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morpeh [17]
4 years ago
9

The Hubble Space Telescope obtains higher resolution images than most ground based telescopes because A) it is closer to the sta

rs. B) it is larger. C) it is above Earth’s atmosphere. D) it is always on the night side of Earth, i.e. it is never affected by the Sun.
Physics
1 answer:
Sergeu [11.5K]4 years ago
7 0

Answer:

option C

Explanation:

The  Hubble Space Telescope is the telescope that is above the atmosphere.  It was launched into low earth orbit in the year 1990.

It is one of the largest and most versatile telescope presents.

The  Hubble Space Telescope obtains a high-resolution image than the telescope on the ground because the obstruction of the telescope on the ground is more as compared to than outer atmosphere .i.e. Hubble Space Telescope obtains higher resolution images.

The correct answer is option C

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An l-c cirucit with a 70 mh inductor and a ,54 F capacitor oscillates. The maximum charge on the capacitor is 11.5 C. What are t
Oxana [17]

This question is incomplete, the complete question is;

An l-c cirucit with a 70 mh inductor and a 0.54 μF capacitor oscillates. The maximum charge on the capacitor is 11.5 μC. What are the oscillation frequency and the maximum current in this circuit ;

Options

a) 1.07 kHz, 63.4 mA

b) 4.38 kHz, 101.3 mA

c) 6.74 kHz, 55.7 mA

d) 2.31 kHz, 93.5 mA

e) 0.82 kHz, 59.1 mA

Answer:

the oscillation frequency and the maximum current in this circuit are; 0.82 kHz and 59.1 mA respectively.

so Option e) 0.82 kHz, 59.1 mA is the correct answer

Explanation:

Given that;

inductor L = 70 mH = 70 × 10⁻³ H

Capacitor C = 0.54 μf = 0.54 × 10⁻⁶ f

Qmax on capacitor = 11.5 μf = 11.5 × 10⁻⁶ c

oscillation frequency in L-C circuit;

f = 1/2π√(LC)

we substitute our values;

f = 1/2π√(70 × 10⁻³ × 0.54 × 10⁻⁶  )

f = 0.0818 × 10⁴ Hz

f = 0.082 × 10³ Hz ≈ 0.82 kHz

Maximum circuit in L-C circuit is given by

I_max = Qmax/√(LC)

we substitute

I_max = 11.5 × 10⁻⁶ / √(70 × 10⁻³ × 0.54 × 10⁻⁶  )

= 59.1 × 10³ A ≈  59.1 mA

Therefore the oscillation frequency and the maximum current in this circuit are; 0.82 kHz and 59.1 mA respectively.

so Option e) 0.82 kHz, 59.1 mA is the correct answer

5 0
3 years ago
A simple pendulum 2 m long swings through a maximum angle of 30 ∘ with the vertical. part a calculate its period assuming a smal
victus00 [196]

Length of the pendulum (l) = 2 m

Acceleration due to gravity = g = 9.8 m/s^2

For small amplitude, the pendulum will undergo simple harmonic motion.

Hence, the time period of the pendulum for small amplitude = 2\pi \sqrt{\frac{l}{g} }

Now, plug the values of l and g

T = 2\pi \sqrt{\frac{2}{9.8} }

T = 2 × 3.14 × 0.451

T = 2.83 seconds

Hence, the time period of the pendulum for small amplitude = 2.83 s

3 0
3 years ago
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Which requires more work, increasing a car's speed from 0 mph to 30 mph or from 50 mph to 60 mph?
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Increasing a car's speed from 0 mph to 30 mph requires more work than that of 50 mph to 60 mph.

<h3>What is work?</h3>

Work is a measure of energy expended in moving an object. It is calculated by multiplying the force by the distance.

According to this question, work is required to move a car from one position to another.

The distance moved from 0 - 30mph is more than that moved from 50 - 60mph, therefore, increasing a car's speed from 0 mph to 30 mph requires more work than that of 50 mph to 60 mph.

Learn more about work at: brainly.com/question/19382352

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2 years ago
Why is electronic balance more accurate?
Citrus2011 [14]
Precision balances have a readability of 0.01 g. They produce steady readings in a wider range of environmental conditions than analytical balances, being less sensitive to temperature fluctuations.
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Which of the following best describes the velocity of an object? *
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D fam hope this helps
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