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Shtirlitz [24]
3 years ago
7

Are my answers correct?

Physics
1 answer:
Tema [17]3 years ago
5 0

Explanation:

Your answers to 1 and 4 are correct, and your answer to 5 is half correct.

2. d = ((vf + vi) / 2) t

To solve for t, divide both sides by what's in the parenthesis:

d / ((vf + vi) / 2) = t

If you wish, you can simplify:

t = 2d / (vf + vi)

3. d = vi t + ½ a t²

To solve for a, first subtract vi t from both sides:

d − vi t = ½ a t²

Multiply both sides by 2:

2d − 2 vi t = a t²

Divide by t²:

a = (2d − 2 vi t) / t²

To solve for t when vi = 0, first substitute 0 for vi:

d = (0) t + ½ a t²

d = ½ a t²

Multiply both sides by 2:

2d = a t²

Divide both sides by a:

t² = 2d / a

Take the square root:

t = √(2d / a)

5. Fg = G m₁ m₂ / r²

To solve for r, first multiply both sides by r²:

Fg r² = G m₁ m₂

Divide both sides by Fg:

r² = G m₁ m₂ / Fg

Take the square root;

r = √(G m₁ m₂ / Fg)

Your other answer is correct.

Overall, good effort!

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Norma-Jean [14]

Answer:

1.\theta=29.84^{0}

2.\theta=60.15^{0}

Explanation:

Polarizes axis can create two possible angles with the vertical.

first we have to find the intensity of  first polarizer

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I=\frac{I_{0} }{2}

I= \frac{655\frac{W}{M^{2} } }{2}

I=327.5\frac{W}{m^{2} }

For a smaller angle for the first polarizer:

According to Malus Law

I_{2} =I_{1} Cos^{2}(90^{0} - \theta)

I_{2} =I_{1} sin^{2}\theta

\frac{I_{2} }{I_{1} }=Sin^{2}\theta

taking square root on both sides

\sqrt{\frac{163}{327.5} } = sin\theta

\theta=Sin^{-1}(0.4977)

\theta=29.84^{0}

For a larger angle for the first polarizer:

According to Malus Law

I_{2} =I_{1} cos^{2}\theta

\frac{I_{2} }{I_{1} }=Cos^{2}\theta

taking square root on both sides

\sqrt{\frac{163}{327.5} } = cos\theta

\theta=Cos^{-1}(0.4977)

\theta=60.15^{0}

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Electrons have a negative charge.
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How much force is needed to accelerate an object of mass 90 kg at a rate of 1.2 m/s2
soldier1979 [14.2K]
∑F = ma = (90 kg)(1.2 m/s²) = 108 N = 100 N (1 significant digit)

3 0
3 years ago
Read 2 more answers
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