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pochemuha
1 year ago
12

A proton is projected into a magnetic field that is directed along the positive x axis. Find the direction of the magnetic force

exerted on the proton for each of the following directions of the proton's velocity: (b) the negative y direction
Physics
1 answer:
bulgar [2K]1 year ago
4 0

Te direction of the magnetic force for the velocity of the proton in the

-ve  y direction will be +ve z direction.

As we know that the right-hand rule is based on the relation of magnetic fields and the forces that they exert on moving charges.When a charged particle moves under a magnetic field, it exerts a force on the particle, which is not in the same direction but different than the direction of the magnetic field.Under the right-hand rule,  if we point our pointer finger in the direction of the charged particle is moving and the middle finger is representing the direction of the magnetic field then our thumb depicts the direction of the magnetic force which is exerted on the charged particle.

So,  we are given that the direction of the velocity of the proton is in the negative y direction and the direction of the magnetic field is in the positive x  direction,  so the magnetic force is acting in the positive z direction.

To know  more about the right-hand rule refer to the link brainly.com/question/9750730?referrer=searchResults.

#SPJ4

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Answer:

The coefficient of static friction between the object and the disk is 0.087.

Explanation:

According to the statement, the object on the disk experiments a centrifugal force due to static friction. From 2nd Newton's Law, we can represent the object by the following formula:

\Sigma F_{r} = \mu_{s}\cdot N = m\cdot \frac{v^{2}}{R} (1)

\Sigma F_{y} = N - m\cdot g = 0 (2)

Where:

N - Normal force from the ground on the object, measured in newtons.

m - Mass of the object, measured in newtons.

g - Gravitational acceleration, measured in meters per square second.

v - Linear speed of rotation of the disk, measured in meters per second.

R - Distance of the object from the center of the disk, measured in meters.

By applying (2) on (1), we obtain the following formula:

\mu_{s}\cdot m\cdot g = m\cdot \frac{v^{2}}{R}

\mu_{s} = \frac{v^{2}}{g\cdot R}

If we know that v = 0.8\,\frac{m}{s}, g = 9.807\,\frac{m}{s^{2}} and R = 0.75\,m, then the coefficient of static friction between the object and the disk is:

\mu_{s} = \frac{\left(0.8\,\frac{m}{s} \right)^{2}}{\left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.75\,m)}

\mu_{s} = 0.087

The coefficient of static friction between the object and the disk is 0.087.

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Answer:

(a) 1, average velocity = -65.6 m/s

   2, average velocity = -64.8 m/s

   3, average velocity = -64.16 m/s

(b) The instantaneous velocity is -96 m/s

Explanation:

(a)

Average velocity is given  by;

y(t_2,t_1) = \frac{y(t_2) - y(t_1)}{t_2-t_1}

(1)

y(2.1,2) = \frac{(235-16*2.1^2) - (235-16*2^2)}{2.1-2}\\\\ y(2.1,2) = -65.6 \ m/s

(2)

y(2.05,2) = \frac{(235-16*2.05^2) - (235-16*2^2)}{2.05-2}\\\\ y(2.05,2) = -64.8 \ m/s

(3)

y(2.01,2) = \frac{(235-16*2.01^2) - (235-16*2^2)}{2.01-2}\\\\ y(2.01,2) = -64.16 \ m/s

b. y = 235 - 16t²

The instantaneous velocity is given by;

v = dy /dt

dy / dt = -32t

when t = 3 s

v = -32(3)

v = -96 m/s

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3 years ago
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