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SVETLANKA909090 [29]
3 years ago
10

2 cycleists, 90 miles apart starting riding toward each other at some times one cycles twice as fast as the other if they meet 2

hrs later at what average speed is each cyclist traveling
Physics
1 answer:
svetlana [45]3 years ago
6 0

Average speed of the first car= 15 miles/hr

Average speed of the second car= 30 miles/hr

Explanations:

distance= d= 90 miles

Time= 2 hrs

let speed of first cyclist= v

speed of second cyclist= 2v

relative speed = v + 2v= 3v

using relative speed= Total distance/Total time

3v= 90/2

3 v=45

v=15 miles/hr

so speed of first cyclist= 15 miles/hr

speed of second cyclist= 2 (15)= 30 miles/hr

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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
If the baseball and the plastic ball were moving at the same speed which ball would hit a bat harder
Lerok [7]
A baseball would hit the bat harder. This is because the baseball is a lot heavier and more dense than the plastic ball. The keyword that you're looking for is density. The baseball is dense.
7 0
3 years ago
The photeselestric effect is observed when light of a sufficiently high frequency is focused onto a polished metal surface, emit
Helga [31]

Answer:

3.4\cdot 10^{-19} J

Explanation:

In order to convert the work function of cesium from electronvolts to Joules, we must use the following conversion factor:

1 eV = 1.6 \cdot 10^{-19} J

In our problem, the work function of cesium is

E=2.1 eV

so, we can convert it into Joules by using the following proportion:

1 eV : 1.6\cdot 10^{-19} J = 2.1 eV : x\\x=\frac{(1.6\cdot 10^{-19} J)(2.1 eV)}{1 eV}=3.4\cdot 10^{-19} J

8 0
3 years ago
Plz help me and thx if you do!
Rina8888 [55]
I think it’s b... not sure tho sorry
3 0
3 years ago
What is the order of the types of nuclear radiation from lowest to highest
tangare [24]
The best and most correct answer to the question is :
Alpha particles
Beta particles
Gamma rays
Cosmic radiation
Neutrons
Hope this helped you :)
8 0
3 years ago
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