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9966 [12]
3 years ago
8

A cosmic-ray proton in interstellar space has an energy of 59 MeV and executes a circular orbit with a radius equal to that of M

ercury’s orbit (5.8 × 1010 m) around the Sun. What is the magnetic field in that region of space? The proton has a charge of 1.60218 × 10−19 C and a mass of 1.67262 × 10−27 kg. Answer in units of T
Physics
1 answer:
makkiz [27]3 years ago
8 0

Answer:

Explanation:

Let mass and velocity of proton be m and v .

1/2 m v² = 59 x 10⁶ e V

= 59 x 10⁶ x 1.6 x 10⁻¹⁹ J

= 94.4 x 10⁻¹³ J

mv² = 188.8 x 10⁻¹³ J

v² =  188.8 x 10⁻¹³ / m

= 188.8 x 10⁻¹³ / 1.67262 x 10⁻²⁷

= 112.8768 x 10¹⁴

v = 10.62 x 10⁷ m / s

In circular path of proton , magnetic force equals centripetal force .

m v² / r = B q v , B is magnetic field , q is charge on proton , r is radius of circular path .

188.8 x 10⁻¹³ / 5.8 x 10¹⁰ = B x 1.6 x 10⁻¹⁹ x  10.62 x 10⁷

B = 1.9157 x 10⁻¹¹ T.

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The engine starter and a headlight of a car are connected in parallel to the 12.0-V car battery. In this situation, the headligh
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The total power they will consume in series is approximately 2.257 W

Explanation:

The connection arrangement of the headlight and the engine starter = Parallel to the battery

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The power at which the kick starter operates in parallel, P_{kick \ starter} = 2.40 kW

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V = The voltage = 12 V (The voltage is the same in parallel circuit)

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R₁ = V²/P_{headlight}  = 12²/38 = 72/19

R₁ = 72/19 Ω

For the kick starter, we have;

R₂ = V²/P_{kick \ starter} = 12²/2.4 = 60

R₂ = 60 Ω

When the headlight and kick starter are rewired to be in series, we have;

Total resistance, R = R₁ + R₂

Therefore;

R = ((72/19) + 60) Ω = (1212/19) Ω

The current flowing, I = V/R

∴ I = 12 V/(1212/19) Ω = (19/101) A

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∴ P_{headlight} = ((19/101) A)² × 72/19 Ω = 1368/10201 W ≈ 0.134 W

The power at which the headlight operates in series, P_{headlight, S} ≈ 0.134 W

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The power at which the kick starter operates in series, P_{kick \ starter, S} ≈ 2.123 W

The total power they will consume, P_{Total} = P_{headlight, S} + P_{kick \ starter, S}

Therefore;

P_{Total} ≈ 0.134 W + 2.123 W = 2.257 W

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