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Sergeu [11.5K]
2 years ago
11

You walk 12.0 m West and then 4.00 m south. What is your displacement?​

Physics
1 answer:
Karo-lina-s [1.5K]2 years ago
7 0

Answer:

12.6 (3 sig. fig.)

Explanation:

(12^2 + 4^2)^1/2 = 12.6 (3 sig. fig.)

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Eduardwww [97]
My teacher said 36m when I asked her
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3 years ago
Who exerts more pressure? a) A girl of 50 kg, wearing heels with an area of 1 cm2. b) An elephant of 4000 kg with foot area of 2
Mrrafil [7]

Answer:

The girl exerts more pressure.

Explanation:

Pressure can be defined as the force exerted normally or perpendicularly per unit area.

i.e P = F/A

<u>Girls</u>

Area of the heel = 1cm² = 10^(-4) m²

Force = mg = 50 × 10 = 500N

Pressure =

\frac{500}{10 ^{ - 4} }

= 5 \times  {10}^{6}

<u>Elephant</u>

<u>Area</u><u> </u><u>=</u><u> </u><u>2</u><u>5</u><u>0</u><u>cm</u><u>²</u><u> </u><u>=</u><u> </u><u>2</u><u>.</u><u>5</u><u> </u><u>x</u><u> </u><u>1</u><u>0</u><u>^</u><u>(</u><u>-</u><u>2</u><u>)</u><u>b</u><u> </u><u>m</u><u>²</u>

<u>Force</u><u> </u><u>=</u><u> </u><u>mg</u><u> </u><u>=</u><u> </u><u>4</u><u>0</u><u>0</u><u>0</u><u>0</u><u>N</u>

<u>Pressure</u><u> </u><u>=</u><u> </u>

<u>\frac{40000}{2.5 \times  {10}^{ - 2} }</u>

<u>= 1.6 \times  {10}^{6}</u>

5 0
3 years ago
A diffraction grating produces a first-order bright fringe that is 0.18 m away from the central bright fringe on a flat screen.
Georgia [21]

Answer:

The wavelength of the light is 562.5 nm

Solution:

As per the question:

Order, n = 1

Slit separation, d = 2.5\times 10^{- 6}\ m

Distance from the bright fringe, y = 0.18 m

Distance between the screen and the grating, D = 0.8 m

Now,

We know from the eqn for diffraction:

n\lambda = dsin\theta

n = 1

\lambda = dsin\theta            (1)

Also,

For very small angle, \theta:

sin\theta ≈ tan\theta = \frac{y}{D} = \frac{0.18}{0.8} = 0.225

Using the above value in eqn (1):

\lambda = 2.5\times 10^{- 6}\times 0.225 = 5.625\times 10^{- 7}\ m = 562.5\ nm

3 0
3 years ago
A vibrating tuning fork is held over a water column with one end closed and the other open. As the water level is allowed to fal
DiKsa [7]

Answer:

630 Hz.

Explanation:

As we are considering the one end open pipe. So for the sound wave there will be a pressure node at the open end of the tube as at that place the molecules can not move back and forth. However on the closed end there will be a flow node as the water molecules their are moving back and forth. So it will produces the resonance at the positions 1/4, 3/4.......

we can find the wavelength by multiplying the levels distance by 2.

λ = 2 × 0.27 m = 0.54

f = Vs/λ

  = 340/0.54

  = 630 Hz

6 0
3 years ago
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Answer:

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Explanation:

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2 years ago
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