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sergiy2304 [10]
2 years ago
8

If the magnetic flux through a loop increases according to the relation; where , is in milliweber (mWb) and t is in seconds. Cal

culate the magnitude of e.m.f induced in the loop when t = 2s​

Physics
1 answer:
Alekssandra [29.7K]2 years ago
6 0

The magnitude of e.m.f induced in the loop when t = 2 s is 31 Volts.

<h3>emf induced in the loop</h3>

The magnitude of e.m.f induced in the loop is calculated as follows;

emf = dФ/dt

Ф = 6t² + 7t

dФ/dt = 12t + 7

at t = 2 seconds

emf = dФ/dt = 12(2) + 7 = 31 V

Thus,  the magnitude of e.m.f induced in the loop when t = 2 s is 31 Volts.

Learn more about emf here: brainly.com/question/24158806

#SPJ1

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A car is initially driving at 30 m/s. It hits a large pothole, after which it is traveling in the same direction but at 25 m/s.
Bogdan [553]

Answer:

The time of Mars is 1.65 times larger on Mars than on Earth

Explanation:

The equation that describes the system is the final speed is equal to the speed minos the speed lost by the collision with the porhole

       Vf = Vo - V pothole

B) let's transform the weight of free groin system and N international system

      1 N = 0.2248 lb

      2.8 lbs (1N / 0.2248lbs) = 12.5 N

c) Kinematic equations are the same in all inertial systems, Mars and Earth, so we can use the height equation, with zero initial velocity

                   

        Y = Vo t - ½ g t²

        Y = - ½ g t²

        t = √ 2Y / g

     

Mars

         gm = 0.37g

         gm = 0.37 9.8

         gm = 3,626 m / s²

         t = √( 2 1.9 / 3.626 )

         t = 1.02 s

Earth

         t = √( 2 1.9 / 9.8)

         t = 0.62 s

To make the comparison of time we are the relationship between the two

         tm / te = 1.02 / 0.62

         tm / te = 1.65

The time of Mars is 1.65 times larger on Mars than on Earth

7 0
3 years ago
Read 2 more answers
Violet light has a frequency of 7.26 × 1014 Hz and travels at a speed of 3.00 × 108 m/s.
Anuta_ua [19.1K]

Answer:

what's the question

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8 0
2 years ago
Please help! Thanks so much! :)
notsponge [240]

Answer: B

Explanation: I'm not 100% sure tho sorry if i'm wrong

7 0
3 years ago
Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.8×106N , one an angle 11degrees west of north an
sp2606 [1]

Answer: 2,9 ×10¹⁰J.

Explanation:

1. The <em><u>work </u></em>is a measure of energy and is calculated as the product of the displacement times the parallel force to such displacement.

That means that the only components of the force that contribute to work are those that result parallel to the displacement.

2. Since it is given that the <em>two tugboats "pull the tanker a distance 0.83km toward the north"</em>, that is the displacement, and you have to calculate the net force toward the north.

3. <u>Tugboat #1</u>.

a) Force magnitude: F₁ = 1.8×10⁶N

b) Angle: α = 11° West of North

c) North component of the force F₁: Fy₁ = F₁cos(α) =  1.8×10⁶N  × cos(11°) = 1.77×10⁶N

4. <u>Tugboat #2</u>:

a) Force magnitude: F₂ = 1.8×10⁶N

b) Angle:  = 11° East of North

c) North component of the force F₂: Fy₂ = F₂cos(β) =  1.8×10⁶N  × cos(11°) = 1.77×10⁶N =

5. <u>Total net force, Fn</u>:

Fn = Fy₁ + Fy₂ = 1.77×10⁶N + 1.77×10⁶N = 3.54×10⁶N

6. <u>Work, W</u>:

Displacement, d = 0.83 km = 8,300 m

W = Fn×d = 3.54×10⁶N×8,300m = 29,000 ×10⁶J = 2,9 ×10¹⁰J

The answer is rounded to two significant figures because both data, Force and displacement, have two significant figures.

7 0
3 years ago
A 4 kilogram box is at rest on a frictionless floor. A net force of 12 newtons acts on it. What is the acceleration of the box?
Rashid [163]

Answer:

<h2>3 m/s^2</h2>

Explanation:

Step one:

given

Mass m= 4kg

Force F= 12N

Required

Acceleration the relation between force, acceleration, and mass is Newton's first equation of motion, which says a body will continue to be at rest or uniform motion unless acted upon by an external force

F=ma

a=F/m

a=12/4

a=3 m/s^2

4 0
3 years ago
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