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fredd [130]
3 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]3 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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Answer:

-75.35°

Explanation:

Let C be the sum of the two vectors A and B. Hence, we can write the following  

A_{x} +B_{x} =C_{x} ......(1)\\A_{y} +B_{y} =C_{y} ......(2)

but since the vector C is in the -y direction, C_{x}  = 0 and C_{y} = —12 m.  

Thus  

B_{x} =-A_{x} =-[-Acos(180-127)]=(8)*cos(53)\\B_{x} =4.81m

similarly, we can determine B_{y} by rearranging equation (1)  

 B_{y} =C_{y} -A_{y} =-12m-[(8)*sin(53)\\B_{y} =-18.4m

so the magnitude of B is

B=\sqrt{B_{x}^2+B_{y}^2  } \\B=19m

Finally, the direction of B can be calculated as follows  

Ф=tan^{-1} (\frac{B_{y} }{B_{x} } )\\=-75.35

hence the vector B makes an angle of 75.35 clockwise with + x axis

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6 0
2 years ago
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An object is thrown upward with a speed of 15 m/s on the surface of planet X where the acceleration due to gravity is 2.5 m/s2.
Alex

Answer:

12s

Explanation:

the formula is

s = u×t + 0.5×a×t²

where:

s is the displacement of the object at this time (so it's equal to 0)

u is the initial speed

t is the time you have to find

a is the acceleration due to gravity

0 = 15 \frac{m}{s}  \times t + 0.5 \times 2.5 \frac{m}{{s}^{2} }  \times  {t}^{2}

15 \frac{m}{s}  \times t = 0.5 \times 2.5 \frac{m}{ {s}^{2} }   \times  {t}^{2}

15 \frac{m}{s} = 1.25 \frac{m}{{s}^{2} }  \times  {t}

t =  \frac{15 \frac{m}{ {s}^{} } }{1.25 \frac{m}{ {s}^{2} } }

t = 12s

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2 years ago
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