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4 years ago

A snowboarder travels 150 m down a mountain slope that is 65 degrees above horizontal. What is his vertical displacement?

1 answer:

3 0

This can be answered using trigonometric analysis. This sloped path that is 150 m long is the hypotenuse of the triangle. The adjacent angle would then be 65 degrees. Given these:

sin 65 = h / 150

Where: h = vertical displacement = 150 (sin 65)
h = 135.95 meters

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If 6.24×10¹⁸ electron pass through a wire in 1s how many pass through it during a time interval of 2hr, 47min and 10s

levacccp [35]

Answer = 6.24x10^18 x ((2 x 3600) + (47 x 60) + 10)

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3 years ago

If the input force on a simple machine is 2 N and the output force is 4 N, what is it’s mechanical advantage?

ivann1987 [24]

Explanation:

Given that,

Input force on a simple machine, F = 2 N

The output force on a simple machines, F' = 4 N

The ratio of the output force to the input force of any simple machines is called its mechanical advantage. Mathematically, it is given by :

$m=\dfrac{F'}{F}$

Putting the value of F and F' in above equation. We get :

$m=\dfrac{4}{2}$

m = 2

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3 years ago

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How do scientist check repeatability of results

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Answer:

If research results can be replicated, it means they are more likely to be correct.

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3 years ago

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Circle the path function(s) which are affected by a detailed path from one state to another: (A) Temperature (B) Boundary work (

Naddika [18.5K]

Answer:

(B) Boundary work

(D) Heat

Explanation:

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3 years ago

A solid cylinder with a mass of 2.46 kg and a radius of 0.049 m starts from rest at a height of 4.80 m and rolls down a 24.3 ◦ s

irina1246 [14]

Answer:

$v_{f}\approx 2.097\,\frac{m}{s}$

Explanation:

Let assume that the solid cylinder rolls down a frictionless incline. The translational speed can be found by using the Principle of Energy Conservation and the Work-Energy Theorem:

$m_{cyl}\cdot g\cdot y_{o} = \frac{1}{2}\cdot m_{cyl} \left( 1+ \frac{1}{R}\right)\cdot v_{f}^{2}$

$g\cdot y_{o} = \frac{1}{2}\cdot\left( 1+ \frac{1}{R}\right)\cdot v_{f}^{2}$

The translational speed is:

$v_{f} = \sqrt{\frac{2\cdot g\cdot y_{o}}{\left(1 + \frac{1}{R} \right)}}$

$v_{f} = \sqrt{\frac{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (4.80\,m)}{\left(1 + \frac{1}{0.049\,m} \right)} }$

$v_{f}\approx 2.097\,\frac{m}{s}$

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3 years ago