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3 years ago

Why there is deflection in galvanometer when magnetic is approachin or withdraw

Physics

1 answer:

spin [16.1K]3 years ago

7 0

Answer:

When the bar magnet is withdrawn from inside the coil, again an induced current is setup in the coil due to change of magnetic field through it. As a result galvanometer gives a deflection in the reverse direction

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Thomson's experiments accounted for: the mass of the electron the fundamental charge the force on a current-carrying wire the el

gogolik [260]

Thomson experiment he calculated the charge to mass ratio just be passing the fundamental charge through a tube

He calculated the charge to mass ratio just by finding the deflection of charge while it is passing through the constant electric field

so here we will use the deflection as following

let say it passes the field of length "L"

so here we have

$t = \frac{L}{v}$

now in the same time if it deflect by some distance

$\delta y = \frac{1}{2}at^2$

$\delta y = \frac{1}{2}{eE}{m}t^2$

now by solving this equation we can find e/m ratio

so here correct answer will be

the electron's charge-to-mass ratio

7 0

3 years ago

Read 2 more answers

A 500 kg rollercoaster car starts from rest at the top of a 10.0 m tall hill. it then travels down the track and up a loop. the

malfutka [58]

Speed of the roller coaster at the top of the loop= 7.67 m/s

Explanation:

using the law of conservation of energy

KEi + PEi= KEf + PEf

KEi= kinetic energy at the top of the hill=0 because the car is at rest there.

PEi= potential energy at the top of the hill

PEf= potential energy at the top of the loop

KEf= kinetic energy at the top of the loop

Also kinetic energy= 1/ 2m v² and potential energy= mgh

m= mass

h= height

v= velocity

so 0+ mghi = 1/2mv² + mg h

500 (9.8)(10)+ 1/2 (500) v²= 500 ( 9.8) (7)

49000+250 v²= 34300

250v²= 14700

v²=58.8

v=7.67 m/s

4 0

3 years ago

Which of the following describes the renewability of utility-scale wind energy?

schepotkina [342]

Unlimited, clean energy plant describes the renewability of utility-scale wind energy.

<h3>What is utility-scale wind energy?</h3>

Utility-scale wind turbines are installed in large, multi-turbine wind farms connected to the nation's transmission system. These turbines produce clean and unlimited energy.

So we can conclude that Unlimited, clean energy plant describes the renewability of utility-scale wind energy.

Learn more about energy here: brainly.com/question/13881533

6 0

2 years ago

The 1.19 kg head of a hammer has a speed of 8.4 m/s just before it strikes a nail (Fig. 14-13) and is brought to rest. Estimate

KATRIN_1 [288]

Answer:

The temperature rise is $143.9\ K$ .

Explanation:

Given the mass of the hammer $(m)$ is 1.19 kg.

And the speed of the hammer $(v)$ is 8.4 m/s.

We need to find the temperature change of iron nail having 20 hammer blows.

First, we will find the kinetic energy of the each blow of hammer.

$K.E=\frac{1}{2}mv^2$

$K.E=\frac{1}{2}\times 1.19\times 8.4^2=41.99\ J$

For 20 such blows, the kinetic energy will be

$20\times K.E=20\times 41.99=839.8\ J$

Let us assume that the nail absorbs all the kinetic energy. So, this 839.8 J will be converted into heat energy $(Q)$ .

Now,

$Q=839.8\ J$

Also, we know that

$Q=mc\Delta_T$

Where,

$m$ is the mass of the nail, which is $13\ g=\frac{13}{1000}=0.013\ kg$

$c$ is the specific heat capacity of iron, which is 449 J/kg.K

$\Delta_T$ is the change in temperature.

Plug these value we get,

$Q=mc\Delta_T\\839.8\ J=(0.013\ kg)(449\ J/kg.K)\times\Delta_T\\\Delta_T=\frac{839.8}{0.013\times 449}\\\\\Delta_T=143.9\ K$

We can see the temperature rise is 143.9 K.

4 0

4 years ago

What is the value of the angle of refraction of light from a crown glass slab when the angle of incidence is 50°? The slab has a

Artemon [7]

The angle of refraction is $30.3^{\circ}$

Explanation:

Refraction is a wave phenomenon that occurs when a light wave passes through the interface between two mediums. When this occurs, the speed and the direction of the light wave change.

We can find the final direction of the refracted light in this problem by applying Snell's law:

$n_i sin \theta_i = n_r sin \theta_r$