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2 years ago

A 800hz tuning fork is vibrating, producing a sound wave in the air.

Physics

1 answer:

blsea [12.9K]2 years ago

3 0

Answer:

The frequency of the sound wave is 800Hz

The speed of sound in a is about 340m/s.

Velocity = frequency x wavelength

making wavelength the subject formula

wavelength = Velocity/frequency.

wavelength = 340/800

wavelength = 0.425m.

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Help me plssssssss cause I’m struggling

krok68 [10]

Answer:

I am pretty sure it is C

Explanation:

It can be found all over the universe

5 0

3 years ago

Read 2 more answers

A parallel-plate, air-gap capacitor has a capacitance of 0.14 mu F. The plates are 0.5 mm apart, What is the area of each plate?

Marysya12 [62]

Answer:

7.9060 m²

8.57 Volts

5.142×10⁻⁶ Joule

1.2×10⁻⁶ Coulomb

Explanation:

C = Capacitance between plates = 0.14 μF = 0.14×10⁻⁶ F

d = Distance between plates = 0.5 mm = 0.5×10⁻³ m

Q = Charge = 1.2 μC = 1.2×10⁻⁶ C

ε₀ = Permittivity = 8.854×10⁻¹² F/m

Capacitance

$C=\frac{\epsilon_{0}A}{d}\\\Rightarrow A=\frac{Cd}{\epsilon_{0}}\\\Rightarrow A=\frac{0.14\times 10^{-6}\times 0.5\times 10^{-3}}{8.854\times 10^{-12}}\\\Rightarrow A=7.9060\ m^2$

∴ Area of each plate is 7.9060 m²

Voltage

$V=\frac{Q}{C}\\\Rightarrow V=\frac{1.2\times 10^{-6}}{0.14\times 10^{-6}}\\\Rightarrow V=8.57\ Volts$

∴ Potential difference between the plates if the capacitor is charged to 1.2 μC is 8.57 Volts.

Energy stored

E=0.5CV²

⇒E = 0.5×0.14×10⁻⁶×8.57²

⇒E = 5.142×10⁻⁶ Joule

∴ Stored energy is 5.142×10⁻⁶ Joule

Charge

Q = CV

⇒Q = 0.14×10⁻⁶×8.57

⇒Q = 1.2×10⁻⁶ C

∴ Charge the capacitor carries before a spark occurs between the two plates is 1.2×10⁻⁶ Coulomb

6 0

3 years ago

Two runners start at the same point on a straight track. The first runs with constant acceleration so that he covers 98 yards in

charle [14.2K]

Answer:

94.13 ft/s

Explanation:

Given:

$t$ = time interval in which the rock hits the opponent = 10 s - 5 s = 5 s

$s$ = distance to be moved by the rock long the horizontal = 98 yards

$y$ = displacement to be moved by the rock during the time of flight along the vertical = 0 yard

Assume:

$u$ = magnitude of initial velocity of the rock

$\theta$ = angle of the initial velocity with the horizontal.

For the motion of the rock along the vertical during the time of flight, the rock has a constant acceleration in the vertically downward direction.

$\therefore y = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow 0 = u\sin \theta 5 +\dfrac{1}{2}(-9.8)\times 5^2\\\Rightarrow u\sin \theta 5 =\dfrac{1}{2}(9.8)\times 5^2......(1)\\$

Now the rock has zero acceleration along the horizontal. This means it has a constant velocity along the horizontal during the time of flight.

$\therefore u\cos \theta t = s\\\Rightarrow u\cos \theta 5 = 98.....(2)\\$

On dividing equation (1) by (2), we have

$\tan \theta = \dfrac{25}{20}\\\Rightarrow \tan \theta = 1.25\\\Rightarrow \theta = \tan^{-1}1.25\\\Rightarrow \theta = 51.34^\circ$

Now, putting this value in equation (2), we have

$u\cos 51.34^\circ\times 5 = 98\\\Rightarrow u = \dfrac{98}{5\cos 51.34^\circ}\\\Rightarrow u =31.38\ yard/s\\\Rightarrow u =31.38\times 3\ ft/s\\\Rightarrow u =94.13\ ft/s$

Hence, the initial velocity of the rock must a magnitude of 94.13 ft/s to hit the opponent exactly at 98 yards.

3 0

3 years ago

Calculate the work efecttuated, when the force is 300 N, and the distance is 300 meters.

Alexxx [7]

Hello!

Use the formula:

W = Fd

Replacing:

W = 300 N * 300 m

Resolving:

W = 90000 J

The work is 90000 Joules.

8 0

3 years ago

Read 2 more answers

It is proposed that a spaceship might be propelled in the solar system by radiation pressure, using a large sail made of foil. W

Anvisha [2.4K]

Answer:

962291.57928 m²

Explanation:

$P_r$ = Pressure = $2\dfrac{I}{c}$ (full reflection)

I = Intensity = $\dfrac{P}{A}=\dfrac{P}{4\pi r^2}$

P = Power = $3.9\times 10^{26}\ W$

c = Speed of light = $3\times 10^8\ m/s$

M = Mass of Sun = $1.99\times 10^{30}\ kg$

m = Mass of ship = 1500 kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

Force of radiation is given by

$F_r=P_rA\\\Rightarrow F_r=2\dfrac{I}{c}\times A\\\Rightarrow F_r=2\dfrac{P}{4\pi r^2c} A$

This force will balance the gravitational force as stated in the question

$\dfrac{GMm}{r^2}=2\dfrac{P}{4\pi r^2c} A\\\Rightarrow A=\dfrac{4\pi cGMm}{2P}\\\Rightarrow A=\dfrac{4\times \pi\times 3\times 10^8\times 6.67\times 10^{-11}\times 1.99\times 10^{30}\times 1500}{2\times 3.9\times 10^{26}}\\\Rightarrow A=962291.57928\ m^2$

The area of the must be 962291.57928 m²

3 0

3 years ago