All categories

3 years ago

What are non examples of a medium?

Physics

1 answer:

Nataliya [291]3 years ago

7 0

Vacuum is an example of a non medium.

Non Medium is anything through which physical forces can not be transmitted.

The only electromagnetic wave that can not travel through a non medium is sound. It can only travel through a medium. Light does not needs a medium to travel through.

So basically mediums can be, solid, liquid and gas and many more substances and the most common example of a non medium thing is vacuum.

You might be interested in

TheE-field strength is 50,000 N/C inside a parallel plate capacitor with 2.0 mmspacing. A proton is released from rest at the po

Ghella [55]

Answer:

The speed is $138852.4 \frac{m}{s}$

Explanation:

Electric field magnitude (E) and electric force magnitude (F) are related by the equation

$F=Eq$

with q the charge of the proton ( $1.61\times10^{-19} C$ ). Because between parallel plates electric field is almost constant, electric force is constant too:

$F=(50000)(1.61\times10^{-19})=8.05\times10^{-15}N$

because electric force is constant, then by Newton's second law acceleration (a) is constant too, it is:

$a=\frac{F}{m}$

with m the mass of the proton ( $1.67\times10^{-27} kg$ ):

$a=\frac{8.05\times10^{-15}}{1.67\times10^{-27}}= 4.82\times10^{12}\frac{m}{s^{2}}$

Now with this constant acceleration we can use the kinematic equation

$v^{2}=v_{0}^{2}+2ad$

with v the final speed, $v_i$ the initial velocity that is zero (because proton starts at rest) and d is the distance between the plates, so:

$v=\sqrt{2ad}=\sqrt{2(4.82\times10^{12}\frac{m}{s^{2}})(2.0\times10^{-3}m)}$

$v=138852.4 \frac{m}{s}$

5 0

3 years ago

What are the units for electric current

Anettt [7]

Answer:

B) Amps

Which are called Ampere or Amp.

Explanation:

3 0

3 years ago

Help with 5 and the question below. BRAINLIEST FOR THE CORRECT ANSWER! Urgent Physics help!

krok68 [10]

Sound level at distance of 15 m is given as 20 dB

so intensity at this distance is given as

$L = 10 Log\frac{I}{I_0}$

$20 = 10 Log{I}{10^{-12}}$

$I_1 = 10^{-10}W/m^2$

now if we move closer to some some distance the sound level is now 50 dB

now the intensity is given as

$L = 10 Log\frac{I}{I_0}$

$50 = 10Log\frac{I}{10^{-12}}$

$I_2 = 10^{-7}W/m^2$

now we know that

$\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2}$

$\frac{10^{-10}}{10^{-7}} = \frac{r_2^2}{15^2}$

$r_2 = 47 cm$

so now the distance from friend must be 47 cm

8 0

2 years ago

A loudspeaker located in air generates sound waves of frequency 1,000 Hz. Some of these sound waves enter a pool of water, where

prisoha [69]

Answer:

1.4 m

Explanation:

v = Speed of sound in water = 1400 m/s

f = Frequency of sound = 1000 Hz

$\lambda$ = Wavelength

When we multiply the frequency and the wavelength of a wave we get the velocity of sound in that medium

$v=f\lambda\\\Rightarrow \lambda=\dfrac{v}{f}\\\Rightarrow \lambda=\dfrac{1400}{1000}\\\Rightarrow \lambda=1.4\ m$

The wavelength of the sound waves in water is 1.4 m

8 0

2 years ago

As you move away from a positive charge distribution, the electric field:

GalinKa [24]

Answer:

The electric field always decreases.

Explanation:

The electric field due to a point charge is given by :

$E=\dfrac{kq}{r^2}$

Where

k = electric constant

q = charge

r = distance from the charge

It is clear from the above equation that as the distance from the charge particle increases the electric field decreases. As you move away from a positive charge distribution, the electric field always decreases. Hence, the correct option is (c) "Always decreases".

3 0

2 years ago