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FinnZ [79.3K]
3 years ago
6

The great contribution of nicholas copernicus was to

Physics
1 answer:
DIA [1.3K]3 years ago
6 0

publish a book in which he explained his heliocentric theory.

Explanation:

The heliocentric theory is the one according which the Sun is in the center of our Solar Sistem and the planets go around it.  

Previously to this theory (before Copernicus) the prevalent theory was geocentrism accordind which it was the earth who stood in the center and all the other celestial bodies went around it.

:-)

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A glass rod with yellow crosses on it touches a sheet of silk covered with negative signs on it. the sheet of silk is sticking t
forsale [732]

The image as shown here can here can be used to describe charging by induction.

<h3>What is a charge?</h3>

A charge may be positive or negative. One of the methods of transferring a charge is by induction.

In this case, an objects induces an opposite charge on a material.  The image as shown here can here can be used to describe charging by induction.

Learn more about charging by induction:brainly.com/question/10254645

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5 0
2 years ago
A string is stretched between a fixed support and a pulley a distance 111 cm apart. The tension on the string is controlled by a
kvv77 [185]

Answer:

88.8 m/s= Speed of wave propagation in the required mode.(3 loops)

Explanation:

When there are 3 loops.  

the total length = L = 3 λ /2

⇒ λ  = 2 L / 3 = 2 ( 1.11 ) / 3 = 0.74 m

Velocity = v = f λ = (120)(0.74) = 88.8 m/s

3 0
4 years ago
Consider a cyclotron in which a beam of particles of positive charge q and mass m is moving along a circular path restricted by
Ulleksa [173]

A) v=\sqrt{\frac{2qV}{m}}

B) r=\frac{mv}{qB}

C) T=\frac{2\pi m}{qB}

D) \omega=\frac{qB}{m}

E) r=\frac{\sqrt{2mK}}{qB}

Explanation:

A)

When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:

\Delta U = qV

where

q is the charge of the particle (positive)

On the other hand, the change (increase) in the kinetic energy of the particle is (assuming it starts from rest):

\Delta K=\frac{1}{2}mv^2

where

m is the mass of the particle

v is its final speed

According to the law of conservation of energy, the change (decrease) in electric potential energy is equal to the increase in kinetic energy, so:

qV=\frac{1}{2}mv^2

And solving for v, we find the speed v at which the particle enters the cyclotron:

v=\sqrt{\frac{2qV}{m}}

B)

When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is

F=qvB

where B is the strength of the magnetic field.

This force acts as centripetal force, so we can write:

F=m\frac{v^2}{r}

where r is the radius of the orbit.

Since the two forces are equal, we can equate them:

qvB=m\frac{v^2}{r}

And solving for r, we find the radius of the orbit:

r=\frac{mv}{qB} (1)

C)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

It can be calculated as the ratio between the length of the circumference (2\pi r) and the velocity of the particle (v):

T=\frac{2\pi r}{v} (2)

From eq.(1), we can rewrite the velocity of the particle as

v=\frac{qBr}{m}

Substituting into(2), we can rewrite the period of revolution of the particle as:

T=\frac{2\pi r}{(\frac{qBr}{m})}=\frac{2\pi m}{qB}

And we see that this period is indepedent on the velocity.

D)

The angular frequency of a particle in circular motion is related to the period by the formula

\omega=\frac{2\pi}{T} (3)

where T is the period.

The period has been found in part C:

T=\frac{2\pi m}{qB}

Therefore, substituting into (3), we find an expression for the angular frequency of motion:

\omega=\frac{2\pi}{(\frac{2\pi m}{qB})}=\frac{qB}{m}

And we see that also the angular frequency does not depend on the velocity.

E)

For this part, we use again the relationship found in part B:

v=\frac{qBr}{m}

which can be rewritten as

r=\frac{mv}{qB} (4)

The kinetic energy of the particle is written as

K=\frac{1}{2}mv^2

So, from this we can find another expression for the velocity:

v=\sqrt{\frac{2K}{m}}

And substitutin into (4), we find:

r=\frac{\sqrt{2mK}}{qB}

So, this is the radius of the cyclotron that we must have in order to accelerate the particles at a kinetic energy of K.

Note that for a cyclotron, the acceleration of the particles is achevied in the gap between the dees, where an electric field is applied (in fact, the magnetic field does zero work on the particle, so it does not provide acceleration).

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When the temperature of air rises, the amount of water needed for saturation
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It Increases. I just took a quiz with the same question.
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