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jasenka [17]
1 year ago
13

Why is the sun not the major factor in the tides?

Physics
1 answer:
nirvana33 [79]1 year ago
5 0

Answer: the moon is closer.

Explanation:

The moon is closer to the earth than the sun, which results in a larger gravitational gradient for the moon even though the sun is considerably more massive and has stronger overall gravity than the moon.

You might be interested in
A rough estimate of the radius of a nucleus is provided by the formula r 5 kA1/3, where k is approximately 1.3 × 10213 cm and A
Sphinxa [80]

Answer:

Density of 127 I = \rm 1.79\times 10^{14}\ g/cm^3.

Also, \rm Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.

Explanation:

Given, the radius of a nucleus is given as

\rm r=kA^{1/3}.

where,

  • \rm k = 1.3\times 10^{-13} cm.
  • A is the mass number of the nucleus.

The density of the nucleus is defined as the mass of the nucleus M per unit volume V.

\rm \rho = \dfrac{M}{V}=\dfrac{M}{\dfrac 43 \pi r^3}=\dfrac{M}{\dfrac 43 \pi (kA^{1/3})^3}=\dfrac{M}{\dfrac 43 \pi k^3A}.

For the nucleus 127 I,

Mass, M = \rm 2.1\times 10^{-22}\ g.

Mass number, A = 127.

Therefore, the density of the 127 I nucleus is given by

\rm \rho = \dfrac{2.1\times 10^{-22}\ g}{\dfrac 43 \times \pi \times (1.3\times 10^{-13})^3\times 127}=1.79\times 10^{14}\ g/cm^3.

On comparing with the density of the solid iodine,

\rm \dfrac{Density\ of\ ^{127}I}{Density\ of\ the\ solid\ iodine}=\dfrac{1.79\times 10^{14}\ g/cm^3}{4.93\ g/cm^3}=3.63\times 10^{13}.\\\\\Rightarrow Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.

7 0
3 years ago
At a given instant, the force on an electron is in the +z-direction (out of the page), which the electron is moving in the +x-di
Hitman42 [59]

Answer:

The direction of the B-field is in the +y-direction.

Explanation:

The corresponding formula is

F_B = qv\times B

This means, we should use right-hand rule.

Our index finger is pointed towards +x-direction (direction of velocity),

our middle finger should point towards the direction of the B-field,

and our thumb should point towards the +z-direction (direction of the force).

Since our middle finger in this situation points towards +y-direction, the B-field should be in +y-direction.

\^{x} \times \^{y} = \^{z}

3 0
3 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).

6 0
3 years ago
) Un círculo de 120 cm de radio gira a 600 rpm. Calcula: a) su velocidad angular
DIA [1.3K]

Responder:

20πrads ^ -1; 24πrads ^ -1; 0,1 seg; 10 Hz

Explicación:

Dado lo siguiente:

Radio (r) del círculo = 120 cm

600 revoluciones por minuto en radianes por segundo

(600 / min) * (2π rad / 1 rev) * (1min / 60seg)

(1200πrad / 60sec) = 20π rad ^ -1

Velocidad angular (w) = 20πrads ^ -1

Velocidad lineal = radio (r) * velocidad angular (w)

Velocidad lineal = (120/100) * 20πrad

Velocidad lineal = 1.2 * 20πrads ^ -1 = 24πrads ^ -1

C.) Período (T):

T = 2π / w = 2π / 20π = 0.1 seg

D.) Frecuencia (f):

f = 1 / T = 1 / 0.1

1 / 0,1 = 10 Hz

5 0
3 years ago
If a box with a mass of 8.0 kg is sitting on a frictionless surface and experiences an acceleration of 4.0 m/s2 to the right, wh
mr Goodwill [35]

The net force acting on a box of mass 8.0kg that experiences an acceleration of 4.0m/s² is 32N. Details about net force can be found below.

<h3>How to calculate net force?</h3>

The net force of a body can be calculated by multiplying the mass of the body by its acceleration as follows:

Force = mass × acceleration

According to this question, a box with a mass of 8.0 kg is sitting on a frictionless surface and experiences an acceleration of 4.0 m/s2 to the right.

Net force = 8kg × 4m/s²

Net force = 32N

Therefore, the net force acting on a box of mass 8.0kg that experiences an acceleration of 4.0m/s² is 32N.

Learn more about net force at: brainly.com/question/18031889

#SPJ1

4 0
2 years ago
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