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Oksi-84 [34.3K]
2 years ago
5

10. A wave of speed 5000 m/s is traveling with a wavelength of 100 meters. Calculate the period and

Physics
1 answer:
tekilochka [14]2 years ago
3 0

Answer:

0.02 sec, 50 Hz

Explanation:

period = 100/5000 = 0.02 sec

frequency = 1/0.02 = 50 Hz

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Which formula is used to find fluctuation of the shape of body
Sladkaya [172]

Answer:

varn=n1+1ehvkT–1

Explanation:

This is Einstein's equation.

5 0
3 years ago
What is the kinetic energy in Joules of a 1,500kg car traveling at 75mph?What is the kinetic energy in Joules of a 1,500kg car t
Sergio039 [100]

Answer: 2812500 joules

Explanation:

Mass of car = 1500kg

Velocity of car = 75mph

Kinetic energy = ?

Recall that kinetic energy is the energy possessed by a moving object, and it depends on its mass M and velocity, V

Thus, Kinetic energy = 1/2 x mv^2

= 1/2 x 1000kg x (75mph)^2

= 0.5 x 1000kg x (75mph)^2

= 500 x 5625

= 2812500 joules

Thus, the car travels with a kinetic energy of 2812500 joules

5 0
3 years ago
PLEASE HELP! WILL GIVE BRAINLIEST!
Irina18 [472]

It is talking about how from a different perspective things look different.

That picture should help.

7 0
3 years ago
Estimate the electric field at a point 2.40 cm perpendicular to the midpoint of a uniformly charged 2.00-m-long thin wire carryi
nadya68 [22]

Answer:

E = 1.85*10^{12}\frac{N}{C}

Explanation:

Hi!

The perpendicular distance 2.4cm, is much less than the distance to both endpoints of the wire, which is aprox 1m. Then the edge effect is negligible at this field point, and we can aproximate the wire as infinitely long.

The electric filed of an infinitely long wire is easy to calculate. Let's call z the axis along the wire. Because of its simmetry (translational and rotational), the electric field E must point in the radial direction,  and it cannot depende on coordinate z. To calculate the field Gauss law is used, as seen in the image, with a cylindrical gaussian surface. The result is:

E = \frac{\lambda}{2\pi \epsilon_0 r}\\\lambda=\text{charge per unit length}=\frac{4.95 \mu C}{2 m} = 2.475 \frac{C}{m}\\r=\text{perpendicular distance to wire}\\\epsilon_0=8.85*10^{-12}\frac{C^2}{Nm^2}

Then the electric field at the point of interest is estimated as:

E = \frac{\22.475}{2\pi*( 8.85*10^{-12})*(2.4*10^{-2})}\frac{N}{C}=1.85*10^{12}\frac{N}{C}

6 0
3 years ago
A violinist is tuning her instrument to concert A (440 Hz). She plays the note while listening to an electronically generated to
Soloha48 [4]

To solve the problem it is necessary to take into account the concepts related to beat frequency, i.e., The number of those wobbles per second.

The equation that describes the beat frequency is

f_{beat} = |f_2-f_1|

For our given case we have that the frequency of the instrument is 440Hz and the Beat frequency is 5Hz therefore,

A) The frequency of the violin would be given by

f_{beat} = |f_2-f_1|

5Hz = |f_2-440Hz|

f_2 = 440 \pm 5

f_2 = 445Hz or 435Hz

B) <em>The violinist must loosen the string.</em> As the tightening increases the frequency, thereby increasing the number of beats from 5 to 6, i. e, on thightening the string, the frequency further increases as high frequency will be produced by short trings.

5 0
3 years ago
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