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maria [59]
3 years ago
6

A charged particle moves in a circular path in a uniform magnetic field.Which of the following would increase the period of the

particle's motion?Check all that apply.Increasing its chargeIncreasing its massIncreasing the field strengthIncreasing its speed
Physics
1 answer:
Bond [772]3 years ago
4 0

Answer:

Increasing its charge

Increasing the field strength

Explanation:

For a charged particle moving in a circular path in a uniform magnetic field, the centripetal force is provided by the magnetic force, so we can write:

qvB = m\frac{v^2}{r}

where

q is the charge

v is the velocity

B is the magnetic field

m is the mass

r is the radius of the orbit

The period of the motion is

T=\frac{2\pi r}{v}

Re-arranging for r

r=\frac{Tv}{2\pi}

And substituting into the previous equation

qvB = m \frac{Tv^3}{2\pi}

Solving for T,

T=\frac{2\pi q B}{m v^2}

So we see that the period is:

- proportional to the charge and the magnetic field

- inversely proportional to the mass and the square of the speed

So the following will increase the period of the particle's motion:

Increasing its charge

Increasing the field strength

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Explanation:

For a charge concentrated nearly at a point, the electric field is directly proportional to the amount of charge; it is inversely proportional to the square of the distance radially away from the centre of the source charge and depends also upon the nature of the medium.

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What two bodily functions are increased by a warm up
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in an effort to reach the store before it closed the motorist increase the speed of his car from 20 m per second to 60 meters pe
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Answer:

10

Explanation:

Givens

vi = 20 m/s

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t = 4 second.

Formula

a = (vf - vi) / t

a = (60 - 20)/4

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7 0
2 years ago
A 2.5-kg ball and a 5.0-kg ball have an elastic collision. Before the collision, the 2.5-kg ball was at rest and the other ball
lesantik [10]

The kinetic energy of 2.5 kg ball after collision is 27.09 J.

Answer:

Explanation:

In elastic collision, the sum of momentum of the objects before collision will be equal to the sum of momentum of the objects after collision.  

We know that momentum is the product of mass and velocity acting on any object.

So, the conservation of energy in elastic collision leads to following equation:

M_{1} u_{1} +M_{2} u_{2}=M_{1}  v_{1}+M_{2}  v_{2}

Since, the momentum is conserved ,the kinetic energy will also be conserved in elastic collision. So

M_{1} u_{1} ^{2}+M_{2} u_{2} ^{2}=M_{1}v_{1} ^{2}+  M_{2}v_{2} ^{2}

Since initial velocity for M1 ball is zero, then

M_{2} u_{2}=M_{1}  v_{1}+M_{2}  v_{2}

and

M_{2} u_{2} ^{2}=M_{1}v_{1} ^{2}+  M_{2}v_{2} ^{2}

So, on solving all the above equation, we get an equation for velocity and that is

\frac{2M_{2}u_{2} }{(M_{1}+M_{2}  }=final velocity of ball with mass 2.5 kg

v = \frac{2(5*3.5)}{2.5+5}=4.67 m/s

So kinetic energy will be 1/2 mv2

Kinetic energy of 2.5 kg ball is \frac{1}{2}*2.5*(4.67)^{2}  =27.09 J

So the kinetic energy of 2.5 kg ball after collision is 27.09 J.

6 0
3 years ago
Two identical conducting spheres are placed 80.0 cm apart. One is given a charge of 5.8 C and the other is given a charge of 6.4
Zepler [3.9K]

Answer: 5.214(10)^{11} N

Explanation:

According to <u>Coulomb's Law:</u>  

<em>"The electrostatic force F_{E} between two point charges q_{1} and q_{2} is proportional to the product of the charges and inversely proportional to the square of the distance d that separates them, and has the direction of the line that joins them".</em>

<em />

Mathematically this law is written as:

F_{E}=K\frac{q_{1}.q_{2}}{d^{2}}  

Where:

F_{E}  is the electrostatic force

K=8.99(10)^{9} Nm^{2}/C^{2} is the Coulomb's constant  

q_{1}=5.8 C and q_{2}=6.4 C are the electric charges

d=80 cm \frac{1 m}{100 cm}=0.8 m is the separation distance between the charges

Solving:

F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{(5.8 C)(6.4 C)}{(0.8 m)^{2}}  

F_{E}=5.214(10)^{11} N    

7 0
3 years ago
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