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vekshin1
3 years ago
15

The hubbles telescopes orbit is 5.6 x10 ^5 meters above earths suface. the telescope has a mass os 1.1 x10^4 kilograms. earth ex

erts a gravitational force of 9.1 x10^4 newtons on the telescope. what is the magnitude of earths gravitational field strength at this location
Physics
1 answer:
Andru [333]3 years ago
7 0
(3) 8.3 N/kg. The gravitational field strength at a point is the force per unit mass exerted on a mass placed at that point. So at the point where the Hubble telescope is, it is (9.1 x 10^4)N/(1.1 x 10^4 kg) = 8.3 N/kg

Fam
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The correct answer is Metals.

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morpeh [17]

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Explanation:

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3 years ago
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Two toy cars (m1 = 0.200 kg and m2 = 0.250 kg) are held together rear to rear with a compressed spring between them. When they a
timurjin [86]

Answer:

Acceleration of the second particle at that moment is given as

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So we will have

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now we have

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m_2 = 0.250 kg

a_1 = 2.65 m/s^2

now we have

0.200(2.65) + 0.250a_2 = 0

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3 years ago
Starting from zero, the electric current takes 2 seconds to reach half its maximum possible value in an RL circuit with a resist
Leno4ka [110]

Answer:

time=4s

Explanation:

we know that in a RL circuit with a resistance R, an inductance L and a battery of emf E, the current (i) will vary in following fashion

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Given that, at i(2)=\frac{imax}{2} =\frac{E}{2R}

⇒\frac{E}{2R}=\frac{E}{R}(1-e^\frac{-2}{\frac{L}{R}})

⇒\frac{1}{2}=1-e^\frac{-2}{\frac{L}{R}}

⇒\frac{1}{2}=e^\frac{-2}{\frac{L}{R}}

Applying logarithm on both sides,

⇒log(\frac{1}{2})=\frac{-2}{\frac{L}{R}}

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Now substitute i(t)=\frac{3}{4}imax=\frac{3E}{4R}

⇒\frac{3E}{4R}=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})

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Applying logarithm on both sides,

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⇒t=log4\frac{2}{log2}

also log4=log2^{2}=2log2

⇒t=2log2\frac{2}{log2}

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