Tom rides his motorcycle at a speed of 15 meters/second for an hour.

The speed of sound in air is 345 m/s. A tuning fork vibrates above the open end of a sound resonance tube. If sound waves have w

Delvig [45]

To develop this problem it is necessary to apply the concept of Frequency based on speed and wavelength.

According to the definition the frequency can be expressed as

$f = \frac{v}{\lambda}$

Where,

v = Velocity

$\lambda =$ Wavelength

Our value are given by,

v = 345m/s

$\lambda = 63cm$

Replacing

$f = \frac{345}{0.63}$

$f = 547.61Hz$

Therefore the frequency of the tuning fork is 547.61Hz

6 0

3 years ago

Light of wavelength 500 nm is incident perpendicularly from air on a film 10-4cm thick and of refractive index 1.375. Part of th

Marysya12 [62]

Answer

given,

wavelength (λ)= 500 n m

thickness of film= 10⁻⁴ cm

refractive index = μ = 1.375

distance traveled is double which is equal to 2 x 10⁻⁴ cm

a) Number of wave

$N = \dfrac{d}{\mu\lambda}$

$N = \dfrac{2 \times 10^{-6}}{1.375\times 500 \times 10^{-9}}$

N = 2.91

N = 3

b) phase difference is equal to

Reflection from the first surface has a 180° (½λ) phase change.

There is no phase change for the 2nd surface reflection and there is no phase difference for the 2nd wave having traveled an exact whole number of waves.

net phase difference = $180^0\times \dfrac{3}{2}$

= 270°

6 0

3 years ago

What are some examples and non-examples of volume

Jlenok [28]

Answer:

Explanation:

Some correct non-examples are: A glass half-empty; Anything in two dimensions; The amount that covers something.

4 0

3 years ago

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A capacitor is charged until its stored energy is 7.54 J. A second capacitor is then connected to it in parallel. If the charge

Ivan

Answer:

2 J

Explanation:

A charged capacitor of capacitance $C_1$ with energy of 7.54 J, is connected in parallel with another capacitor $C_2$ , so the charge is equally distributed between them.

(a) The energy stored in the capacitor before it being connected to the other capacitor is:

$U_O=q_0^2/2C_1=7.54 J\\$

The energy stored in the electric field is the sum of the energies of the two capacitors:

$U=U_1+U_2\\U=q_1^2/2C_1+q_2^2/2C_2$

since the charge equally distributed, $q_1$ = $q_2$ = $q_o/2$ . and since they are connected in parallel the potential difference on both of them is the same :