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2 years ago

Four ways to increase magnitude of current in dynamo

Physics

1 answer:

8090 [49]2 years ago

5 0

Answer:

hmm

Explanation:

By increasing the number of turns in the coil, strength of magnetic field, speed of rotation of the coil in the magnetic field and by decreasing the distance between the coil and the magnet the magnitude of the induced e.m.f. can be increased in generator/dynamo.

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How can we magnetise a piece of iron.

liq [111]

Place the magnet at one end of the piece of metal. The magnet must make as much contact with the metal as possible. Place light pressure on the magnet and rub the metal in one direction only. Magnetization will take some time to accomplish so continue rubbing until the iron or steel attracts other pieces of metal.

4 0

3 years ago

What is the torque τa about axis a due to the force f⃗ ? express the torque about axis a at cartesian coordinates (0,0)?

pentagon [3]

Answer:

$\tau_a = F a sin \theta$

Explanation:

The torque of a force is given by:

$\tau = F d sin \theta$

where

F is the magnitude of the force

d is the distance between the point of application of the force and the centre of rotation of the system

$\theta$ is the angle between the direction of the force and d

In this problem, we have:

$F$ , the force

$a$ , the distance of application of the force from the centre (0,0)

$\theta$ , the angle between the direction of the force and a

Therefore, the torque is

$\tau_a = F a sin \theta$

5 0

3 years ago

Help me please (*￣(ｴ)￣*)

Len [333]

Answer:

1) Are always conservative

Explanation:

Elastic forces are always conservative.

Hope it helps you.

please mark as the brainliest answer.

3 0

2 years ago

Read 2 more answers

(a) If a proton with a kinetic energy of 6.2 MeV is traveling in a particle accelerator in a circular orbit with a radius of 0.5

Tju [1.3M]

Answer:

The fraction of its energy that it radiates every second is $3.02\times10^{-11}$ .

Explanation:

Suppose Electromagnetic radiation is emitted by accelerating charges. The rate at which energy is emitted from an accelerating charge that has charge q and acceleration a is given by

$\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}$

Given that,

Kinetic energy = 6.2 MeV

Radius = 0.500 m

We need to calculate the acceleration

Using formula of acceleration

$a=\dfrac{v^2}{r}$

Put the value into the formula

$a=\dfrac{\dfrac{1}{2}mv^2}{\dfrac{1}{2}mr}$

Put the value into the formula

$a=\dfrac{6.2\times10^{6}\times1.6\times10^{-19}}{\dfrac{1}{2}\times1.67\times10^{-27}\times0.51}$

$a=2.32\times10^{15}\ m/s^2$

We need to calculate the rate at which it emits energy because of its acceleration is

$\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}$

Put the value into the formula

$\dfrac{dE}{dt}=\dfrac{(1.6\times10^{-19})^2\times(2.3\times10^{15})^2}{6\pi\times8.85\times10^{-12}\times(3\times10^{8})^3}$

$\dfrac{dE}{dt}=3.00\times10^{-23}\ J/s$

The energy in ev/s

$\dfrac{dE}{dt}=\dfrac{3.00\times10^{-23}}{1.6\times10^{-19}}\ J/s$

$\dfrac{dE}{dt}=1.875\times10^{-4}\ ev/s$

We need to calculate the fraction of its energy that it radiates every second

$\dfrac{\dfrac{dE}{dt}}{E}=\dfrac{1.875\times10^{-4}}{6.2\times10^{6}}$

$\dfrac{\dfrac{dE}{dt}}{E}=3.02\times10^{-11}$

Hence, The fraction of its energy that it radiates every second is $3.02\times10^{-11}$ .

5 0

3 years ago

07. How do scientists use spectroscopy (the study of light frequencies released

Vera_Pavlovna [14]

Every element is able to be recognized individually in many different ways. A very easy and common way is using light absorption also known as spectroscopy. Every atom has electrons, and these electrons like to stay in their lowest-energy configuration. However, when photons collide with an electron it can increase it to a higher energy level.. This is absorption, and each element’s electrons absorb light at specific wavelengths related to the difference between energy levels in that atom. But the electrons want to return to their original levels, so they don’t hold onto the energy for long. When they emit the energy, they release photons with exactly the same wavelengths of light that were absorbed in the first place. An electron can release this light in any direction, so most of the light is emitted in directions away from our line of sight. Therefore, a dark line appears in the spectrum at that particular wavelength.

Because the wavelengths at which absorption lines occur are unique for each element, astronomers can measure the position of the lines to determine which elements are present in a target. The amount of light that is absorbed can also provide information about how much of each element is present.

8 0

3 years ago

Four ways to increase magnitude of current in dynamo​

Four ways to increase magnitude of current in dynamo