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ANEK [815]
3 years ago
11

Please help ASAP. There are 3 question it would be a big help if you can answer any.

Physics
2 answers:
sesenic [268]3 years ago
7 0

Answer:

number 1 is falling freely number 2 I think the 2nd option number 3 3rd option

erastova [34]3 years ago
7 0

1).  If you're standing on a scale in an elevator, and the scale reads much less than your real weight, then the elevator must be<em> accelerating downward.</em>

(Think of the extreme limit: If the elevator is in free-fall, accelerating down at 9.8 m/s², then you'll apear to be weightless ... the scale will read zero.)

2).  A 100 kg man is standing in an elevator, accelerating up at 2.0 m/s².

The net force acting on him is F = m·a = (100 kg) x (2.0 m/s²) = 200 N up.

The forces acting on  him are:

-- gravity, m·g = (100 kg) x (9.8 m/s²) = 980 N down

-- elevator floor force . . . EF up

Net force = (980N down) + (EF up)

(200N up) = (980N down) + (EF up)

Add (980N up) to each side:

(200N up + 980N up) = EF up

<em>Elevator force = 1180N up</em>.

Pick the second choice:  <em>1.2 x 10³ N up</em>

<em></em>

3).  My computer is resting on my desk.

<em>There are many forces acting on my computer, and they are all balanced.</em>

That means the NET force acting on my computer is zero.

If the net force were NOT zero, then my computer would be accelerating in the direction of the net force. But it's at rest. So the net force must be zero.

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Visible light passes through a diffraction grating that has 900 slits per centimeter, and the interference pattern is observed o
kobusy [5.1K]

Answer:

\Delta \lambda=14.3\ nm

Explanation:

It is given that,

The number of lines per unit length, N = 900 slits per cm

Distance between the formed pattern and the grating, l = 2.3 m

n the first-order spectrum, maxima for two different wavelengths are separated on the screen by 2.98 mm, \Delta Y=2.98\ mm = 0.00298\ m

Let d is the slit width of the grating,

d=\dfrac{1}{N}

d=\dfrac{1}{900\ cm}

d=1.11\times 10^{-5}\ m

For the first wavelength, the position of maxima is given by :

y_1=\dfrac{L\lambda_1}{d}

For the other wavelength, the position of maxima is given by :

y_2=\dfrac{L\lambda_2}{d}

So,

\Delta \lambda=\dfrac{\Delta y d}{l}

\Delta \lambda=\dfrac{0.00298\times 1.11\times 10^{-5}}{2.3}

\Delta \lambda=1.43\times 10^{-8}\ m

or

\Delta \lambda=14.3\ nm

So, the difference between these wavelengths is 14.3 nm. Hence, this is the required solution.

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