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makkiz [27]
3 years ago
5

At a location near the equator, the earth’s magnetic field is horizontal and points north. An electron is moving vertically upwa

rd from the ground. What is the direction of the magnetic force that acts on the electron?1.North
2.West
3.The magnetic force is zero.
4.East
5.South
Physics
1 answer:
Elan Coil [88]3 years ago
5 0

Answer:

2. west

Explanation:

Given an electron is moving vertically upward from ground.

Now Fleming right hand rule state that: make L shape with thumb and index finger then point middle finger perpendicular to index and thumb.Then index finger points in the direction of moving charge , middle finger points in the direction of the magnetic field and thumb points in the direction of the magnetic force.

According to Fleming right hand rule the direction of the magnetic that acts on the electron is west.

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Answer:

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Betty (mass 40 kg), standing on slippery ice, catches her leaping dog (mass 15 kg) moving horizontally at 3.0 m/s. Show that the
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Answer:

v = 2.18m/s

Explanation:

In order to calculate the speed of Betty and her dog you take into account the law of momentum conservation. The total momentum before Betty catches her dog must be equal to the total momentum after.

Then you have:

Mv_{1o}+mv_{2o}=(M+m)v        (1)

M: mass Betty = 40kg

m: mass of the dog = 15kg

v1o: initial speed of Betty = 3.0m/s

v2o: initial speed of the dog = 0 m/s

v: speed of both Betty and her dog = ?

You solve the equation (1) for v:

v=\frac{Mv_{1o}+mv_{2o}}{M+m}=\frac{(40kg)(3.0m/s)+(15kg)(0m/s)}{40kg+15kg}\\\\v=2.18m/s

The speed fo both Betty and her dog is 2.18m/s

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2 years ago
HELP!!! Why is plasma considered the most common state of matter in the universe?
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A wad of clay of mass m1 = 0.49 kg with an initial horizontal velocity v1 = 1.89 m/s hits and adheres to the massless rigid bar
notka56 [123]

Answer:

<h2>The angular velocity just after collision is given as</h2><h2>\omega = 0.23 rad/s</h2><h2>At the time of collision the hinge point will exert net external force on it so linear momentum is not conserved</h2>

Explanation:

As per given figure we know that there is no external torque about hinge point on the system of given mass

So here we will have

L_i = L_f

now we can say

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so we will have

0.49(1.89)(0.45) = (2.13(0.90)^2 + 0.49(0.45)^2)\omega

\omega = 0.23 rad/s

Linear momentum of the system is not conserved because at the time of collision the hinge point will exert net external force on the system of mass

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