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Setler [38]
3 years ago
8

If a 200 turn, 10-3 m2 cross-sectional area coil is immersed in a magnetic field such that the plane of the coil is perpendicula

r to the field. If the magnetic field increases by 6 T/s, then how much voltage is induced
Physics
1 answer:
Stels [109]3 years ago
8 0

Answer:

<h3>1.2 volt induced in coil.</h3>

Explanation:

Given:

Number of turns N =  200

Cross sectional area A = 10^{-3} m^{2}

Rate of increasing magnetic field \frac{dB}{dt} = 6 \frac{T}{s}

From the faraday's law,

Induced emf is given by,

   \epsilon = -N\frac{d \phi}{dt }

Where \phi = magnetic flux

  \phi = AdB \cos(0)          ( because angle between normal coil and field is zero)

Where A = area of coil

Put the value of \phi in above equation,

Here we neglect minus sign

   \epsilon = NA\frac{dB}{dt}

   \epsilon = 200 \times 10^{-3} \times 6

   \epsilon = 1.2 V

Therefore, 1.2 volt induced in coil

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To what potential should you charge a 3.0 μf capacitor to store 1.0 j of energy?
Nimfa-mama [501]
The energy stored in a capacitor is given by:
U= \frac{1}{2}CV^2
where
U is the energy
C is the capacitance
V is the potential difference

The capacitor in this problem has capacitance
C=3.0 \mu F = 3.0 \cdot 10^{-6} F
So if we re-arrange the previous equation, we can calculate the potential V that should be applied to the capacitor to store U=1.0 J of energy on it:
V= \sqrt{ \frac{2U}{C} }= \sqrt{ \frac{2 \cdot 1.0 J}{3.0 \cdot 10^{-6}F} }=816 V
8 0
3 years ago
Water moves through a constricted pipe in steady, ideal flow. At the
Irina-Kira [14]

A) Speed in the lower section: 0.638 m/s

B) Speed in the higher section: 2.55 m/s

C) Volume flow rate: 1.8\cdot 10^{-3} m^3/s

Explanation:

A)

To solve the problem, we can use Bernoulli's equation, which states that

p_1 + \rho g h_1 + \frac{1}{2}\rho v_1^2 = p_2 + \rho g h_2 + \frac{1}{2}\rho v_2^2

where

p_1=1.75\cdot 10^4 Pa is the pressure in the lower section of the tube

h_1 = 0 is the heigth of the lower section

\rho=1000 kg/m^3 is the density of water

g=9.8 m/s^2 is the acceleration of gravity

v_1 is the speed of the water in the lower pipe

p_2 is the pressure in the higher section

h_2 = 0.250 m is the height in the higher pipe

v_2 is hte speed in the higher section

We can re-write the equation as

v_1^2-v_2^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho} (1)

Also we can use the continuity equation, which state that the volume flow rate is constant:

A_1 v_1 = A_2 v_2

where

A_1 = \pi r_1^2 is the cross-section of the lower pipe, with

r_1 = 3.00 cm =0.03 m is the radius of the lower pipe (half the diameter)

A_2 = \pi r_2^2 is the cross-section of the higher pipe, with

r_2 = 1.50 cm = 0.015 m (radius of the higher pipe)

So we get

r_1^2 v_1 = r_2^2 v_2

And so

v_2 = \frac{r_1^2}{r_2^2}v_1 (2)

Substituting into (1), we find the speed in the lower section:

v_1^2-(\frac{r_1^2}{r_2^2})^2v_1^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho}\\v_1=\sqrt{\frac{2(p_2-p_1+\rho g h_2)}{\rho(1-\frac{r_1^4}{r_2^4})}}=0.638 m/s

B)

Now we can use equation (2) to find the speed in the lower section:

v_2 = \frac{r_1^2}{r_2^2}v_1

Substituting

v1 = 0.775 m/s

And the values of the radii, we find:

v_2=\frac{0.03^2}{0.015^2}(0.638)=2.55 m/s

C)

The volume flow rate of the water passing through the pipe is given by

V=Av

where

A is the cross-sectional area

v is the speed of the water

We can take any point along the pipe since the volume  flow rate is constant, so

r_1=0.03 cm

v_1=0.638 m/s

Therefore, the volume flow rate is

V=\pi r_1^2 v_1 = \pi (0.03)^2 (0.638)=1.8\cdot 10^{-3} m^3/s

Learn more about pressure in a liquid:

brainly.com/question/9805263

#LearnwithBrainly

0 0
3 years ago
2. A brick is sitting on a building 22 m high. It has a mass of 7.9 kg. What amount of potential
Citrus2011 [14]

Answer:

1703.24J

Explanation:

Given parameters:

Mass of brick  = 7.9kg

Height of building  = 22m

Unknown:

Potential energy of the brick  = ?

Solution:

The potential energy of a body is the energy at rest of the body. Mathematically;

 

         P.E = mgh

m is the mass of the brick

g is the acceleration due to gravity

h is the height of the building

  Insert the given parameters and solve;

        P.E  = 7.9 x 9.8 x 22  = 1703.24J

6 0
3 years ago
A machine gun fires 50-g bullets at the rate of 4 bullets per second. The bullets leave the gun at a speed of 1000 m/s. What is
TEA [102]

Answer:

Average recoil force experienced by machine will be 200 N

Explanation:

We have give mass of each bullet m = 50 gram = 0.05 kg

There are 4 bullets

So mass of 4 bullets = 4×0.05 = 0.2 kg

Initial speed of the bullet u = 0 m/sec

And final speed of the bullet v = 1000 m/sec

So change in momentum P=m(v-u)=0.2\times (1000-0)=200kgm/sec

Time is given per second so t = 1 sec

We know that force is equal to rate of change of momentum

So force will be equal to F=\frac{200}{1}=200N

So average recoil force experienced by machine will be 200 N

5 0
3 years ago
PLEASE ANSWER QUICKLY
Rom4ik [11]

Answer:

a luminous ball of plasma

6 0
3 years ago
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