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stiks02 [169]
3 years ago
5

An airplane is heading due east. The airspeed indicator shows that the plane is moving at a speed of 370 km/h relative to the ai

r. If the wind is blowing from the north at 92.5 km/h, the velocity of the airplane relative to the ground is?
Physics
1 answer:
PolarNik [594]3 years ago
5 0

Answer:

Vp = 358.3m/s

Explanation:

Taking

North: UP

South: DOWN

East: RIGHT

West: LEFT

In the attachment, the velocity vector triangle is shown for the scenario. the triangle formed is a right angled triangle.

We can apply the Pythagora's Theorem.

370 ^ 2 = 92.5^2 + Vp^2    (Vp: plane velocity relative to ground)

Vp = sqr.rt(370^2 - 92.5^2)

Vp = 358.3m/s

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Suppose that you are swimming in a river while a friend watches from the shore. In calm water, you swim at a speed of 1.25 m/s .
aliya0001 [1]

Answer: The observing friend will the swimmer moving at a speed of 0.25 m/s.

Explanation:

  • Let <em>S</em> be the speed of the swimmer, given as 1.25 m/s
  • Let S_{0} be the speed of the river's current given as 1.00 m/s.

  • Note that this speed is the magnitude of the velocity which is a vector quantity.
  • The direction of the swimmer is upstream.

Hence the resultant velocity is given as, S_{R} = S — S 0S_{0}

S_{R} = 1.25 — 1

S_{R} = 0.25 m/s.

Therefore, the observing friend will see the swimmer moving at a speed of 0.25 m/s due to resistance produced by the current of the river.

6 0
3 years ago
Coherent light of wavelength 525 nm passes through two thin slits that are 4.15×10^(−2) mm apart and then falls on a screen 7
IRINA_888 [86]

A) 4.7 cm

The formula for the angular spread of the nth-maximum from the central bright fringe for a diffraction from two slits is

sin \theta=\frac{n \lambda}{d}

where

n is the order of the maximum

\lambda is the wavelength

a is the distance between the slits

In this problem,

n = 5

\lambda=525 nm =5.25\cdot 10^{-7} m

a=4.15\cdot 10^{-2} mm=4.15\cdot 10^{-5} m

So we find

\theta=sin^{-1} (\frac{(5)(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=3.62^{\circ}

And given the distance of the screen from the slits,

D=75.0 cm = 0.75 m

The distance of the 5th  bright fringe from the central bright fringe will be given by

y=D tan \theta = (0.75 m)tan 3.62^{\circ}=0.047 m = 4.7 cm

B) 8.1 cm

The formula to find the nth-minimum (dark fringe) in a diffraction pattern from double slit is a bit differente from the previous one:

sin \theta=\frac{(n+\frac{1}{2}) \lambda}{d}

To find the angle corresponding to the 8th dark fringe, we substitute n=8:

\theta=sin^{-1} (\frac{(8+\frac{1}{2})(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=6.17^{\circ}

And the distance of the 8th dark fringe from the central bright fringe will be given by

y=D tan \theta = (0.75 m)tan 6.17^{\circ}=0.081 m = 8.1 cm

5 0
3 years ago
A boy finds an abandoned mine shaft in the woods, and wants to know how deep the hole is. He drops in a stone, and counts 4 seco
oee [108]
S=(0x4)+(0.5x4.81x4x4)
S=0.78.48

The depth is approximately 78 meters.
(My brain hurts now) :P Good Luck!
4 0
3 years ago
Sodium and phosphorus combine to form Na3P. The name of this compound is .
g100num [7]
Binary ionic compound
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3 years ago
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The visible spectrum refers to the tiny portion of the electromagnetic spectrum that we ________.
marusya05 [52]

Answer:

are able to see/observe

Explanation:

Humans are not able to see most wavelengths in the universe--but there is a select range that is visible to our eyes. This (which is usually shown on an electromagnetic spectrum diagram/chart/depiction by a small portion of rainbow) is the visible light spectrum

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2 years ago
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