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3241004551 [841]
3 years ago
12

A charged particle enters a uniform magnetic field B_vec with a velocity v_vec at right angles to the field.It moves in a circle

with period T.1. If a second particle, with the same electric charge but tentimes as massive, enters the field with the same velocity v_vec, what is its period?A. T/10B. TC. 5TD. 10T
Physics
1 answer:
Finger [1]3 years ago
8 0

Answer:

D. 10 T

Explanation:

When a particle is moving in a magnetic field, the magnetic force provides the centripetal force that keeps the particle in circular motion.

The cyclotron period (the period the particle takes to complete one orbit) can be found to be

T=\frac{2\pi m}{qB}

where

m is the mass of the particle

q is its charge

B is the magnetic field

As we see, the period is directly proportional to the mass of the particle.

In this problem, the second particle is ten times as massive as the first one:

m' = 10 m

while the speed is the same. So, the period of the second particle is

T'=\frac{2\pi m'}{qB}=\frac{2\pi (10m)}{qB}=10 \frac{2\pi m}{qB}=10 T

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When a material in liquid state is given energy, it changes its phase from liquid to vapor; the energy absorbed in this process is called heat of vaporization. The heat of vaporization of water is about 2,260 kJ/kg, which is equal to 40.8 kJ/mol. The vaporization is the opposite process of condensation.

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3 years ago
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CaHeK987 [17]

Explanation:

What is the weight of a 2.00-kilogram object on the surface of Earth?

2.00 N

4.91 N

9.81 N

19.6 N

Given parameters:

Mass of the object = 2kg

Unknown:

Weight of the object  = ?

Solution:

The weight of an object is the force of gravity acting on the object;

 Weight  =  mass x acceleration due to gravity

Acceleration due to gravity  = 9.8m/s²

 Now insert the parameters and solve;

    Weight  = 2 x 9.8 = 19.6N

A person weighing 785 Newtons on the surface of the Earth would weigh 47 Newtons on the surface of Pluto. What is the magnitude of the gravitational acceleration on the surface of Pluto?

1.7 m/s²

0.59 m/s²

0.29 m/s²

9.8 m/s²

Given parameters:

Weight on Earth  = 785N

Weight on Pluto = 47N

Unknown:

Acceleration due to gravity on Pluto = ?

Solution

The mass of the body both on Earth and Pluto is the same.

  Weight = mass x acceleration due to gravity

Now find the mass on Earth;

  Acceleration due to gravity on Earth  = 9.8m/s²

    785   = mass x 9.8

         mass  = \frac{785}{9.8}   = 80.1kg

So;

  Acceleration due to gravity on Pluto = \frac{Weight on Pluto}{mass }  

  Acceleration due to gravity  = \frac{47}{80.1}   = 0.59m/s²

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3 years ago
Certain gases in the atmosphere – water vapor, carbon dioxide, methane and nitrous oxIde – help maintain the Earth’s temperature
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Answer:

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Explanation:

8 0
3 years ago
Three charges, Q1, Q2, and Q3 are located in a straight line. The position of Q2 is 0.301 m to the right of Q1. Q3 is located 0.
Alexxx [7]

Answer and Explanation: A charge exerts a force over another charge even if they are very far apart. This force is called <u>Electrostatic</u> <u>Force</u>.

If the two charges have the same sign, e.g. both aare positive, the force between them is opposite. If they have opposite sign, the force is towards each other. In other words, for electrostatic force, equal charges repel and different charges attract.

So,

1. If Q2 and Q3 have opposite signs, it is TRUE force in Q2 will go the left;

2. If the 2 are negative, they have the same sign, so it's FALSE force is to the right;

Sentences 3 and 4 are also TRUE due to the reasons described above;

5. If the charges have opposite signs, it means force is towards each other, or, to the right, so the sentence is TRUE;

1. Force is directly proportional to charges in Coulomb [C] and inversely proportional to distance squared in [m]:

F=\frac{k.q.Q}{r^{2}}

where k is a constant that equals 9 x 10⁹ N.m²/C²

Calculating force between 1 and 2:

F_{12}=\frac{9.10^{9}(1.9.10^{-6})(2.84.10^{-6})}{(0.301)^{2}}

F_{12}=536.02.10^{-3} N

Force between 2 and 3:

F_{23}=\frac{9.10^{9}(2.84.10^{-6})(3.03.10^{-6})}{(0.169)^{2}}

F_{23}=2711.63.10^{-3} N

Total force is the net force. Since Q2 is negative and the others are positive, force of 2 related to 1 is to left and related to 3 is to the right. Therefore, total force is the difference between those two forces:

F_{T}=2711.63.10^{-3}-536.02.10^{-3}

F_{T}=2175.61.10^{-3} N

The total force on Q2 is 2175.61 x 10⁻³ N

2. For net force to be 0, F_{13}=F_{23}. Suppose distance from 1 to 3 is x, then from 2 to 3 is x-0.301

Calculating:

\frac{k(1.90.10^{-6})(3.03.10^{-6})}{x^{2}}=\frac{k(2.84.10^{-6})(3.03.10^{-6})}{(x-0.301)^{2}}

\frac{5.757.10^{-12}}{x^{2}} =\frac{8.6052.10^{-12}}{x^{2}-0.602x+0.090601}

\frac{5.757.10^{-12}}{8.6052.10^{-12}}=\frac{x^{2}}{x^{2}-0.602x+0.090601}

x^{2}=0.67x^{2}-0.40x+0.061

0.33x^{2}+0.40x-0.061=0

roots = 0.14 or -1.35

Solving quadratic equation gives 2 roots, but one of the roots is negative. As distance is a measure that cannot be negative, the solution is x = 0.14.

The distance of Q3 relative to Q1 is 0.14 m

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elena-s [515]
The answer I received was 1561 kg

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