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photoshop1234 [79]
2 years ago
12

Define uniform and non uniform

Physics
2 answers:
cricket20 [7]2 years ago
7 0

Answer:

When a body moves along a straight line with uniform speed or steady speed is called Uniform motion. When a body moves along a straight line but with variable or change in speed is called non-uniform motion.Hope this answer helps.

FinnZ [79.3K]2 years ago
7 0

Answer:

. When a body moves along a straight line with uniform speed or steady speed is called Uniform motion. When a body moves along a straight line but with variable or change in speed is called non-uniform motion

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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
At time t1 = 14 s, a car is located at 99, 80, 27 m and has velocity 4, 0, −3 m/s. At time t2 = 18 s, what is the position of th
Korvikt [17]

Answer:

115, 80, 15m

Explanation

t1 = 14s

t2 = 18s

change in time = 4s (18-14)

r(final) = r(initial) + (average velocity) x (change in time)

multiply the average velocity with the change in time

= (4, 0, -3) x 4 = 16, 0, -12

now we'll add this value to the initial position of the car

(99, 80, 27)m + (16, 0, -12)m = (115, 80, 15)m

8 0
3 years ago
A uniform plank of length 5.0 m and weight 225 N rests horizontally on two supports, with 1.1 m of the plank hanging over the ri
lawyer [7]

Answer:

x = 0.6034 m

Explanation:

Given

L = 5 m

Wplank = 225 N

Wman = 522 N

d = 1.1 m

x = ?

We have to take sum of torques about the right support point.  If the board is just about to tip, the normal force from the left support will be going to zero.  So the only torques come from the weight of the plank and the weight of the man.

∑τ = 0  ⇒     τ₁ + τ₂ = 0  

Torque come from the weight of the plank = τ₁

Torque come from the weight of the man = τ₂

⇒  τ₁ = + (5 - 1.1)*(225/5)*((5 - 1.1)/2) - (1.1)*(225/5)*((1.1)/2) = 315 N-m (counterclockwise)

⇒  τ₂ = Wman*x = 522 N*x   (clockwise)

then

τ₁ + τ₂ = (315 N-m) + (- 522 N*x) = 0

⇒  x = 0.6034 m

7 0
3 years ago
What is the definition of physical change
jeka94
They are used to separate mixtures into their component compounds but can no usually be used to deprecate compounds into chemical elements or simpler compounds
7 0
3 years ago
As the solar system formed, most of its mass became concentrated in one part of it. What is that that part?
sertanlavr [38]
That was sun as some smaller masses formed planets and other remaining formed sun
4 0
3 years ago
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