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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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Licemer1 [7]

The kinematic energy of the positive charge is 2 10⁻⁸ J

This electrostatics exercise must be done in parts, the first part: let's start by finding the charge of the capacitor, the capacitance is defined by

        C = \frac{Q}{\Delta V}

        C = ε₀ \frac{A}{d}

we solve for the charge (Q)

        \frac{Q}{\Delta V} = \epsilon_o \frac{A}{d}

indicates that for the initial point d₁ = 3 mm = 0.003 m and the voltage is DV₁ = 12

         Q = \epsilon_o \  \frac{A \ \Delta V_1 }{d_1}

Now the voltage source is disconnected so the charge remains constant across the ideal capacitor.

For the second part, the condenser is separated at d₂ = 5mm = 0.005 m

         Q = \epsilon_o \  \frac{A \ \Delta V_2 }{d_2}

we match the expressions of the charge and look for the voltage

          \frac{\Delta V_1}{d_1} = \frac{\Delta V_2}{d_2}

          ΔV₂ = \frac{d_2}{d_1 } \ \Delta V_1

The third part we use the concepts of conservation of energy

starting point. With the test load (q = 1 nC = 1 10⁻⁹ C) next to the left plate

          Em₀ = U = q DV₂

          Em₀ = q  \frac{d_2}{d_1 } \ \Delta V_1

           

final point. Proof load on the right plate

         Em_f = K

energy is conserved

         Em₀ = em_f

         q  \frac{d_2}{d_1 } \ \Delta V_1 = K

   

we calculate

         K = 1 10⁻⁹  12  \frac{0.005}{0.003}  

         K = 20 10⁻⁹ J

In this exercise, as the conditions at two different points of separation give, the area of ​​the condenser is not necessary and with conservation of energy we find the final kinetic energy of 2 10⁻⁸ J

3 0
2 years ago
Two balloons (m = 0.021 kg) are separated by a distance of d = 16 m. They are released from rest and observed to have an instant
evablogger [386]

(a) 2.56\cdot 10^{-5} C

According to Newton's second law, the force experienced by each balloon is given by:

F = ma

where

m = 0.021 kg is the mass

a = 1.1 m/s^2 is the acceleration

Substituting, we found:

F=(0.021)(1.1)=0.0231 N

The electrostatic force between the two balloons can be also written as

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

where

k is the Coulomb's constant

Q is the charge on each balloon

r = 16 m is their separation

Since we know the value of F, we can find Q, the magnitude of the charge on each balloon:

Q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(0.0231)(16)^2}{9\cdot 10^9}}=2.56\cdot 10^{-5} C

(b) 1.6\cdot 10^{14} electrons

The magnitude of the charge of one electron is

e=1.6\cdot 10^{-19}C

While the magnitude of the charge on one balloon is

Q=2.56\cdot 10^{-5} C

This charge can be written as

Q=Ne

where N is the number of electrons that are responsible for this charge. Solving for N, we find:

N=\frac{Q}{e}=\frac{2.56\cdot 10^{-5}}{1.6\cdot 10^{-19}}=1.6\cdot 10^{14}

5 0
3 years ago
PLEASE HELP!! 13 POINTS! SCIENCE!
pogonyaev

Answer:

A.

Explanation:

3 0
3 years ago
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Paha777 [63]

Answer:

A) V = 4.92 \cdot 10^{-4} m^{3} = 492 cm^{3}

B) d = 4181.49 kg/m^{3} = 4.18 g/cm^{3}

Explanation:

A) Using the Archimedes' force we can find the weight of water displaced:

W_{d} = W_{a} - W_{w}

Where:

W_{a}: is the weight of the block in the air = 20.1 N

W_{w}: is the weight of the block in the water = 15.3 N

W_{d} = W_{a} - W_{w} = 20.1 N - 15.3 N = 4.8 N

Now, the mass of the water displaced is:

m = \frac{W_{d}}{g} = \frac{4.8 N}{9.81 m/s^{2}} = 0.49 kg

The volume of the block can be found using the mass of water displaced and the density of the water:

V = \frac{m}{d} = \frac{0.49 kg}{997 kg/m^{3}} = 4.92 \cdot 10^{-4} m^{3} = 492 cm^{3}

B) The density of the block can be found as follows:

d = \frac{W_{a}}{g*V} = \frac{20.1 N}{9.81 m/s^{2}*4.92 \cdot 10^{-4} m^{3}} = 4181.49 kg/m^{3} = 4.18 g/cm^{3}

I hope it helps you!            

6 0
3 years ago
Which of the following statements is true regarding the constructive interference diagram shown below? Select all that apply.
yulyashka [42]

Answer:

I believe its B and C

Explanation:

<u><em>If I'm wrong please tell me so I can correct my answer.</em></u>

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6 0
2 years ago
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