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Step2247 [10]
3 years ago
12

A 5.2 cm diameter circular loop of wire is in a 1.35 T magnetic field. The loop is removed from the field in 0.29 sec. Assume th

at the loop is perpendicular to the magnetic field. What is the average induced emf?
Physics
1 answer:
Katyanochek1 [597]3 years ago
3 0

Answer:

9.88 milivolt

Explanation:

Given: diameter d = 5.2 cm

magnetic field B_1 = 1.35 T, final magnetic field B_2 =0 T

t = 0.29 sec.

we know emf =  - dΦ/dt

and flux Φ = BA

A= area

therefore emf ε = -A(B_2-B_1)/Δt

=-\pi(d/2)^2\frac{B_2-B_1}{\Delta t} \\=-\pi(0.052/2)^2\frac{0-1.35}{0.29} \\=98.8\times10^4\\=9.88 mV

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SashulF [63]

Given data

*The value of battery voltage is V = 10 V

*The current flows through the resistor is I = 5 A

The formula for the resistor is given by the Ohm's law as

R=\frac{V}{I}

Substitute the values in the above expression as

\begin{gathered} R=\frac{10}{5} \\ =2\text{ ohm} \end{gathered}

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3 years ago
How much tension must a rope withstand if it is used to accelerate a 6526 kg car vertically upwards at 8.9 m/s^2?
ExtremeBDS [4]

Answer:

The value is T =  122036.2 \  N

Explanation:

From the question we are told that

     The mass of the car is  m  = 6526 \  kg

      The acceleration  is  a=  8.9 \  m/s^2

Generally the net force applied on the rope is mathematically represented as

          F_{net}   =  T -  W

Here W is the weight of the car which is evaluated as

         W =  m * g

=>      W =  6526  * 9.8

=>       W =  63954.8 \  N

Generally the net force can also be mathematically represented as

       F =  m * a

So

        m * a  =  T  -  63954.8

=>     6526  *  8.9 =  T  -  63954.8

=>      T =  122036.2 \  N

7 0
3 years ago
Assume that the radius ????r of a sphere is expanding at a rate of 70 cm/min.70 cm/min. The volume of a sphere is ????=43???????
rodikova [14]

Answer:

the rate of change in volume with time is 280πr² cm³/min

Explanation:

Data provided in the question:

Radius of the sphere as 'r'

\frac{d\textup{r}}{\textup{dt}}  = 70 cm/min

Volume of the sphere, V = \frac{\textup{4}}{\textup{3}}\pi r^3

Surface area of the sphere as 4πr²

Now,

Rate of change in volume with time, \frac{d\textup{V}}{\textup{dt}}

 = \frac{d(\frac{\textup{4}}{\textup{3}}\pi r^3)}{dt}

= 3\times\frac{\textup{4}}{\textup{3}}\pi r^2}\times\frac{dr}{dt}

Substituting the value of \frac{dr}{dt}

= 3\times\frac{\textup{4}}{\textup{3}}\pi r^2}\times70

= 280πr² cm³/min

Hence, the rate of change in volume with time is 280πr² cm³/min

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3 years ago
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