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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 must a psychological researcher debrief human test subjects?
myrzilka [38]
<span>Psychological researchers must debrief human test subjects </span><span>at the end of every experiment.

The current code of ethics in p</span>sychological research states that researchers absolutely must debrief human test subjects at the end of every study regardless or whether or not harm or deception was involved.

Debriefing a subject after a study is an essential opportunity for the researcher to explain the purpose and aim of the study to the subject, make sure the subject is not harmed or mentally disturbed, clarify why deception was used (if deception was involved) and overall, to clarify any questions or doubts the subject might have.
3 0
3 years ago
Using the right amount of significant figures, calculate the answer to the following problem, 15.33879 + 15.555
MAXImum [283]

Answer:

there are 7 significant figures

Explanation:

15.33879+15.555

=30.89379

there are 7 significant figures

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7 0
4 years ago
A block with mass m 2.00 kg is placed against a spring on a frictionless incline with angle 30 degrees (Fig. B-43). (The block i
guajiro [1.7K]

Answer:

Explanation:

a )

The stored elastic energy of compressed spring

= 1 / 2 k X²

= .5 x 19.6 x (.20)²

= .392 J

b ) The stored potential energy will be converted into gravitational potential energy of the block earth system when the block will ascend along the incline . So change in the gravitational potential energy will be same as stored elastic potential energy of the spring that is .392 J .

c ) Let h be the distance along the incline which the block ascends.

vertical height attained ( H ) =h sin30

= .5 h

elastic potential energy = gravitational energy

.392 = mg H

.392 = 2 x 9.8 x .5 h

h = .04 m

4 cm .

=

7 0
3 years ago
A 15 kilogram mass is traveling with a velocity of 2.7m/s. What is the object's kinetic energy?
Usimov [2.4K]

Answer:

KE=1/2*m*v^2

KE=1/2*15*2.7^2

KE=54.675 J

Explanation:

4 0
3 years ago
A proton is projected in the positive x direction into a region of a uniform electric field E S 5 126.00 3 105 2 i^ N/C at t 5 0
Dafna11 [192]

Answer:

(a). The magnitude of the acceleration of the proton is 5.74\times10^{13}\ m/s^2

(b). The initial peed of the protion is 2.83\times10^{6}\ m/s

(c). The time is 0.493\times10^{-7}\ sec.

Explanation:

Given that,

Electric field E=-6.00\times10^{5}i\ N/C

Time = 5.0 sec

Distance 7.00 cm

(a). We need to calculate the acceleration

Using formula of force

F=F_{e}

ma=qE

a=\dfrac{Eq}{m}

Where, E = electric field

m = mass of proton

q = charge of proton

Put the value into the formula

a=\dfrac{-6.00\times10^{5}\times1.6\times10^{-19}}{1.67\times10^{-27}}

a=-5.74\times10^{13}\ m/s62

The magnitude of the acceleration of the proton is 5.74\times10^{13}\ m/s^2

(b). We need to calculate the initial peed

Using equation of motion

v^2-u^2=2as

Where, s = distance

Put the value into the formula

0-u^2=2\times(-5.74\times10^{13})\times7.00\times10^{-2}

u=\sqrt{2\times5.74\times10^{13}\times7.00\times10^{-2}}

u=2834783.94

u=2.83\times10^{6}\ m/s

The initial peed of the protion is 2.83\times10^{6}\ m/s

(c). We need to calculate the time interval over which the proton comes to rest

Using formula

t=\dfrac{u}{a}

Where, u = initial velocity

a = acceleration

Put the value into the formula

t=\dfrac{2.83\times10^{6}}{5.74\times10^{13}}

t=0.493\times10^{-7}\ sec

Hence, (a). The magnitude of the acceleration of the proton is 5.74\times10^{13}\ m/s^2

(b). The initial peed of the protion is 2.83\times10^{6}\ m/s

(c). The time is 0.493\times10^{-7}\ sec.

5 0
3 years ago
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