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Over [174]
3 years ago
11

6)the speed of light is approximately​ 186,000 mi/sec. It takes light from a particular star approximately 9 yrs to reach Earth.

How many miles away is the star from​ Earth? Express the answer in scientific notation. Use 365 days in 1 year. The star is nothing miles away from Earth.
Physics
1 answer:
NeTakaya3 years ago
5 0

Answer:

5.2791264*10¹³

Explanation:

Convert the 9 years to seconds and then multiple it by 186000

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A proton is 0.9 meters away from a 1.4 C charge. What is the magnitude of the electric force between the proton and the charge
Digiron [165]

Answer:

F = 2.49 x 10⁻⁹ N

Explanation:

The electrostatic force between two charged bodies is given by Colomb's Law:

F = \frac{kq_1q_2}{r^2}\\

where,

F = Electrostatic Force = ?

k = colomb's constant = 9 x 10⁹ N.m²/C²

q₁ = charge on proton = 1.6 x 10⁻¹⁹ C

q₂ = second charge = 1.4 C

r = distace between charges = 0.9 m

Therefore,

F = \frac{(9\ x\ 10^9\ N.m^2/C^2)(1.6\ x\ 10^{-19}\ C)(1.4\ C)}{(0.9\ m)^2}

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3 years ago
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Explain with one example that different types of organisums live in one habitation​
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Answer:

When you look at a simple koi pond you can find Koi (the secondary consumer) that feeds off of the zooplankton (first consumer), they eat the phytoplankton (producers). All in a simple food chain

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Basically, Koi eat the little animal plankton (zooplankton) that then eats the plant plankton (phytoplankton) that can only end when a part of that habitat is removed. If you got rid of the plant plankton then the whole chain would collapse and most likely die.

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4 years ago
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Space debris left from old satellites and their launchers is becoming a hazard to other satellites. (a) Calculate the speed of a
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Answer:

Part a)

v = 7407.1 m/s

Part b)

v_{rel} = 1.05 \times 10^4 m/s

Explanation:

Part a)

As we know that orbital velocity at certain height from the surface of Earth is given as

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here we know that

M = 5.98 \times 10^{24} kg

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now we have

v = \sqrt{\frac{(6.67 \times 10^{-11})(5.98 \times 10^{24})}{6.37 \times 10^6 + 9.0 \times 10^5}}

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Part b)

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v_{rel} = \sqrt{2} v

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