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Brut [27]
3 years ago
6

Radio astronomers detect electromagnetic radiation at a frequency of 42MHz from an interstellar gas cloud. If the radiation resu

lts from electrons spiraling in a magnetic field, What is the magnetic field strength inside the gas cloud?
Physics
1 answer:
ExtremeBDS [4]3 years ago
5 0

Answer:

Explanation:

The electrons will also spiral with frequency of 42 x 10⁶

n = 42 x 10⁶

For circular motion of electron

= Bqv = m ω² r ( centripetal force )

v / r = m ω² / Bq

= 4π² n² m / Bq

T =  2πr / v

1 / n  = 2πr / v

r / v = 1 / 2π n

2π n = ω²m / Bq

ω = ω² m/ Bq

B =  ω m / q

B = 2π nm / q

= 2  x 3.14 x 42 x 10⁶ x 9.1 x 10⁻³¹ / 1.6 x 10⁻¹⁹

= 15 x 10⁻⁴  T

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astraxan [27]

Answer: (a)F=7(10)^{-7}N

              (b)F=1.344(10)^{-6}N  

              (c) The force of Jupiter on the baby is slightly greater than the the force of the father on the baby.

Explanation:

According to the law of universal gravitation, which is a classical physical law that describes the gravitational interaction between different bodies with mass:

F=G\frac{m_{1}m_{2}}{r^2}   (1)

Where:

F is the module of the force exerted between both bodies

G is the universal gravitation constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

m_{1} and m_{2} are the masses of both bodies.

r is the distance between both bodies

Knowing this, let's begin with the answers:

<h2 /><h2>(a) Gravitational force Father exertes on baby</h2>

Using equation (1) and taking into account the mass of the father m_{1}=100kg, the mass of the baby m_{2}=4.20kg and the distance between them r=0.2m, the force F_{F}  exerted by the father is:

F_{F}=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(100kg)(4.20kg)}{(0.2m)^2}   (2)

F_{F}=0.0000007N=7(10)^{-7}N   (3)

<h2>(b) Gravitational force Jupiter exertes on baby</h2>

Using again equation (1) but this time taking into account the mass of Jupiter m_{J}=1.898(10)^{27}kg, the mass of the baby m_{2}=4.20kg and the distance between Jupiter and Earth (where the baby is) r_{E}=6.29(10)^{11}m, the force F_{J}  exerted by the Jupiter is:

F_{J}=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(1.898(10)^{27}kg)(4.20kg)}{(6.29(10)^{11}m)^2}   (4)

F_{J}=0.000001344N=1.344(10)^{-6}N   (5)

<h2>(c) Comparison</h2>

Now, comparing both forces:

F_{J}=0.000001344N=1.344(10)^{-6}N   and F_{F}=0.0000007N=7(10)^{-7}N  we can see F_{J} is greater than F_{F}. However, the difference is quite small as well as the force exerted on the baby.

7 0
3 years ago
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