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alexandr1967 [171]
3 years ago
14

PLEASE HELP !!!   100

Physics
1 answer:
Vlad [161]3 years ago
7 0

A line that's 'normal' to the surface is perpendicular to the surface.

The angle of incidence is the angle between the normal and the light ray ARRIVING at the surface.

The angle of reflection is the angle between the normal and the light ray LEAVING the surface after it's reflected.

The law of reflection says these two angles are equal.

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Machmer Hall is 400 m North and 180 m West of Witless.
yan [13]

Answer:

The distance from Witless to Machmer is 438.63 m.

Explanation:

Given that,

Machmer Hall is 400 m North and 180 m West of Witless.

We need to calculate the distance

Using Pythagorean theorem

D = \sqrt{(d_{m})^2+(d_{w})^2}

Where, d_{m} =distance of Machmer Hall

d_{w} =distance of Witless

Put the value into the formula

D = \sqrt{(400)^2+(180)^2}

D=438.63\ m

Hence, The distance from Witless to Machmer is 438.63 m.

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Un acróbata de 60.0 kg está unido a un cordón de bungee con un resorte de 10.0 m de longitud . Salta de un puente que abarca un
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Answer:

What kind of languege is this please?

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2 years ago
:) What is practical machine? what is the reda<br>tion between MA and VR in a practical<br>machine?​
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3 years ago
The speed of a certain electron is 995 km s−1 . If the uncertainty in its momentum is to be reduced to 0.0010 per cent, what unc
Tpy6a [65]

Answer:

The uncertainty in the location that must be tolerated is 1.163 * 10^{-5} m

Explanation:

From the uncertainty Principle,

Δ_{y} Δ_{p} = \frac{h}{2\pi }

The momentum P_{y} = (mass of electron)(speed of electron)

                                = (9.109 * 10^{-31}kg)(995 * 10^{3} m/s)

                                = 9.0638 * 10^{-25}kgm/s

If the uncertainty is reduced to a 0.0010%, then momentum

                              = 9.068 * 10^{-30}kgm/s

Thus the uncertainty in the position would be:

                              Δ_{y} = \frac{h}{2\pi } * \frac{1}{9.068 * 10^{-30} }

                              Δ_{y} \geq  1.163 * 10^{-5}m

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How much time would it take for an
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36 and a half hours

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