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Sunny_sXe [5.5K]
3 years ago
13

If the sun suddenly ceased to shine, how long would it take earth to become dark? you will have to look up the speed of light in

a vacuum and the distance from the sun to earth.
Physics
1 answer:
sleet_krkn [62]3 years ago
3 0
We are 8 light minutes from the sun. That means two things, we see the sun as it was 8 minutes ago, and we WOULD continue to see the sun for 8 minutes after it disappeared.
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How does the Coriolis effect impact ocean currents in the Northern and Southern Hemispheres?
Strike441 [17]
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3 years ago
What is one way you can state how you feel?
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3 0
3 years ago
Read 2 more answers
1. A signal source has an open-circuit voltage of 1V, and a short-circuit current of10mA. What is the source resistance
LiRa [457]

Answer:

100 Ω

Explanation:

Given that

Open circuit voltage, V = 1 V

Short circuit current, I = 10 mA

Source resistance R, = ?

This is rather a straight forward question. Remember Ohms Law? Current being directly proportional to the voltage and inversely proportional to the resistance?

Yeah, that's the formula we'd be using.

Ohms Law states that V = IR, and thus, if we make R subject of the formula, we have

R = V / I, on substituting the values, we have

R = 1 / 10*10^-3

R = 1 / 0.01

R = 100 Ω

8 0
3 years ago
A cube is 4.4 cm on a side, with one corner at the origin. Part 1 (a) What is the unit vector pointing from the origin to the di
Sidana [21]

Answer:

(a) \hat{A} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}

(b) \theta = 85.44^{\circ}

Solution:

As per the question:

Side of the cube, a = 4.4 cm

Coordinates of the diagonally opposite corner, A = <4.4, 4.4, 4.4> cm

Now,

(a) To calculate the unit vector:

\hat{A} = \frac{\vec{A}}{|A|}

\hat{A} = \frac{4.4\hat{i} + 4.4\hat{j} + 4.4\hat{k}}{\sqrt{()4.4}^{2} + (4.4)^{2} + (4.4)^{2}}

\hat{A} = \frac{4.4(\hat{i} + \hat{j} + \hat{k})}{4.4\sqrt{3}}

\hat{A} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}

(b) To calculate the angle between the two vectors say A and A' is given by:

\vec{A}\cdot \vec{A'} = \vec{A}\vec{A'}cos\theta                      

\theta = cos^{- 1}(\frac{\vec{A}\cdot \vec{A'}}{\vec{A}\vec{A'}})        (1)

Now,

The coordinates of the diagonally opposite corner, A' is <0, 0, 1> cm

Thus

\vec{A'} = 0\hat{i} + 0\hat{j} + 1\hat{k} = \hat{k}

Now,

Using equation (1) :

\theta = cos^{- 1}(\frac{(\frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}})\cdot \hat{k}}{|A||A'|})

|A||A'| = (\sqrt{4.4^{2} +4.4^{2} + 4.4^{2}})(\sqrt{0^{2} + 0^{2} + 0^{2}}) = 7.261

Thus

\theta = cos^{- 1}(\frac{\frac{1}{\sqrt{3}}}{7.261})

\theta = cos^{- 1}(0.07946) = 85.44^{\circ}

4 0
3 years ago
A car accelerates uniformly in a straight line
julia-pushkina [17]

The car travels a distance <em>d</em> from rest with acceleration <em>a</em> after time <em>t</em> of

<em>d</em> = 1/2 <em>a</em> <em>t</em>²

It covers 69 m with 2.8 m/s² acceleration, so that

69 m = 1/2 (2.8 m/s²) <em>t</em>²

<em>t</em>² = 2 (69 m) / (2.8 m/s²)

<em>t</em> ≈ 7.02 s

where we take the positive square root because we're talking about time *after* the car begins accelerating.

8 0
3 years ago
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