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nekit [7.7K]
3 years ago
11

Which of the following is a category of mechanical wave?

Physics
2 answers:
givi [52]3 years ago
8 0

Answer:

a

because the mechanical wave is when it goes over and over again

castortr0y [4]3 years ago
6 0

Answer:

The answer is a like i said 3hrs ago i dont know if this guy copied me tbh

Explanation:

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Sound waves are converted by
Rainbow [258]

Answer:

<h3>sound waves enter the ear canal and cause the ear drum vibrate </h3>

vibration pass through 3 connected bones in the middle ear. this motion set FUILD MOVING bends thousands of delicate hair-like cells which convert the vibrations into NERVE IMPULSES.

8 0
2 years ago
Why is it important that we learn about our solar system, the sun, and planets in our solar system other than Earth?
tangare [24]

Answer:

To understand all the different kind of elements and how they work. Nature is a beautiful thing and it is important we understand how they survive and how we can help. The solar system is a mandatory nature. Learning about other planets also helps us feel safe and more knowledgable.

8 0
3 years ago
I need the answer asap plz!!
Vesnalui [34]

Answer:

266.7k

Explanation:

Given parameters:

Initial pressure  = 36Pa

Initial temperature  = 300k

New pressure  = 32Pa

Unknown:

New temperature  = ?

Solution:

To solve this problem, we have to apply the combined gas law when the volume is constant;

in this instant, when the volume is constant, the pressure of a given mass of gas varies directly with the absolute temperature.

          \frac{P_{1} }{T_{1} }   = \frac{P_{2} }{T_{2} }  

P and T are pressure and temperature

1 and 2 are initial and final states

  Insert the parameters and solve;

       \frac{36}{300}   = \frac{32}{T_{2} }  

      T₂  = 266.7k

7 0
3 years ago
A car moving at a speed of 20 m/s has a kinetic energy of 200,000 J from that motion. What is the car's mass?
zhenek [66]

Answer:

i think the answer is 1000 kg but i might be wrong

8 0
3 years ago
Consider a uniformly volume‑charged sphere of radius R and charge Q . What is the electric potential on the surface of the spher
VARVARA [1.3K]

To solve this problem we will start by applying the given load ratio, and we will rely on the two types of distances given. Later we will use Gauss's law and through its integrals, in which it is equivalent to the potential we will obtain its value in the center of the sphere. Since it is uniformly charged we have to,

\frac{Q'}{Q} = \frac{4\pi r^3}{4\pi R^3}

Q' = \frac{r^3}{R^3} Q

By Gauss Law

\phi = E \cdot 4\pi r^2

Here, E is the electric Field and is equal to

E = \frac{Q'}{\epsilon_0}

For \epsilon_0 being the Permeability constant at free space

Replacing with the previous value we have,

\phi = \frac{Qr^3}{\epsilon_0 R^3}

Then the value of the electric field is,

E = \frac{QR}{4\pi \epsilon_0 R^3}

Now potential

V = \int_0^R E\cdot dr'

V = - \frac{QR^2}{8\pi \epsilon_0 R^3}

V = -\frac{Q}{8 \pi \epsilon_0 R}

6 0
3 years ago
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