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3 years ago

What causes sounds that are lower than 0 decibels?

Physics

1 answer:

vichka [17]3 years ago

4 0

Answer:

So 0 dB does not mean no sound, it means a sound level where the sound pressure is equal to that of the reference level. This is a small pressure, but not zero. It is also possible to have negative sound levels: - 20 dB would mean a sound with pressure 10 times smaller than the reference pressure, i.e. 2 μPa.

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What is the difference between magnitude and intensity

MArishka [77]

The difference is intensity is assessed using a special scale and magnitude is measure of the size ( for example a earthquake)

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3 years ago

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sweet-ann [11.9K]

Answer:

it may be efficient at first but over time there will be a lot of wasted energy.

Explanation:

I'm pretty sure

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3 years ago

Which of the following is NOT a characteristic of a healthy diet?

anzhelika [568]

Answer: Drinking pop

Explanation:

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3 years ago

What current, in units of milliAmperes, would produce a potential difference of 4.5

Bess [88]

Answer;

<u>= 56.96 mA</u>

Explanation;

From the Ohm's law;

V = IR, where V is the potential difference in volts, I is the electric current in amperes and R is the resistance in ohms.

V = 4.5 volts, R = 79 Ω

Making I the subject of the formula we get;

I =V/R

= 4.5 /79

= 0.05696 A

But, 1 A = 1000 mA

Thus; 0.05696 × 1000

<u>= 56.96 mA</u>

5 0

3 years ago

The practical limit to an electric field in air is about 3.00 × 10^6 N/C . Above this strength, sparking takes place because air

AleksAgata [21]

Answer:

(a) x=157 m

(b) No

Explanation:

Given Data

Mass of proton m=1.67×10⁻²⁷kg

Charge of proton e=1.6×10⁻¹⁹C

Electric field E=3.00×10⁶ N/C

Speed of light c=3×10⁸ m/s

For part (a) distance would proton travel

Apply the third equation of motion

$(v_{f})^{2} =(v_{i})^{2}+2ax$

In this case vi=0 m/s and vf=c

$c^{2}=(0)^{2}+2ax\\ c^{2}=2ax\\x=\frac{c^{2} }{2a}$

$x=\frac{c^{2}}{2a}--------Equation (i)$

From the electric force on proton

$F=qE\\where\\ F=ma\\so\\ma=qE\\a=\frac{qE}{m}\\$

put this a(acceleration) in Equation (i)

$x=\frac{c^{2} }{2(qE/m)}\\ x=\frac{mc^{2}}{2qE} \\x=\frac{(1.67*10^{-27})*(3*10^{8})^{2} }{2*(1.6*10^{-19})*(3*10^{6})}\\ x=157m$

For part (b)

No the proton would collide with air molecule

7 0

3 years ago