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fredd [130]
4 years ago
11

he magnetic flux through a loop of wire decreases from 1.7 Wb to 0.3 Wb in a time of 0.4 s. What was the average value of the in

duced emf, in units of volts?
Physics
1 answer:
Pani-rosa [81]4 years ago
5 0

Answer:

Induced emf through a loop of wire is 3.5 V.

Explanation:

It is given that,

Initial magnetic flux, \phi_i=1.7\ Wb

Final magnetic flux, \phi_f=0.3\ Wb

The magnetic flux through a loop of wire decreases in a time of 0.4 s, t = 0.4 s

We need to find the average value of the induced emf. It is equivalent to the rate of change of magnetic flux i.e.

\epsilon=-\dfrac{\phi_f-\phi_i}{t}

\epsilon=-\dfrac{0.3\ Wb-1.7\ Wb}{0.4\ s}

\epsilon=3.5\ V

So, the value of the induced emf through a loop of wire is 3.5 V.

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What does the octect rule state that explains why atoms bond ?
Simora [160]

Answer:

Explanation:

Atoms naturally bond with each other in an attempt to have 8 or zero valence electrons. This occurs in salts, covalent bonds, and mettalic bonds. This is how the octet rule demonstrates why atoms bond.

Hope this helps!

8 0
3 years ago
A race card drives one lap around a race track that is 500 meters in length. How is this displacement different from the distanc
gulaghasi [49]

Answer:

Distance is 500 m, displacement is 0

Explanation:

Distance and displacement are defined in two different ways:

- Distance is the total length of the path covered by an object in motion - so it depends on the path taken. In this problem, the distance travelled by the car corresponds to the length of one lap, which is the length of the track, so 500 m

- Displacement is the distance in a straight line between the final point and the initial point of the motion. This means that displacement does not depend on the path taken, but only on the starting and ending point of the motion. In this problem, the car completes one lap, so the final position of the car is equal to its starting position - therefore the displacement is zero, since the distance between these two points is zero.

5 0
3 years ago
A long distance runner running a 5.0km track is pacing himself by running 4.5km at 9.0km/h and the rest at 12.5km/h. What is his
sweet [91]

Answer:

9.26 km/h

Explanation:

Applying,

V' = D'/t'............... Equation 1

Where V' = Average speed, D' = Total distance, t' = total time.

Given: D' = 5 km

But,

v = d/t............ Equation 2

Where v = speed , d = distance, t = time

t = d/v............ Equation 3

Given: d = 4.5 km, v = 9 km/h, and d = 0.5 km, v = 12.5 km/h

Therefore,

t₁ = 4.5/9 = 0.5 hours

t₂ = 0.5/12.5

t₂ = 0.04 hours

Therefore,

V' = 5/(0.5+0.04)

V' = 5/0.54

V' = 9.26 km/h

6 0
3 years ago
The self-referencing effect refers to ________.
kykrilka [37]
The self-reference effect is the tendency an individual to have better memory for information that relates to oneself than information that is not personally relevant.
5 0
3 years ago
The question is in the attachment!!​
Bingel [31]

Explanation:

|F|= B/2

B/2 = √(A²+B²+2ABcosθ) ------(1)

Since the resultant of A and B is perpendicular to Vector A

tan90°= BSinθ/(A+BCosθ)

(A+BCosθ)=0

Cosθ=-A/B ----(2)

Using equation (1)

B/2 = √(A²+B²+2ABcosθ)

B/2 = √(A²+B²+2AB×-A/B)

B/2=√(A²+B²-2A²)

B/2=√(B²-A²)

B²/4=B²-A²

A²=B²-B²/4

A²=3B²/4

A=√3B/2

Using equation (2)

Cosθ=-A/B

Cosθ=-[√3B/2]/B

Cosθ=-√3/2

θ= cos^-1 (-√3/2)

θ= 150°

7 0
3 years ago
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