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qwelly
1 month ago
9

The length of stereocilia actually vary from 10 to 50 micrometers. Again, assuming that they behave like simple pendula, over wh

at frequency range of sound waves would they resonate. (The actual frequency range of human hearing is 20 Hz 20,000 Hz, so might there be other mechanisms involved and/or might the pendular model be rather oversimplified?)
A. About 70 Hz -160 Hz.
B. About 440 Hz - 1000 Hz.
C. About 20 Hz - 50 Hz.
D. About 0.07 to 0.16 Hz.
Physics
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A dipole of moment 0.5 e·nm is placed in a uniform electric field with a magnitude of 8 times 104 N/C. What is the magnitude of
Ostrovityanka [3204]

Answer:

The resultant torque is zero.

Explanation:

Assuming the dipole consists of two equal but opposite charges e, it can be represented by a rod with one end featuring a charge e and the other end with -e. Since the dipole is aligned with the electric field, both charges experience forces that are parallel to this electric field. Consequently, there are no force components that act perpendicular to the rod, which is necessary for torque to occur.

8 0
3 months ago
Two resistors ( 3 ohms & 6 ohms) in a series circuit with a power supply = 12 volts. The current through resistor 6 ohms is
Ostrovityanka [3204]

In a series circuit...

-- The overall resistance equals the sum of the individual resistances.

-- The current remains identical throughout the circuit.

The total resistance in this circuit is (3Ω + 6Ω )  =  9Ω

<pThe current at every point measures (V/R) = (12v / 9Ω ) = 1.33 A.

Select choice (a).

6 0
3 months ago
A pressure vessel that has a volume of 10m3 is used to store high-pressure air for operating a supersonic wind tunnel. If the ai
serg [3582]
The vessel contains a total air mass of 235.34 kilograms. Assuming the air behaves like an ideal gas, the air's density (in kilograms per cubic meter) can be expressed using the following formula: (1) Where: - Pressure is in kilopascals. - Molar mass is expressed in kilomoles per kilogram. - The ideal gas constant is measured in kilopascal-cubic meters per kilomole-Kelvin. - Temperature is indicated in Kelvin. Given the values of pressure, molar mass, the gas constant, and temperature, we can find the air density. The mass of the air inside the vessel is calculated from the density definition as follows: (2) Where m represents the mass, given in kilograms. With the known values of density and volume in mind, we derive the mass of air stored in the vessel as 235.34 kilograms.
7 0
2 months ago
A BMX bicycle rider takes off from a ramp at a point 2.4 m above the ground. The ramp is angled at 40 degrees from the horizonta
Ostrovityanka [3204]

Answer:

The BMX rider lands 5.4 meters horizontally away from the ramp's end.

Explanation:

The BMX position vector is represented as:

r = (x0 + v0 × t × cos α, y0 + v0 × t × sin α + ½ × g × t²)

Where:

r = position at time t

x0 = initial horizontal position

v0 = initial speed

α = angle of jump

y0 = initial vertical height

g = gravitational acceleration (-9.8 m/s², upward positive)

Refer to the diagram for clarity. At the landing time, the vertical coordinate of the position vector is -2.4 m, measured from the ramp's edge as the origin. Using the vertical component equation for y, one can solve for t, then substitute t to find the horizontal distance.

The vertical position equation:

-2.4 m = 0 + 5.9 m/s × t × sin 40° - ½ × 9.8 m/s² × t²

Rearranged:

0 = -4.9 t² + 5.9 t × sin 40° + 2.4

Solving this quadratic yields:

t = 1.2 seconds

Then, calculate horizontal distance:

x = 0 + 5.9 m/s × 1.2 s × cos 40° = 5.4 m

This means the BMX lands 5.4 meters from the ramp's edge.

Have a great day!

8 0
4 months ago
A vacuum tube diode consists of concentric cylindrical electrodes, the negative cathode and the positive anode. Because of the a
serg [3582]

Answer:

   C = 4,174 10³ V / m^{3/4},  E = 7.19 10² / ∛x,    E = 1.5  10³ N/C

Explanation:

In this problem, we are tasked with determining the constant value and the generated electric field.

We will begin with computing the constant C:

           V = C x^{4/3}

           C = V / x^{4/3}

            C = 220 / (11 10⁻²)^{4/3}

            C = 4,174 10³ V / m^{3/4}

Next, we will find the electric field by utilizing the formula:

            V = E dx

             E = dx / V

             E = ∫ dx / C x^{4/3}

            E = 1 / C  x^{-1/3} / (- 1/3)

            E = 1 / C (-3 / x^{1/3})

We consider the evaluation from the lower limit x = 0 where E = E₀ = 0 to the upper limit x = x, resulting in E = E:

            E = 3 / C     (0- (-1 / x^{1/3}))

            E = 3 / 4,174 10³   (1 / x^{1/3})

           E = 7.19 10² / ∛x

Substituting x = 0.110 cm:

          E = 7.19 10² /∛0.11

          E = 1.5  10³ N/C

6 0
3 months ago
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