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disa [49]
3 years ago
14

A 600-turn solenoid, 25 cm long, has a diameter of 2.5 cm. A 14-turn coil is wound tightly around the center of the solenoid. If

the current in the solenoid increases uniformly from 0 to 5.0 A in 0.60 s, what will be the induced emf in the short coil during this time
Physics
1 answer:
Delvig [45]3 years ago
8 0

Answer:

The induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

Explanation:

The magnetic field at the center of the solenoid is given by;

B = μ(N/L)I

Where;

μ is permeability of free space

N is the number of turn

L is the length of the solenoid

I is the current in the solenoid

The rate of change of the field is given by;

\frac{\delta B}{\delta t} = \frac{\mu N \frac{\delta i}{\delta t} }{L} \\\\\frac{\delta B}{\delta t} = \frac{4\pi *10^{-7} *600* \frac{5}{0.6} }{0.25}\\\\\frac{\delta B}{\delta t} =0.02514 \ T/s

The induced emf in the shorter coil is calculated as;

E = NA\frac{\delta B}{\delta t}

where;

N is the number of turns in the shorter coil

A is the area of the shorter coil

Area of the shorter coil = πr²

The radius of the coil = 2.5cm / 2 = 1.25 cm = 0.0125 m

Area of the shorter coil = πr² = π(0.0125)² = 0.000491 m²

E = NA\frac{\delta B}{\delta t}

E = 14 x 0.000491 x 0.02514

E = 1.728 x 10⁻⁴ V

Therefore, the induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

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Light from two lasers is incident on an opaque barrier with a single slit of width 4.0 x 10^-4 m. One laser emits light of wavel
Sergio [31]

Answer:

a) y = 2.4 x 10⁻³ m = 0.24 cm

b) y = 3.2 x 10⁻³ m = 0.32 cm

Explanation:

The formula of Young's Double Slit experiment will be used here:

y = \frac{\lambda L}{d}\\\\

where,

y = distance between dark spots = ?

λ = wavelength

L = distance of screen = 2 m

d = slit width = 4 x 10⁻⁴ m

a) FOR λ = 480 nm = 4.8 x 10⁻⁷ m:

y = \frac{(4.8\ x\ 10^{-7}\ m)(2\ m)}{4\ x\ 10^{-4}\ m}

<u>y = 2.4 x 10⁻³ m = 0.24 cm</u>

<u></u>

a) FOR λ = 640 nm = 6.4 x 10⁻⁷ m:

y = \frac{(6.4\ x\ 10^{-7}\ m)(2\ m)}{4\ x\ 10^{-4}\ m}

<u>y = 3.2 x 10⁻³ m = 0.32 cm</u>

7 0
2 years ago
A tank is full of water. Find the work W required to pump the water out of the spout. (Use 9.8 m/s2 for g. Use 1000 kg/m3 as the
Sergio039 [100]

Answer:

W = 1.06 MJ

Explanation:

- We will use differential calculus to solve this problem.

- Make a differential volume of water in the tank with thickness dx. We see as we traverse up or down the differential volume of water the side length is always constant, hence, its always 8.

- As for the width of the part w we see that it varies as we move up and down the differential element. We will draw a rectangle whose base axis is x and vertical axis is y. we will find the equation of the slant line that comes out to be y = 0.5*x. And the width spans towards both of the sides its going to be 2*y = x.

- Now develop and expression of Force required:

                                             F = p*V*g

                                             F = 1000*(2*0.5*x*8*dx)*g

                                             F = 78480*x*dx

- Now, the work done is given by:

                                             W = F.s

- Where, s is the distance from top of hose to the differential volume:

                                             s = (5 - x)

- We have the work as follows:

                                            dW = 78400*x*(5-x)dx

- Now integrate the following express from 0 to 3 till the tank is empty:

                                           W = 78400*(2.5*x^2 - (1/3)*x^3)

                                           W = 78400*(2.5*3^2 - (1/3)*3^3)

                                           W = 78400*13.5 = 1058400 J

 

5 0
3 years ago
If a puddle of water is frozen, do the particles in the ice have kinetic energy? Explain.
TEA [102]
They have some but not very much, the particles in the ice are still vibrating just not as much as in water. the only time a substance would have 0 kinetic energy is when that substance is at 0 degrees kelvin(absolute zero) so far no place in the universe has been recorded at absolute zero though
8 0
3 years ago
A block is resting on a platform that is rotating at an angular speed of 2.4 rad/s. The coefficient of static friction between t
Sloan [31]

Answer:

r = 0m is the Minimum distance from the axis at which the block can remain in place wothout skidding.

Explanation:

From a sum of forces:

Ff = m*a   where Ff = μ * N    and a = \frac{V^2}{r}=\omega^2*r

N - m*g = 0   So, N = m*g.   Replacing everything on the original equation:

\mu*m*g = m*\omega^2*r   (eq2)

Solving for r:

r = \frac{\mu*g}{\omega^2}=1.41m

If we analyze eq2 you can conclude that as r grows, the friction has to grow (assuming that ω is constant), so the smallest distance would be 0 and the greatest 1.41m. Beyond that distance, μ has to be greater than 0.83.

4 0
2 years ago
1. An object has a mass of 10g and a volume of 50mL. What is the density?
Alona [7]

1. 0.2 g/mL

The relationship between mass, density and volume of an object is

d=\frac{m}{V}

where

d is the density

m is the mass

V is the volume

For the object in this problem, we have

m = 10 g

V = 50 mL

Substituting into the equation,

d=\frac{10}{50}=0.2 g/mL

2. 10 mL

In this exercise we know:

- The density of the object: d = 2 g/mL

- The mass of the object: m = 20 g

Therefore, we can re-arrange the previous equation to find the volume:

V=\frac{m}{d}

And substituting values into the equation, we find

V=\frac{20}{2}=10 mL

7 0
3 years ago
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