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Novosadov [1.4K]
2 years ago
9

What is the answer to 895cm a Km

Physics
1 answer:
victus00 [196]2 years ago
8 0

Answer:

if you're converting then the answer is 0.00895

Explanation:

895 centimetres converted into kilometres= 0.00895

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A feather, a tennis ball, and a bowling ball are dropped from a high tower on
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In the absence of air resistance. I think no. D ) The bowling ball.

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3 years ago
Visible light passes through a diffraction grating that has 900 slits per centimeter, and the interference pattern is observed o
love history [14]

Answer:

13.51 nm

Explanation:

To solve this problem, we are going to use angle approximation that sin θ ≈ tan θ ≈ θ where our θ is in radians

y/L=tan θ ≈ θ

and ∆θ ≈∆y/L

Where ∆y= wavelength distance= 2.92 mm =0.00292m

L=screen distance= 2.40 m

=0.00292m/2.40m

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The grating spacing is d = (90000 lines/m)^−1

=1.11 × 10−5 m.

the small-angle

approx. Using difraction formula with m = 1 gives:

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∆λ ≈ d∆θ = =1.11 × 10^-5 m×0.001217 rad

=0.000000001351m

= 13.51 nm

6 0
3 years ago
5. Why does a Ntion try to influence the value of its currency?
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Answer:

If inflation is relatively lower than competitors, then the countries goods will become more attractive and demand will rise. Lower inflation tends to increase the value of the currency in the long term.

Explanation:

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2 years ago
What happened to the speed of light if it travels from air into glass?
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Answer:

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3 years ago
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A stone is dropped into a river from a bridge at a height h above the water. Another stone is thrown vertically down at a time t
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Answer:

v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}

Explanation:

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First stone:

Let us denote the time spent after the second stone is thrown as 'T'.

y - y_0 = v_{y_0}(t+T) + \frac{1}{2}a(t+T)^2\\0 - h = 0 + \frac{1}{2}(-g)(t+T)^2\\(t+T)^2 = \frac{2h}{g}\\T = \sqrt{\frac{2h}{g}}-t

Second stone:

y - y_0 = v_{y_0}T + \frac{1}{2}aT^2\\0 - h = v_{y_0}T -\frac{1}{2}gT^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\sqrt{\frac{2h}{g}} - t)^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\frac{2h}{g} + t^2 - 2t\sqrt{\frac{2h}{g}})\\-h = v_{y_0}\sqrt{\frac{2h}{g}} - v_{y_0}t - h -\frac{g}{2}t^2 + gt\sqrt{\frac{2h}{g}}\\v_{y_0}(\sqrt{\frac{2h}{g}} - t) = \frac{g}{2}t^2 - gt\sqrt{\frac{2h}{g}}\\v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}

6 0
3 years ago
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