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Morgarella [4.7K]
3 years ago
5

A circular loop of wire with a radius of 20 cm lies in the xy plane and carries a current of 2 A, counterclockwise when viewed f

rom a point on the positive z axis. Its magnetic dipole moment is
Physics
1 answer:
Zina [86]3 years ago
6 0

To solve this problem we will apply the concept related to the magnetic dipole moment that is defined as the product between the current and the object area. In our case we have the radius so we will get the area, which would be

A=\pi r^2

A =\pi (0.2)^2

A =0.1256 m^2

Once the area is obtained, it is possible to calculate the magnetic dipole moment considering the previously given definition:

\mu=IA

\mu=2(0.1256)

\mu=0.25 A\cdot m^2

Therefore the magnetic dipole moment is 0.25A\cdot m^2

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When a 4.20-kg object is hung vertically on a certain light spring that obeys Hooke's law, the spring stretches 2.00 cm. (a) If
mr_godi [17]

(a) 0.714 cm

First of all, we need to find the spring constant of the spring. This can be done by using Hooke's law:

F=kx

where

F is the force applied on the spring

k is the spring constant

x is the stretching of the spring

At the beginning, the force applied is the weight of the block of m = 4.20 kg hanging on the spring, therefore:

F=mg=(4.20)(9.8)=41.2 N

The stretching of the spring due to this force is

x = 2.00 cm = 0.02 m

Therefore, the spring constant is

k=\frac{F}{x}=\frac{41.2}{0.02}=2060 N/m

Now, a new object of 1.50 kg is hanging on the spring instead of the previous one. So, the weight of this object is

F=mg=(1.50)(9.8)=14.7 N

And so, the stretching of th spring in this case is

x=\frac{F}{k}=\frac{14.7}{2060}=0.00714 m = 0.714 cm

(b) 1.65 J

The work done on a spring is given by:

W=\frac{1}{2}kx^2

where

k is the spring constant

x is the stretching of the spring

In this situation,

k = 2060 N/m

x = 4.00 cm = 0.04 m is the stretching due to the external agent

So, the work done is

W=\frac{1}{2}(2060)(0.04)^2=1.65 J

8 0
4 years ago
What is the mass of a neutron?
Anastaziya [24]

Answer:

1

Explanation:

8 0
3 years ago
Read 2 more answers
1. if an egg has a mass of .1k and its dropped from a height of 5 meters, calculate its potential energy at the top
Nataly_w [17]

Answer: (a) 5Joules

(b) 10m/s

(c) 1.0kgm/s

(d) 33.3N

Explanation:

(a) P E = mass * gravity * height

= 0.1*10*5 = 5Joules

(b) All PE is converted to KE

Therefore;

KE = 1/2 * mass * velocity^2

5 = 1/2 * 0.1 * v^2

V = sqrt(100)

v = 10m/s

(c) change in momentum = mass * velocity

= 0.1 * 10

= 1.0kgm/s

(d) impulse = change in momentum

Change in momentum = force * time

1.0 = force * 0.03

Force = 1.0/0.03

Force = 33.3N

Therefore, the force acting on the egg is 33.3N

8 0
3 years ago
25 points! Please someone help me, this seems like a pretty easy question but I can’t seem to get it right, please list the give
elixir [45]

Answer:

1 my brother say that

Explanation:

i know my brother said it

4 0
3 years ago
In the effects of kinetic energy, the slower an object goes, the longer that it will take to bring that object to a stop.
horsena [70]
False                                 
the faster an object goes the longer it'll take to stop the object. 
5 0
3 years ago
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