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aliya0001 [1]
3 years ago
7

An airplane flies with a velocity of 55.0 m/s [ 35° N of W] with respect to the air (this is

Physics
1 answer:
rodikova [14]3 years ago
6 0

Answer:

21 m/s.  

Explanation:

The computation of the wind velocity is shown below:

But before that, we need to find out the angles between the vectors

53° - 35° = 18°

Now we have to sqaure it i.e given below

v^2 = 55^2 + 40^2 - 2 · 55 · 40 · cos 18°

v^2 = 3025 + 1600 - 2 · 55 · 40 · 0.951

v^2 = 440.6

v = √440.6

v = 20.99

≈ 21 m/s

Hence, The wind velocity is 21 m/s.  

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Aristotle created and it’s credited as the creator.
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3 years ago
A psychopathologist records the number of criminal offenses among teenage drug users in a nationwide sample of 1,201 participant
lara [203]

Answer:

Standard deviation = 3

Explanation:

Given

N = 1201

SS = 10800

Required

Determine the standard deviation

First, we need to determine the variance;

Variance = \frac{SS}{N - 1}

This gives:

Variance = \frac{10800}{1201 - 1}

Variance = \frac{10800}{1200}

Variance = 9

Know that:

Variance = SD^2

Where SD represents standard deviation

This gives

9 = SD^2

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3 years ago
How many Fahrenheit degrees are needed to equal one Celsius (or Kelvin) degree?
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3 years ago
The tuning fork has a frequency of 426.7 is found to cause resonance in a closed column of air measuring 0.186 (the diameter of
lidiya [134]

Answer:

The velocity of the tuning fork sound is 317.4648 m/s

Explanation:

The given parameters are;

The frequency of the fork, f = 426.7 Hz

The length of the closed air column, L = 0.186 m

The diameter of the tube, d = 0.15 m

At the fundamental frequency in a closed tube, we have;

λ = 4·L

Where;

λ = The wavelength of the wave

L = The length of the tube

From the equation for the velocity of a wave, we have;

v = f·λ

Where;

v = The velocity of the (sound) wave

f = The frequency of the wave = 426.7 Hz

λ = 4·L = 4 × 0.186 m = 0.744 m

∴ v = 426.7 Hz × 0.744 m = 317.4648 m/s

Therefore, the velocity of the sound produced by the tuning fork, v = 317.4648 m/s

3 0
3 years ago
A glider with mass 0.24 kg sits on a frictionless horizontal air track, connected to a spring of negligible mass with force cons
shepuryov [24]

Answer:

v=2.556m/s

Explanation:

From the conservation of mechanical energy

K_{E1}+U_1=K_{E2}+U_2

\frac{1}{2}m*v_1^2+\frac{1}{2}*K*x_1^2=\frac{1}{2}m*v_2^2+\frac{1}{2}*K*x_2^2

x_2=0.08m

v_1=0 m/s

Solve to velocity v2

m*v_2^2=k*x_1^2-k*x_2^2

v^2=\frac{k}{m}*(x_1^2-x_2^2)

v^2=\frac{5.5N/m}{0.24kg}*(0.54m^2-0.080^2)

v=\sqrt{6.54m^2/s^2}=2.556m/s

4 0
3 years ago
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