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wariber [46]
3 years ago
7

An individual electron can be thought of as a spinning sphere of negative charge. A charged spinning sphere will generate a magn

etic field, whose direction is indicated by the magnetic moment of the object, vector mu. In what direction will the electron rotate, based on the direction of its magnetic moment and the direction of the uniform magnetic field that it is immersed in?
Physics
1 answer:
Zinaida [17]3 years ago
8 0

Answer:

based on the direction of its magnetic moment and the direction of the uniform magnetic field that it is immersed in, the electron will rotate in a clockwise direction

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i like honey but when u put it on your feet it sticks to the ground. how do i avoid the stickyness of honey when i want to clean
Gre4nikov [31]

Answer:

please stop asking questions

Explanation:

6 0
3 years ago
Read 2 more answers
65. The weight of a body when totally immersed in a liquid is 4.2N if he weight of the liquid displaced is 2.5N. Find the weight
Anna35 [415]

Answer:

Given, Apparent weight(W₂)=4.2N

          Weight of liquid displaced (u)=2.5N

          Let weight of body in air = W₁

Solution,

             U=W₁-W₂

              W₁=4.2=2.5=6.7N

∴Weight of body in air is 6.7N

5 0
2 years ago
Lloyd is standing on a scaffolding 12 meters above the ground to clean the windows of a tall building. His bucket, which has a m
Sever21 [200]

Answer:

U₂ = 20 J

KE₂ = 40 J

v= 12.64 m/s

Explanation:  

Given that

H= 12 m

m = 0.5 kg

h= 4 m

The potential energy at position 1

U₁ = m g H

U₁ = 0.5 x 10 x 12        ( take g= 10 m/s²)

U₁ = 60 J

The potential energy at position 2

U₂ = m g h

U ₂= 0.5 x 10 x 4        ( take g= 10 m/s²)

U₂ = 20 J

The kinetic energy at position 1

KE= 0

The kinetic energy at position 2

KE= 1/2 m V²

From energy conservation

U₁+KE₁=U₂+KE₂

By putting the values

60 - 20 = KE₂

KE₂ = 40 J

lets take final velocity is v m/s

KE₂= 1/2 m v²

By putting the values

40 = 1/2 x 0.5 x v²

160 = v²

v= 12.64 m/s

3 0
2 years ago
Read 2 more answers
You are driving a 2400.0-kg car at a constant speed of 14.0 m/s along a wet, but straight, level road. As you approach an inters
olya-2409 [2.1K]

Answer:0.43

Explanation:

Given

mass of car m=2400 kg

Speed of car u=14 m/s

Distance traveled before coming to halt s=23.2 m

Let \muthe coefficient of friction

Maximum deceleration road can provide during motion is

a=\mu g

using v^2-u^2=2 as

0-14^2=2\cdot (-\mu g)\cdot 23.2

\mu =\frac{196}{454.72}

\mu =0.431

7 0
2 years ago
A merry-go-round with a a radius of R = 1.9 m and moment of inertia I = 209 kg-m2 is spinning with an initial angular speed of ω
RideAnS [48]

Answer:

340.67 kgm²/s

Explanation:

R = Radius of merry-go-round = 1.9 m

I = Moment of inertia = 209 kgm²

\omega_i = Initial angular velocity = 1.63 rad/s

m = Mass of person = 73 kg

v = Velocity = 4.8 m/s

Initial angular momentum is given by

L=I\omega_i\\\Rightarrow L=209\times 1.63\\\Rightarrow L=340.67\ kgm^2/s

The initial angular momentum of the merry-go-round is 340.67 kgm²/s

8 0
3 years ago
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