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oee [108]
2 years ago
7

The frequency of the given sound is 1.5khz then how many vibration it is completing in one second ?​

Physics
2 answers:
Amanda [17]2 years ago
8 0

Answer: The answer is 1500 per second.

Explanation: Each vibration consist 1.5 kilohertz, so I multiply 1000 and the number of the kilohertz that looks like this, 1.5*1000=1500 per second, I hoped my answer is very helpful to your own physics question and have a Happy Halloween! :D

Sincerely,

Jason Ta,

The ambitious of Brainly and the role of the TDSB and WHCI Student of the high school.

kompoz [17]2 years ago
6 0

Answer:

1500 per second.

Explanation:

vibrations = 1.5 kilohertz

1.5×1000=1500

the answer is 1500 per second.

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Which of Newton's laws of motion describes the motion of an object that has a net<br> force of ON?
algol [13]

Newton's first and second laws of motion both do, but I think the one you're looking for is: <em>The First Law of Motion</em>.  That description is a little more direct.

It says that if an object is not acted on by a net external force, then it continues in "constant, uniform motion".

3 0
2 years ago
Which statement best describes the weather shown by the purple combination of semicircles and triangles on a line on a weather m
Vaselesa [24]

Answer:

D) A warm front brings drizzly weather.

Explanation:

8 0
3 years ago
Read 2 more answers
Explanation A 5000 kg rocket is at rest in deep space. The rocket burns fuel pushing 10kg of exhaust gases rearward at 4000 m/s.
katovenus [111]

Answer:

F = 4000 N

Explanation:

given,

mass of rocket (M)= 5000 Kg

10 Kg gas burns at speed (m)= 4000 m/s

time = 10 s

average force = ?

at the end the rocket is at rest

by conservation of momentum

 M v + m v' = 0

 5000 x v - 10 x 4000 = 0

 5000 v = 40000

    v = 8 m/s

speed of rocket = 8 m/s

now,

we know

change in momentum = F x Δ t

F = \dfrac{m(v_i-v_f)}{\Delta t}

F = \dfrac{5000(8-0)}{10}

      F = 4000 N

Hence, the average force applied to the rocket is equal to F = 4000 N

4 0
3 years ago
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What type of tectonic plate boundary exists along the edge of the North American plate near the coast of Northern California, Or
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Answer:

-transform plate boundary

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4 0
3 years ago
A uniformly charged rod (length = 2.0 m, charge per unit length = 3.0 nc/m) is bent to form a semicircle. What is the magnitude
Artist 52 [7]

Answer:

84.82N/C.

Explanation:

The x-components of the electric field cancel; therefore, we only care about the y-components.

The y-component of the differential electric field at the center is

$dE = \frac{kdQ }{R^2} sin(\theta )$.

Now, let us call \lambda the charge per unit length, then we know that

dQ = \lambda Rd\theta;

therefore,

$dE = \frac{k \lambda R d\theta }{R^2} sin(\theta )$

$dE = \frac{k \lambda  d\theta }{R} sin(\theta )$

Integrating

$E = \frac{k \lambda   }{R}\int_0^\pi sin(\theta )d\theta$

$E = \frac{k \lambda   }{R}*[-cos(\pi )+cos(0) ]$

$E = \frac{2k \lambda   }{R}.$

Now, we know that

\lambda = 3.0*10^{-9}C/m,

k = 9*10^9kg\cdot m^3\cdot s^{-4}\cdot A^{-2},

and the radius of the semicircle is

\pi R = 2.0m,\\\\R = \dfrac{2.0m}{\pi };

therefore,

$E = \frac{2(9*10^9) (3.0*10^{-9})   }{\dfrac{2.0}{\pi } }.$

$\boxed{E = 84.82N/C.}$

7 0
3 years ago
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