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mariarad [96]
3 years ago
5

The area vector of a square loop of 5 turns of a conductor each with side length of 0.2 m carrying a current of 2 A is antiparal

lel to a uniform magnetic field of 50.0 T in which it lies.
Physics
1 answer:
lord [1]3 years ago
3 0

Answer:

the magnitude and the direction of the total magnetic field is 0.4 Am²  antiparallel to the area vector

Explanation:

Given that:

The area vector of a square loop  has 5 numbers of turns i.e n = 5

each with side length = 0.2 m

Current I = 2 A

uniform magnetic field =  50.0 T

Now; the magnitude of the total magnetic field B is calculated as :

B = IA

where;

I = current

A = area ( n × l²)  

B = I ( n × l²)

B = 2 ×  5 ×  0.2²

B = 0.4 Am²

The direction of the magnetic moment is antiparallel to the area vector;

Hence ; the magnitude and the direction of the total magnetic field is 0.4 Am²  antiparallel to the area vector

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sammy [17]

To solve this problem we will apply the concepts related to electric potential and electric potential energy. By definition we know that the electric potential is determined under the function:

V = \frac{k_e q}{r}

k_e = Coulomb's constant

q = Charge

r = Radius

At the same time

U = \frac{k_e q_1q_2}{r}

The values of variables are the same, then if we replace in a single equation we have this expression,

U  = Vq

If we replace the values, we have finally that the charge is,

V = 800V

q = 1\mu C

U = (800V)(1*10^{-6}C)

U = 8*10^{-4}J

Therefore the potential energy of the system is 8*10^{-4} J

7 0
3 years ago
A 200. kg object is pushed 12.0 m to the top of an incline to a height of 6.0 m. If the force applied along the incline is 3000.
Nataliya [291]

Answer:

Approximately 1.2 \times 10^{4}\; {\rm J} (assuming that g = 9.81\; {\rm N \cdot kg^{-1}}.)

Explanation:

The strength of the gravitational field near the surface of the earth is approximately constant: g = 9.81\; {\rm N \cdot kg^{-1}}.

The change in the gravitational potential energy ({\rm GPE}) of an object near the surface of the earth is proportional to the change in the height of this object. If the height of an object of mass m increased by \Delta h, the {\rm GPE} of that object would have increased by m\, g\, \Delta h.

In this question, the height of this object increased by \Delta h = 6.0\; {\rm m}. The mass of this object is m = 200\; {\rm kg}. Thus, the {\rm GPE} of this object would have increased by:

\begin{aligned}& (\text{Change in GPE}) \\ =\; & m\, g\, \Delta h \\ =\; & 200\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times 6.0\; {\rm m} \\ \approx\; & 1.2 \times 10^{4}\; {\rm J}\end{aligned}.

(Note that 1\; {\rm N \cdot m} = 1\; {\rm J}.)

3 0
2 years ago
according to a rule-of-thumb, every 5 seconds, between a lightning flash and the following thunder gives the distance to the fla
jenyasd209 [6]
Given the speed of the sound in the problem which is 1 mile per 5 seconds. 

The speed is calculated by:

Speed = distance/time = (1mi/5s) (1610 m/1mi) = 300 m/s

Note that only 1 significant figure is given which is 5 second and so only 1 significant figure is justified in the result. The speed of sound is 343 m/s. therefore the rule of thumb is fairly close.
3 0
3 years ago
At a rock concert, a dB meter registered 131 dB when placed 2.6 m in front of a loudspeaker on the stage. The intensity of the r
Pavlova-9 [17]

Answer:

Explanation:

A) 131 dB = 10*log(I / 1e-12W/m²)

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13.1 = log(I / 1e-12W/m²

1.25e13= I / 1e-12W/m²

I = 1.25 x10^1W/m²

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B) 86 dB = 10*log(I / 1e-12W/m²)

8.6 = log(I / 1e-12W/m²)

3.98e8 = I / 1e-12W/m²

I = 3.98e-4 W/m²

area A = P / I = 1061W / 3.98e-4W/m² = 2.66e6 m²

A = 4πr²

2.66e6 m² = 4πr²

r = 14.5m ◄

6 0
3 years ago
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Advocard [28]

Explanation:

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Sun's own motion is not that evident to us because of which the impact of that motion is nullified. But, motion of planet is clearly visible and it also shows in their annual round around the Sun. Here, Mars also shows retrograde motion which means it will show a back and forth motion in the sky. Because of which Mars might be visible in the same zodiac for a longer Duration as compared to the Sun.

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