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hichkok12 [17]
3 years ago
14

A metal ring 4.30 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpend

icular to the magnetic field. These magnets produce an initial uniform field of 1.12 T between them but are gradually pulled apart, causing this field to remain uniform but decrease steadily at 0.230 T/s.(a) What is the magnitude of the electric field induced in the ring? (b) In which direction (clockwise or counterclockwise) does the current flow as viewed by someone on the south pole of the magnet?
Physics
1 answer:
CaHeK987 [17]3 years ago
6 0

Answer:

A)0.00966 N/C

B) counterclockwise direction

Explanation:

We are given;

Diameter of the metal ring; d = 4.3 cm

Radius;r = 2.15 cm = 0.021- m

Initial magnetic field, B = 1.12 T

Rate of decrease of the magnetic field;dB/dt = 0.23 T/s

Now, as a result of change in magnetic field, an emf will be induced in it. Thus, , electric field is induced and given by the formula :

∫E•dr = d/dt∫B.A •dA

This gives;

E(2πr) = dB/dt(πr²)

Gives;. 2E = dB/dt(r)

E = dB/dt × 2r

We are given;

E = 0.23 × 2(0.021)

E = 0.00966 N/C

The magnitude of the electric field induced in the ring has a magnitude of 0.00966 N/C

B) The direction of electric field will be in a counterclock wise direction when viewed by someone on the south pole of the magnet

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so your saying the start is 0 N and when he/she hits the ball its inertia is 3 N. if that is so m*v=

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7 0
3 years ago
A diffraction grating with 750 slits per mm is illuminated by light which gives a first-order diffraction angle of 34.0°. What i
lesya [120]
<h2>Answer: 745.59 nm</h2>

Explanation:

The diffraction angles \theta_{n} when we have a slit divided into n parts are obtained by the following equation:

dsin\theta_{n}=n\lambda (1)

Where:

d is the width of the slit

\lambda is the wavelength of the light  

n is an integer different from zero

Now, the first-order diffraction angle is given when n=1, hence equation (1) becomes:

dsin\theta_{1}=\lambda (2)

We know:

\theta_{1}=34\°

In addition we are told the diffraction grating has 750 slits per mm, this means:

d=\frac{1mm}{750}

Solving (2) with the known values we will find \lambda:

\lambda=(\frac{1mm}{750})sin(34\°) (3)

\lambda=0.00074559mm (4)

Knowing 1mm=10^{6}nm:

\lambda=745.59nm  >>>This is the wavelength of the light, wich corresponds to red.

6 0
4 years ago
What is the moment of inertia of a 2.0 kg, 20-cm-diameter disk for rotation about an axis (a) through the center, and (b) throug
FinnZ [79.3K]

Answer:

(a) I=0.01 kg.m²

(b) I=0.03 kg.m²

Explanation:

Given data

Mass of disk M=2.0 kg

Diameter of disk d=20 cm=0.20 m

To Find

(a) Moment of inertia through the center of disk

(b) Moment of inertia through the edge of disk

Solution

For (a) Moment of inertia through the center of disk

Using the equation of moment  of Inertia

I=\frac{1}{2}MR^{2}\\  I=\frac{1}{2}(2.0kg)(0.20m/2)^{2}\\  I=0.01 kg m^{2}

For (b) Moment of inertia through the edge of disk

We can apply parallel axis theorem for calculating moment of inertia

I=(1/2)MR^{2}+MD\\ Here\\D=R\\I=(1/2)(2.0kg)(0.20m/2)^{2}+(2.0kg)(0.20m/2)^{2}\\  I=0.03kgm^{2}

8 0
3 years ago
PLEASE HELP FOR PHYSICS!
Stolb23 [73]

Hi there!

Recall Newton's Law of Universal Gravitation:

\large\boxed{F_g = G\frac{m_1m_2}{r^2}}

Where:

Fg = Force of gravity (N)

G = Gravitational Constant

m1, m2 = masses of objects (kg)

r = distance between objects (m)

Plug in the given values stated in the problem:

F_g = (6.673*10^{-11})\frac{50 * 50}{25^2} = \boxed{2.669 * 10^{-10} N}

8 0
2 years ago
"Calculate the speed of this sound wave by recording how much time it takes to travel 5 meters. Use the (old school) velocity eq
lianna [129]
1) sound velocity reported by you : 292.39 m /s

2) time to travel 1620m at that velocity: t = d / v = 1620 m / 292.39 m/s = 5.54 s, since the moment the sound wave started.

3) You might wanted to tell the time since you watched the lightning.

Then you can calculate the time since the lighting was generated,1620 m away from you, until you saw it, using the speed of light:

 speed of light = 3*10^8 m/s => t = 1620 m / (3*10^8m/s) =0.0000054 s

Then, this time is completely neglectible, and yet the answer is 5.54 s, as calculated in the step 2.
7 0
3 years ago
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