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Zepler [3.9K]
3 years ago
10

Help?

Physics
1 answer:
ankoles [38]3 years ago
8 0

center, power forward, small forward, point guard and shooting guard

1896

the University of Chicago and the University of Iowa

James Naismith was a physical education teacher

Sprinfield College

Upon the request of his boss, Naismith was tasked to create an indoor sports game to help athletes keep in shape as they endured the cold New England winters. Naismith's boss also stipulated that this new game should be "fair for all players and not too rough."

I just looked it up hope this helps tho.

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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
Enumerate the viscosity of magma in different conditions.
Firlakuza [10]

Answer:

The nature of volcanic eruptions is highly dependent on magma viscosity and also on dissolved gas content. ... long it takes the treacle to flow from one end of a boiling tube to the other.

3 0
3 years ago
A hypothetical planet has a mass of one-half that of the earth and a radius of twice that of the earth.
Fed [463]
<h2>Option A is the correct answer.</h2>

Explanation:

Acceleration due to gravity

                  g=\frac{GM}{r^2}

         G = 6.67 × 10⁻¹¹ m² kg⁻¹ s⁻²

  Let mass of earth be M and radius of earth be r.

  We have

               g=\frac{GM}{r^2}

Now

         A hypothetical planet has a mass of one-half that of the earth and a radius of twice that of the earth.

       Mass of hypothetical planet, M' = M/2

       Radius of hypothetical planet, r' = 2r

  Substituting

              g'=\frac{GM'}{r'^2}\\\\g'=\frac{G\times \frac{M}{2}}{(2r)^2}\\\\g'=\frac{\frac{GM}{r^2}}{8}\\\\g'=\frac{g}{8}

Option A is the correct answer.

6 0
3 years ago
A brand new corvette can go from 0 to 85 miles per hour in 4.8 seconds.
klasskru [66]

Answer:

(A) 7.9 m/s^{2}  

(B) 19 m/s

(C) 91 m

Explanation:

initial velocity (U) = 0 mph = 0 m/s

final velocity (V) = 85 mph = 85 x 0.447 = 38 m/s

initial time (ti) = 0 s

final time (t) = 4.8 s

(A) acceleration = \frac{V-U}{t}

         = \frac{38-0}{4.8} = 7.9 m/s^{2}  

(B) average velocity = \frac{V+U}{2}

     =\frac{38+0}{2} = 19 m/s

(C) distance travelled (S) = ut + 0.5at^{2}

  =  (0 x 4.8) + 0.5 x 7.9 x 4.8^{2} = 91 m

5 0
3 years ago
Consider a 150 turn square loop of wire 17.5 cm on a side that carries a 42 A current in a 1.7 T. a) What is the maximum torque
ankoles [38]

Answer:

(a) 328 Nm

(b) 79.35 Nm

Explanation:

N = =150, side = 17.5 cm = 0.175 m, i = 42 A, B = 1.7 T

A = side^2 = 0.175^2 = 0.030625 m^2

(a) Torque = N x i x A x B x Sinθ

For maximum torque, θ = 90 degree

Torque = 150 x 42 x 0.030625 x 1.7 x Sin 90

Torque = 328 Nm

(b) θ = 14 degree

Torque =  150 x 42 x 0.030625 x 1.7 x Sin 14

Torque = 79.35 Nm

7 0
3 years ago
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