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frutty [35]
2 years ago
7

A particle is constrained to move round a circle radius 382400km and makes a single revolution in 27.3 days. (i). Find the veloc

ity in ms-1. (ii). Find its acceleration
Physics
1 answer:
andreyandreev [35.5K]2 years ago
5 0

The velocity and acceleration of the particle moving round the circle is;

<em><u>Velocity = 162.12 m/s</u></em>

<em><u>Velocity = 162.12 m/sAcceleration = 6.873 × 10^(-5) m/s²</u></em>

We are given;

Radius of circle; 382400 km = 382400000 m

Time; t = 27.3 days = 27.3 × 86400 s = 2358720 s

Now, formula for velocity is;

Velocity = distance/time

Thus;

I) velocity = 382400000/2358720

Velocity = 162.12 m/s

II) Acceleration is centripetal acceleration and is given by the formula;

a = v²/r

a = 162.12²/382400000

a = 6.873 × 10^(-5) m/s²

Read more at; brainly.com/question/12199398

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A linear accelerator produces a pulsed beam of electrons. The pulse current is 0.50 A, and the pulse duration is 0.10 μs. (a) Ho
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Answer:

a)N = 3.125 * 10¹¹

b) I(avg)  = 2.5 × 10⁻⁵A

c)P(avg) = 1250W

d)P = 2.5 × 10⁷W

Explanation:

Given that,

pulse current is 0.50 A

duration of pulse Δt = 0.1 × 10⁻⁶s

a) The number of particles equal to the amount of charge in a single pulse divided by the charge of a single particles

N = Δq/e

charge is given by Δq = IΔt

so,

N = IΔt / e

N = \frac{(0.5)(0.1 * 10^-^6)}{(1.6 * 10^-^1^9)} \\= 3.125 * 10^1^1

N = 3.125 * 10¹¹

b) Q = nqt

where q is the charge of 1puse

n = number of pulse

the average current is given as I(avg) = Q/t

I(avg) = nq

I(avg) = nIΔt

         = (500)(0.5)(0.1 × 10⁻⁶)

         = 2.5 × 10⁻⁵A

C)  If the electrons are accelerated to an energy of 50 MeV, the acceleration voltage must,

eV = K

V = K/e

the power is given by

P = IV

P(avg) = I(avg)K / e

P(avg) = \frac{(2.5 * 10^-^5)(50 * 10^6 . 1.6 * 10^-^1^9)}{1.6 * 10^-^1^9}

= 1250W

d) Final peak=

P= Ik/e

= = P(avg) = \frac{(0.5)(50 * 10^6 . 1.6 * 10^-^1^9)}{1.6 * 10^-^1^9}\\2.5 * 10^7W

P = 2.5 × 10⁷W

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3 years ago
Read 2 more answers
Calculate the phase angle (in radians) for a circuit with a maximum voltage of 12 V and w-50 Hz. The voltage source is connected
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Answer:

The phase angle is 0.0180 rad.

(c) is correct option.

Explanation:

Given that,

Voltage = 12 V

Angular velocity = 50 Hz

Capacitance C= 20\times10^{-2}\ F

Inductance L=20\times10^{-3}\ H

Resistance R=  50\ Omega

We need to calculate the impedance

Using formula of impedance

z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})^2}

z=\sqrt{50^2+(50\times20\times10^{-3}-\dfrac{1}{50\times20\times10^{-2}})^2}

z=50.00

We need to calculate the phase angle

Using formula of phase angle

\theta=\cos^{-1}(\dfrac{R}{z})

\theta=\cos^{-1}(\dfrac{50}{50.00})

\theta=0.0180\ rad

Hence, The phase angle is 0.0180 rad.

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