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OLEGan [10]
4 years ago
14

The thickness of a flange on an aircraft component is uniformly distributed between 0.95 and 1.05 millimeters. (a) Determine the

proportion of flanges that exceeds 0.99 millimeters.
Physics
1 answer:
Igoryamba4 years ago
8 0

Answer:

The proportion of flanges that exceeds 0.99 millimeters is 0.6

Explanation:

Given;

integral range of [0.95,  1.05]

Let X be a variable with uniform distribution over the given range.

Then, f(x) = \frac{1}{1.05 -0.95} =10

1 - 0.95 = 0.05, 1.05 - 1 = 0.05

\ minimum, f_(x) = \frac{0.05}{1.05 -0.95} =0.5\\\\

interval = 10 - 0.5 = 9.5

F(x) = 10x + 9.5

When, X exceeds 0.99 millimeters, then the proportion of flanges will be;

P (X > x) = 1 - F(x)

P( X > 0.99 ) = 1 - 10(0.99) + 9.5

P( X > 0.99 ) = 0.6

Therefore, the proportion of flanges that exceeds 0.99 millimeters is 0.6

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spin [16.1K]

The answer for the following answer is answered below.

  • <u><em>Therefore the time period of the wave is 0.01 seconds.</em></u>
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Explanation:

Frequency (f):

The number of  waves that pass a fixed place in a given amount of time.

The SI unit of frequency is Hertz (Hz)

Time period (T):

The time taken for one complete cycle of vibration to pass a given point.

The SI unit of time period is seconds (s)

Given:

frequency (f) = 100 Hz

wavelength (λ) = 2.0 m

To calculate:

Time period (T)

We know;

According to the formula;

<u>f =</u>\frac{1}{T}<u></u>

Where,

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T represents the time period

from the formula;

  T = \frac{1}{f}

 T = \frac{1}{100}

  T = 0.01 seconds

<u><em>Therefore the time period of the wave is 0.01 seconds.</em></u>

5 0
3 years ago
Read 2 more answers
A solenoidal coil with 22 turns of wire is wound tightly around another coil with 330 turns. The inner solenoid is 21.0 cm long
Airida [17]

Answer:

0.00027646\ T

2.33\times 10^{-5}\ H

-0.04194 V

Explanation:

N_2 = Number of turns in outer solenoid = 330

N_1 = Number of turns in inner solenoid = 22

I_1 = Current in inner solenoid = 0.14 A

\dfrac{dI_2}{dt} = Rate of change of current = 1800 A/s

\mu_0 = Vacuum permeability = 4\pi \times 10^{-7}\ H/m

r = Radius = 0.0115 m

Magnetic field is given by

B=\mu_0\dfrac{N_2}{l}I\\\Rightarrow B=4\pi \times 10^{-7}\times \dfrac{330}{0.21}\times 0.14\\\Rightarrow B=0.00027646\ T

The  average magnetic flux through each turn of the inner solenoid is 0.00027646\ T

Magnetic flux is given by

\phi=BA\\\Rightarrow \phi=0.00027646\times \pi 0.0115^2\\\Rightarrow \phi=1.14862\times 10^{-7}\ wb

Mutual inductance is given by

M=\dfrac{N_1\phi}{I}\\\Rightarrow M=\dfrac{22\times 1.14862\times 10^{-7}}{0.14}\\\Rightarrow M=2.33\times 10^{-5}\ H

The mutual inductance of the two solenoids is 2.33\times 10^{-5}\ H

Induced emf is given by

\epsilon=-M\dfrac{dI_2}{dt}\\\Rightarrow \epsilon=-2.33\times 10^{-5}\times 1800\\\Rightarrow \epsilon=-0.04194\ V

The emf induced in the outer solenoid by the changing current inthe inner solenoid is -0.04194 V

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Answer:

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Among the choices above, the one that is most closely related to an activated complex is the transition state. The answer is letter D. This formation forms quickly and does not stay in a way compound is. It usually forms during the enzyme – substrate reaction.

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Answer:

what that guy said

Explanation:

because he provides evidence

4 0
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