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2 years ago

7

onsidering light at the two ends of the visible light spectrum, violet light has a _ wavelength and a _ photon energy th

an red light. Shorter, lower Longer, lower Longer, higher Shorter, higher

1 answer:

Alenkasestr [34]2 years ago

6 0

Answer:

Shorter, higher

Explanation:

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The golden gate bridge in san francisco, california is 227.4 meters above mean sea level (msl). what is the maximum distance awa

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A normal human being can rotate his neck at maximum angle of 70 degree at one stretch

So here the maximum angle is 70 degree upto which he can see the height

now we will have

$tan\theta = \frac{H}{d}$

$tan70 = \frac{227.4 - 4.3}{d}$

now we have

$d = 81.2 m$

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3 years ago

A 700-kg car, driving at 29 m/s, hits a brick wall and rebounds with a speed of 4. 5 m/s. What is the car’s change in momentum d

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Answer:

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2 years ago

How do you find the volume of an irregular shaped object

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3 years ago

Which of these could NOT be used to find the mechanical advantage of an inclined plane?

34kurt

The "c) percent efficiency" could not be used to find the mechanical advantage of an inclined plane. There are two formulae that could be used to determine the mechanical advantage of an inclined plane which stated as MA = Length/rise and Wout=Win. MA is the mechanical advantage, Wout is the output force, Win is the input force, and "rise" is the height of the inclined plane<span>.</span>

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3 years ago

Calculate the acceleration of gravity as a function of depth in the earth (assume it is a sphere). You may use an average densit

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Solution :

Acceleration due to gravity of the earth, g $=\frac{GM}{R^2}$

$g=\frac{G(4/3 \pi R^2 \rho)}{R^2}=G(4/3 \pi R \rho)$

Acceleration due to gravity at 1000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-1000) \times 5.5 \times 10^3\right)$

$= 822486 \times 10^{-8}$

$=0.822 \times 10^{-2} \ km/s$

= 8.23 m/s

Acceleration due to gravity at 2000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-2000) \times 5.5 \times 10^3\right)$

$= 673552 \times 10^{-8}$

$=0.673 \times 10^{-2} \ km/s$

= 6.73 m/s

Acceleration due to gravity at 3000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-3000) \times 5.5 \times 10^3\right)$

$= 3371 \times 153.86 \times 10^{-8}$

= 5.18 m/s

Acceleration due to gravity at 4000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-4000) \times 5.5 \times 10^3\right)$

$= 153.84 \times 2371 \times 10^{-8}$

$=0.364 \times 10^{-2} \ km/s$

= 3.64 m/s

3 0

3 years ago