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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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#6 - Lower frequencies correspond to <em>longer</em> wavelengths.

#7 - To change the pitch of a sound, you have to change its <em>frequency</em>.

6 0
4 years ago
A motorboat approaches you head-on and sounds one short blast. how do you signal that you agree to let it pass on your port (lef
nadya68 [22]
The blast made by the vessel's driver is telling you that he intends to pass by your left side. If you agree with his intention, you will respond by sounding one short blast in return. If you are not in agreement with this move or you do not really understand the intention of the driver, you will give the danger signal which is five short rapid blasts.
6 0
3 years ago
The speed of light in vinegar is 2.30 x 10^8 m/s. Determine the index of refraction. (2​
pantera1 [17]

Answer:

n _{v} =  \frac{c}{v}  \\  =  \frac{3 \times  {10}^{8} }{2.30 \times  {10}^{8} }  \\  = 1.30

6 0
3 years ago
A motorcycle is capable of accelerating at 5.1 m/s starting feom rest how far can it travel im 1.5 seconds
saul85 [17]

5.1 m/ s


~Multiply 1.5 on both sides since you’re trying to finding out how far it can travel in 1.5 seconds


5.1 x 1.5= 7.65

s x 1.5= 1.5


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6 0
3 years ago
A block of mass 5.6 kg is attached to a horizontal spring on a rough floor. Initially the spring is compressed 3.5 cm. The sprin
Mariana [72]

Answer:

The velocity of block = 0.188 \frac{m}{s}

Explanation:

Mass m = 5.6 kg

k = 1040 \frac{N}{m}

\mu = 0.26

x_{1} = 0.035 m  , v_{1} = 0

x_{2} = 0.02 m

From work energy theorem

K_{1} + U_{1} + W_{other} = K_{2} + U_{2}  --------- (1)

Kinetic energy

K = \frac{1}{2} k x^{2}  ------- (1)

Potential energy

U = \frac{1}{2} k x^{2} ------- (2)

Work done

W = F.s ------ (3)

From Newton's second law

R_{N} = mg

R_{N}  = 5.6 × 9.81 = 54.9 N

Friction  force = 0.4 × 54.9 = 21.9 N

Now the work done by the friction

W_{f} = - 21.9 × 0.015

W_{f} = - 0.329 J

Now kinetic energy

At point 1

K_{1} = \frac{1}{2} m v^{2} _{1}

K_{1} = 0

U_{1} = \frac{1}{2} k x^{2}

U_{1} = \frac{1}{2}  (1040) 0.035^{2}

U_{1} = 0.637 J

At point 2

K_{2} = \frac{1}{2} (5.6) v^{2} _{2}

K_{2} = 2.8 v_{2} ^{2}

Potential energy

U_{2} = \frac{1}{2}  k x_2^{2}

U_{2} = \frac{1}{2}  (1040) 0.02^{2}

U_{2} = 0.208 J

From equation (1) we get

0 + 0.637 - 0.329 = 2.8 v_{2} ^{2} + 0.208

2.8 v_{2} ^{2} = 0.1

v_{2} = 0.188 \frac{m}{s}

This is the velocity of block.

6 0
4 years ago
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