey player = 110 Kg

initial speed = 3 m/s

final speed = 0 m/s

distance, d = 0.3 m

change in kinetic energy

$\Delta K E = \dfrac{1}{2}mv_f^2 - \dfrac{1}{2}mv_i^2$

$\Delta K E = \dfrac{1}{2}mv(0)^2 - \dfrac{1}{2}\times 110\times 3^2$

Δ KE = -495 J

Hence, the change in kinetic energy is equal to Δ KE = -495 J

7 0

3 years ago

What conversion takes place in a motor?

LenaWriter [7]

C should be the right answer! hopefully this helps!

4 0

3 years ago

The cabinet is mounted on coasters and has a mass of 45 kg. The casters are locked to prevent the tires from rotating. The coeff

stira [4]

Answer:

the force P required for impending motion is 132.3 N

the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m

Explanation:

Given that:

mass of the cabinet m = 45 kg

coefficient of static friction μ = 0.30

A free flow body diagram illustrating what the question represents is attached in the file below;

The given condition from the question let us realize that ; the casters are locked to prevent the tires from rotating.

Thus; considering the forces along the vertical axis ; we have :

$\sum f_y =0$

The upward force and the downward force is :

$N_A+N_B = mg$

where;

$\mathbf { N_A \ and \ N_B}$ are the normal contact force at center point A and B respectively .

$N_A+N_B = 45*9.8$

$N_A+N_B = 441$ ------- equation (1)

Considering the forces on the horizontal axis:

$\sum f_x = 0$

$F_A +F_B = P$

where ;

$\mathbf{ F_A \ and \ F_B }$ are the static friction at center point A and B respectively.

which can be written also as:

$\mu_s N_A + \mu_s N_B = P$

$\mu_s( N_A + N_B) = P$

replacing our value from equation (1)

P = 0.30 ( 441)

P = 132.3 N

Thus; the force P required for impending motion is 132.3 N

b) Since the horizontal distance between the casters A and B is 480 mm; Then half the distance = 480 mm/2 = 240 mm = 0.24 cm

the largest value of "h" allowed for the cabinet is not to tip over is calculated by determining the limiting condition of the unbalanced torque whose effect is canceled by the normal reaction at $N_A$ and it is shifted to $N_B$ :

Then:

$\sum M _B = 0$

$P*h = mg*0.24$

$h =\frac{45*9.8*0.24}{132.3}$

h = 0.8 m

Thus; the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m

6 0

3 years ago

A 1,250 kg car is moving due to 6,500 N engine force. If the kinetic friction coefficient between the car and the road is 0.32,

Yuliya22 [10]

Answer:

b i hope this is correct answee

7 0

3 years ago

Generators convert mechanical energy to electrical energy by using _________.