ANSWER IN LESS THAN A MINUTE!! EASYY!

g The magnetic force on a charged particle A. depends on the sign of the charge on the particle. B. depends on the velocity of t

Leona [35]

Answer:

E) is described by all of these

Explanation:

The magnetic force on a charged particle is expressed as:

F = qv * B = qvBsinθ

Where,

q = charge on particle

θ = angle between the magnetic field and the particle velocity.

v = velocity of the particle

B = magnitude of field vector

From here, we could denote that magnetic force, F depends on charge on particle, velocity of particle, magnitude of field vector.

The magnetic force on a charged particle is at right angles to both the velocity of the particle. The magnetic force and magnetic field in a charged particle are perpendicular to each other, the right hand rule is used to determine the direction of force.

The correct option is E.

3 0

2 years ago

Read 2 more answers

Plane polarized light with intensity I0 is incident on a polarizer. What angle should the principle axis make with respenct to t

Nataliya [291]

Answer:

Q = 47.06 degrees

Explanation:

Given:

- The transmitted intensity I = 0.464 I_o

- Incident Intensity I = I_o

Find:

What angle should the principle axis make with respect to the incident polarization

Solution:

- The relation of transmitted Intensity I to to the incident intensity I_o on a plane paper with its principle axis is given by:

I = I_o * cos^2 (Q)

- Where Q is the angle between the Incident polarized Light and its angle with the principle axis. Hence, Using the relation given above:

Q = cos ^-1 (sqrt (I / I_o))

- Plug the values in:

Q = cos^-1 ( sqrt (0.464))

Q = cos^-1 (0.6811754546)

Q = 47.06 degrees

8 0

3 years ago

The flywheel of an engine has moment of inertia 2.50 kg m2 about its rotation axis. What constant torque is required to bring it

MrRissso [65]

Answer:

Explanation:

From the question we are told that

The moment of inertia is $I = 2.50 \ kg \cdot m^2$

The final angular speed is $w_f = 400 rev/min = \frac{400 * 2\pi}{60} = 41.89 \ rad/s$

The time taken is $t = 8.0 s$

The initial angular speed is $w_i = 0\ rad/s$

Generally the average angular acceleration is mathematically represented as

$\alpha = \frac{w_f - w_i }{t}$

=> $\alpha = \frac{41.89}{8}$

=> $\alpha = 5.24 \ rad/s^2$

Generally the torque is mathematically represented as

$\tau = I * \alpha$

=> $\tau = 5.24 * 2.50$

=> $\tau = 13.09 \ N \cdot m$

5 0

3 years ago

work done as mass 1 moves to mass 2. the gravitational force between two point masses separated by a distance r is proportional

pav-90 [236]

Answer:

gravitational force

3 0

2 years ago

A 1.0-cm-tall object is 13 cm in front of a converging lens that has a 40 cm focal length.

kicyunya [14]

A) Image position: -19.3 cm

B) Image height: 1.5 cm, upright

Explanation:

A)

In order to calculate the image position, we can use the lens equation:

$\frac{1}{p}+\frac{1}{q}=\frac{1}{f}$

where

p is the distance of the object from the lens

q is the distance of the image from the lens

f is the focal length

In this problem, we have:

p = 13 cm (object distance)

f = 40 cm (focal length, positive for a converging lens)

So the image distance is

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{40}-\frac{1}{13}=-0.0519\\q=\frac{1}{-0.0519}=-19.3 cm$

The negative sign means that the image is virtual.

B)

In order to calculate the image height, we use the magnification equation:

$\frac{y'}{y}=-\frac{q}{p}$

where

y' is the image height

y is the object height

In this problem, we have:

y = 1.0 cm (object height)

p = 13 cm

q = -19.3 cm

Therefore, the image heigth is

$y'=-\frac{qy}{p}=-\frac{(-19.3)(1.0)}{13}=1.5 cm$

And the positive sign means the image is upright.

6 0

3 years ago

ANSWER IN LESS THAN A MINUTE!! EASYY!​

ANSWER IN LESS THAN A MINUTE!! EASYY!