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saw5 [17]
2 years ago
11

If someone travels 3000 feet in one second, how far will it travel in 18 minutes

Physics
2 answers:
MaRussiya [10]2 years ago
8 0

Answer:3240000 feet's

Explanation:

Since 60seconds equal 1minute therefore 1080 seconds equal 18minutes

Therefore the total number of feet traveled for 18minutes(1080seconds)=3000×1080=3240000 feet

34kurt2 years ago
3 0

Answer:

3240000 ft

Explanation:

There are 60 seconds in a minute so, multiply 60 seconds by 18 minutes. That equals 1080 seconds.

Since there are 3000 ft in each second, just multiply 1080 seconds by 3000 ft.

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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
Do sponges have of appendages jointed, not Jointed, or absent
anygoal [31]
Sponges have appendages jointed 
7 0
3 years ago
Please, I need help with this question.
Jet001 [13]

The x and y components of the velocity vector is 17.32 m/s and 10 m/s respectively.

<h3>What is the x - component of the velocity?</h3>

The x-component of the ball's velocity is the velocity of the ball in the horizontal direction or x-axis.

The velocity of the ball in x-direction is calculated as follows;

Vx = V cosθ

where;

  • Vx is the horizontal velocity of the ball
  • V is the speed of the ball
  • θ is the angle of inclination of the speed

Vx = (20 m/s) x (cos 30)

Vx = 17.32 m/s

The velocity of the ball in y-direction is calculated as follows;

Vy = V sinθ

where;

  • Vy is the vertical velocity of the ball
  • V is the speed of the ball
  • θ is the angle of inclination of the speed

Vy = 20 m/s x sin(30)

Vy = 10 m/s

Learn more about x and y components of velocity here: brainly.com/question/18090230

#SPJ1

3 0
11 months ago
Describe Kinetic Energy.​
mina [271]
In physics, the kinetic energy of an object is the energy that it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. The same amount of work is done by the body when decelerating from its current speed to a state of rest.

4 0
2 years ago
An iron wire has length 8.0m and a diameter 0.50mm. The sir has a resistance R.
Rudik [331]
The re<span>sistance of the second wire is 16 R.
where R is the resistance of the first wire.

R = </span>ρ\frac{l}{A}
where l = length of the wire
A = area of the wire
A = \pi r^{2} where, r = \frac{diameter of wire}{2}

Thus, on finding the ratio of resistance of the two wires, we get,

\frac{R1}{R2} =  \frac{l1A2}{l2A1}

here, R1 = R
l1 = 8m
l2 = 2m
A1=π0.25^{2}
A1=π0.50^{2}

we get. R2 = 16R
7 0
3 years ago
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