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Aleks [24]
3 years ago
7

A charged particle with charge of 2 (uC) and mass 10-20 (kg) is traveling with velocity of 108 (m/s) in space. The charge reache

s to a region in space with magnetic field of 0.05 (T) and experience a force of 8 (N) exerted by the magnetic field.
A- What is the angle between velocity of particle and magnetic field direction?
B- What is acceleration of charged particle while experiencing the force?
Physics
1 answer:
-BARSIC- [3]3 years ago
3 0

Answer:

Explanation:

A ) Let the angle be θ between magnetic field and velocity of charged particle

Force created on charged particle F

= Bqv sinθ, B is magnetic field , q is charge , v is velocity of charged particle

F = .05 x 2 x 10⁻⁶ x 10⁸ x sinθ

8 = 10 sinθ

sinθ = .8

θ =  53°.

B )

acceleration = force / mass

= 8 / 10⁻²⁰

= 8 x 10²⁰ m / s²

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8 0
2 years ago
You have 2 resistors of unknown values you label Ra and Rb. You have an old battery and a multimeter you bought years ago for 7$
Veseljchak [2.6K]

Answer:

the answers, the correct one is D,   Rb₂ = 29.97 ohm

Explanation:

For this exercise we use ohm's law and the equivalent resistance ratio for series and parallel circuits.

Serial circuit

         (Ra + Rb) is = V

         (Ra + Rb) 0.111 = 10

         (Ra + Rb) = 10 / 0.111 = 90.09

parallel circuit

         \frac{1}{R} = \frac{1}{Ra} + \frac{1}{Rb}

           R = \frac{Ra \ Rb}{Ra + Rb}

          \frac{Ra \ Rb}{Ra + Rb} i_p = V

         \frac{Ra \ Rb}{Ra + Rb} 0.5 = 10

         \frac{Ra \ Rb}{Ra + Rb} = 10 / 0.5 = 20

we write and solve our system of equations

         Ra + Rb = 90.09

         \frac{Ra \ Rb}{Ra + Rb} = 20

         

we solve for Ra in the first equation

          Ra = 90.09 - Rb

           RaRb = 20 (Ra + Rb)

we substitute Ra in the second equation

           (90.09-Rb) Rb = 20 [(90.09-Rb) + Rb]

            90.09 Rb - Rb² = 20 90.09

           

            Rb² - 90.09 Rb + 1801.8 = 0

we solve the quadratic equation

            Rb = [90.09 ±\sqrt{90.09^2 - 4 \ 1801.8} ] / 2

            Rb = [90.09 ± 30.15] / 2

            Rb₁ = 60.12 ohm

            Rb₂ = 29.97 ohm

the smallest value is Rb = 30 ohm

When checking the answers, the correct one is D

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Hi, to answer this question we have to apply the next formula:

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