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prisoha [69]
3 years ago
14

A negative charge is moving across a rook from south to north. A magnetic field runs from East to west. In what direction is the

magnetic force on the moving negative charge?
A. Toward the East
B. Toward the west
C. Toward the floor
D. Toward the ceiling
Physics
1 answer:
Zinaida [17]3 years ago
5 0

Answer:

The magnetic force on the moving negative charge is towards the ceiling.

Explanation:

The magnetic force acting on a wire is given by the relation as follows :

F=q(v\times B)

q is charged particle

v is velocity of charged particle

B is magnetic field

In this case, a negative charge is moving across a rook from south to north. A magnetic field runs from East to west.

q = -q

v =+j

B = -i

So,

F=-q(vj\times B(-i))

Since, j\times i=-k

So,

F=q(v\times B)(-k)

So, the magnetic force on the moving negative charge is towards the ceiling.

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Perfectly inelastic collision is type of collision during which two objects collide, stay connected and momentum is conserved. Formula used for conservation of momentum is:

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In case of perfectly inelastic collision v'1 and v'2 are same.

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Now we insert given information and solve for m₂:
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3 years ago
1. A student lifts a box of books that weighs 185 N. The box is
aksik [14]

1)  148 J

When lifting an object, the work done on the object is equal to its change in gravitational potential energy. Mathematically:

W = \Delta U = (mg) \Delta h

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Substituting into the equation, we find:

W=(185)(0.800)=148 J

2) (a) 28875 J

The work done by a force applied parallel to the direction of motion of the object is given by

W=Fd

where

F is the magnitude of the force

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In this problem,

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Substituting,

W=(825)(35)=28875 J

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W=Fd

We see that the work done is proportional to the magnitude of the force: therefore, if the force is doubled, then the work done is also doubled.

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Now the force is doubled, so the new work done will be

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In this case, the force acting on the ball is the force of gravity, whose magnitude is:

F = mg

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g=9.8 m/s^2 is the acceleration of gravity

Solving the equation,

F=(0.180)(9.8)=1.76 N

Now we find the work done by gravity using the same formula applied before:

W=Fd

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W = 7.0 kJ = 7000 J

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\Delta h = 1.2 m

As we stated in part a), the work done on the box is equal to its change in gravitational potential energy:

W=mg \Delta h

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This means that the work done on the box depends only on its initial and final position, not on the path taken. The two men carry the box along different paths, however the reach at the end the same position, and they started from the same position: this means that the value of \Delta h is the same for both of them, so the work they have done is exactly the same.

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