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Alexxandr [17]
2 years ago
7

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. Question 5. The magnitude and direction of the total magnetic moment is ……to the area vector.
Physics
1 answer:
Anon25 [30]2 years ago
3 0

Answer:

M= 0.4 Am^2

Explanation:

From the question we are told that:

Number of turns  N=5N=5

Conductor each with side length L=0.2m

Current I=2A

Magnetic field  B=50.0T

Generally the equation for the total magnetic moment  M is mathematically given by

M = current * area

M= I * A

M = 2 * (5* 0.2*0.2)  

M = 2 * 0.2

M= 0.4 Am^2

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the answer is D) The air in the aquarium is saturated

none of the other answers were realistic AND I took this before so I know the right answer.

5 0
3 years ago
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If the final speed of an object as is strikes the ground is 77 m/s and it was in the air for 6.5 seconds. What was the initial d
azamat

Answer:

The answer to your question is: 13.2 m/s

Explanation:

final speed (fs) = 77 m/s

t = 6.5 s

gravity (g) = 9.81 m/s2

initial speed (is) = ?

Formula

fs = is + gt     from this equation we clear "is" = fs - gt

Substitution                         is = 77 - (9,81)(6.5)

Process                               is = 77 - 63.8

                                            is = 13.2 m/s

8 0
2 years ago
A vertical scale on a spring balance reads from 0 to 155 N . The scale has a length of 10.0 cm from the 0 to 155 N reading. A fi
Harrizon [31]

Answer:

mass of the fish is 8.11 kg

Explanation:

As we know that the frequency of oscillation of spring block system is given as

f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}

here we know that the reading of scale varies from 0 to 155 N from length varies from x = 0 to x = 10 cm

Now we have

k = \frac{155}{0.10} N/m

k = 1550 N/m

so now we have

2.20 = \frac{1}{2\pi}\sqrt{\frac{1550}{m}}

m = 8.11 kg

so mass of the fish is 8.11 kg

4 0
2 years ago
A wheel initially has an angular velocity of 18 rad/s, but it is slowing at a constant rate of 2 rad/s 2 . By the time it stops,
ZanzabumX [31]

Answer:

5) 13 revolutions (approximately)

Explanation:

We apply the equations of circular motion uniformly accelerated :

ωf²= ω₀² + 2α*θ Formula (1)

Where:

θ : angle that the body has rotated in a given time interval (rad)

α : angular acceleration (rad/s²)

ω₀ : initial angular speed ( rad/s)

ωf : final angular speed  ( rad/s)

Data:

ω₀ = 18 rad/s

ωf = 0

α = -2 rad/s²  ; (-) indicates that the wheel is slowing

Revolutions calculation that turns the wheel until it stops

We apply the formula (1)

ωf²= ω₀² + 2α*θ

0 = (18)² + 2( -2)*θ

4*θ =  (18)²

θ =  (18)²/4 = 81 rad

1 revolution = 2π rad

θ = 81 rad * 1 revolution / 2πrad

θ =  13 revolutions approximately

8 0
3 years ago
Which is the correct equation for the elastic potential energy stored in a spring
salantis [7]

(B)

Explanation:

pe =  \frac{1}{2} k {x}^{2}

3 0
2 years ago
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