3 Below, someone is trying to balance a plank with

Physics

1 answer:

belka [17]3 years ago

8 0

Answer:

a. The moment of the 4 N force is 16 N·m clockwise

b. The moment of the 6 N force is 12 N·m anticlockwise

Explanation:

In the figure, we have;

The distance from the point 'O', to the 6 N force = 2 m

The position of the 6 N force relative to the point 'O' = To the left of 'O'

The distance from the point 'O', to the 4 N force = 4 m

The position of the 4 N force relative to the point 'O' = To the right of 'O'

a. The moment of a force about a point, M = The force, F × The perpendicular distance of the force from the point

a. The moment of the 4 N force = 4 N × 4 m = 16 N·m clockwise

b. The moment of the 6 N force = 6 N × 2 m = 12 N·m anticlockwise.

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schepotkina [342]

The answer is oil and water. If you want I could give you the reason too

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3 years ago

Read 2 more answers

A 6.72 particle moves through a region of space where an electric field of magnitude 1270 N/C points in the positive direction,

Romashka [77]

Answer:

$v_{y} = -104 m/s$

Explanation:

Using:

Force = electric field * charge

$F=e*q$

Force = magnitude of charge * velocity * magnetic field * sin tither

$F_{x2}= |q|*v*B*sin \alpha$

Force on particle due to electric field:

$F_{x1}= E*q = (1270N/C)*(-6.72*10^{-6} ) = -8.53*10^{-3}$

Force on particle due to magnetic field:

$F_{x2}= |q|*v*B*sin \alpha = (6.72*10^{-6} )*(1.15)*(sin90)*v = (7.728*10^{-6})*(v)$

$F_{x2}$ is in the positive x direction as $F_{x1}$ is in the negative x direction while net force is in the positive x direction.

Magnetic field is in the positive Z direction, net force is in the positive x direction.

According to right hand rule, Force acting on particle is perpendicular to the direction of magnetic field and velocity of particle. This would mean the force is along the y-axis. As this is a negatively charged particle, the direction of the velocity of the particle is reversed. Therefore velocity of particle, v, has to be in the negative y direction.

Now,

$F_{xnet}- F_{x1 } = F_{x2 }$