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3 years ago

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After a big snowfall, you take your favorite rocket‑powered sled out to a wide field. The field is 223 m223 m across, and you kn

ow that your sled accelerates at a rate of 3.25 m/s23.25 m/s2 when the rocket is on. How much time will it take the sled to cross the field starting from rest, assuming the rocket is on the whole time?

1 answer:

pogonyaev3 years ago

6 0

Answer:

11.7 s

Explanation:

In this problem, the rocket is moving in a uniform accelerated motion. We have the following data:

d = 223 m, the distance that the sled has to cover

$a=3.25 m/s^2$ , the acceleration of the rocket

We can use therefore the following SUVAT equation:

$d=ut+\frac{1}{2}at^2$

where

d is the distance

u = 0 is the initial velocity of the sled (it starts from rest)

t is the time

a is the acceleration

Re-arranging the equation and substituting the numbers, we find the time it takes for the rocket to cross the field:

$d=\frac{1}{2}at^2\\t=\sqrt{\frac{2d}{a}}=\sqrt{\frac{2(223)}{3.25}}=11.7 s$

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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

Pleaseeeee HELPPPP THIS IS TIMED ALSO,

TEA [102]

Answer:

Friction, normal force, and weight

Explanation:

If the book slows down, it means that there must be friction acting in the opposite direction of the direction the book is moving in.

Weight is caused by the gravitational pull of the Earth on the book, and normal force is the table pushing the book up because the book is pushing down on the table (3rd law.)

Note that weight and normal force is not the 3rd law action-reaction pair. The pair is the force of the book on the table and the force of the table on the book.

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2 years ago

After a great many contacts with the charged ball, how is the charge on the rod arranged (when the charged ball is far away)?

faust18 [17]

Answer: Option (b) is the correct answer.

Explanation:

Since, there is a negative charge present on the ball and a positive charge present on the rod. So, when the negatively charged metal ball will come in contact with the rod then positive charges from rod get conducted towards the metal ball.

Hence, the rod gets neutralized. But towards the metal ball there is a continuous supply of negative charges. Therefore, after the neutralization of positive charge from the rod there will be flow of negative charges from the metal ball towards the rod.

Thus, we can conclude that negative charge spread evenly on both ends.

8 0

3 years ago

A 10.2-kg mass is located at the origin, and a 4.6-kg mass is located at x = 8.1 cm. Assuming g is constant, what is the locatio

goldfiish [28.3K]

Answer:

center of mass of the two masses will lie at x = 2.52 cm

center of gravity of the two masses will lie at x = 2.52 cm

So center of mass is same as center of gravity because value of gravity is constant here

Explanation:

Position of centre of mass is given as

$r_{cm} = \frac{m_1r_1 + m_2r_2}{m_1 + m_2}$

here we have

$m_1 = 10.2 kg$

$m_2 = 4.6 kg$

$r_1 = (0, 0)$

$r_2 = (8.1cm, 0)$

now we have

$r_{cm} = \frac{10.2 (0,0) + 4.6 (8.1 , 0)}{10.2 + 4.6}$

$r_{cm} = {(37.26, 0)}{14.8}$

$r_{cm} = (2.52 cm, 0)$

so center of mass of the two masses will lie at x = 2.52 cm

now for center of gravity we can use

$r_g_{cm} = \frac{m_1gr_1 + m_2gr_2}{m_1g + m_2g}$

here we have

$m_1 = 10.2 kg$

$m_2 = 4.6 kg$

$r_1 = (0, 0)$

$r_2 = (8.1cm, 0)$

now we have

$r_g_{cm} = \frac{10.2(9.8) (0,0) + 4.6(9.8) (8.1 , 0)}{10.2(9.8) + 4.6(9.8)}$

$r_g_{cm} = {(37.26, 0)}{14.8}$

$r_g_{cm} = (2.52 cm, 0)$

So center of mass is same as center of gravity because value of gravity is constant here

3 0

3 years ago

The two blocks in oscillate on a frictionless surface with a period of 1.5 s. The upper block just begins to slip when the ampli

Setler79 [48]

Answer:

0.72

Explanation:

$T$ = Time period of oscillation = 1.5 s

Angular frequency is given as

$w = \frac{2\pi }{T}\\w = \frac{2(3.14) }{1.5}\\w = 4.2 rad/s$

$A$ = Amplitude of oscillation = 40 cm = 0.40 m

$\mu$ = Coefficient of static friction = ?

$a$ = acceleration of the block

$m$ = mass of the block

Maximum acceleration of the block is given as

$a = Aw^{2}$

frictional force is given as

$f = \mu mg$

As per newton's second law

$f = ma \\\\\mu mg = ma \\\mu g = a\\\mu g = Aw^{2}\\\mu (9.8) = (0.40)(4.2)^{2}\\\mu = 0.72$

8 0

2 years ago