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3 years ago

Addition of a lubricant such as oil to a surface will __ friction ?

Physics

1 answer:

Mashcka [7]3 years ago

5 0

Answer:

decrease

Explanation:

Addition of a lubricant such as oil to a surface will decrease friction. This makes the surface more greasy or slippery. This is very much efficient in machines, for example.

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Which wave has a greater frequency

larisa86 [58]

Answer:

A I think

Explanation:

because what is the most frequency a because it has more frequency I think I'm not that sure

5 0

3 years ago

Read 2 more answers

URGENT HELP PLEASE, GIVING BRAINLIEST IF YOU ANSWER CORRECTLY!! (20 pts!!)

Kobotan [32]

The answer is c 1386j

This calculator is very helpful I use it on my homework

https://www.omnicalculator.com/physics/specific-heat

8 0

3 years ago

Which of the following is true of high clouds?

Wewaii [24]

I’m pretty sure it’s B

3 0

2 years ago

The pressure in a traveling sound wave is given by the equation ΔP = (1.78 Pa) sin [ (0.888 m-1)x - (500 s-1)t] Find (a) the pre

frozen [14]

Answer:

a) $P_m=1.78\ Pa$

b) $f=79.5775\ Hz$

c) $\lambda=7.076\ m$

d) $v=563.06\ m.s^{-1}$

Explanation:

<u>Given equation of pressure variation:</u>

$\Delta P= (1.78\ Pa)\ sin\ [(0.888\ m^{-1})x-(500\ s^{-1})t]$

We have the standard equation of periodic oscillations:

$\Delta P=P_m\ sin\ (kx-\omega.t)$

<em>By comparing, we deduce:</em>

(a)

amplitude:

$P_m=1.78\ Pa$

(b)

angular frequency:

$\omega=2\pi.f$

$2\pi.f=500$

∴Frequency of oscillations:

$f=\frac{500}{2\pi}$

$f=79.5775\ Hz$

(c)

wavelength is given by:

$\lambda=\frac{2\pi}{k}$

$\lambda=\frac{2\pi}{0.888}$

$\lambda=7.076\ m$

(d)

Speed of the wave is gives by:

$v=\frac{\omega}{k}$

$v=\frac{500}{0.888}$

$v=563.06\ m.s^{-1}$

8 0

3 years ago

An oil slick on water is 120 nm thick and illuminated by white light incident perpendicular to its surface. What color does the

gregori [183]

Answer:

$\lambda = 672 nm$

so this is nearly red colour light

Explanation:

As we know that the interference of light from reflected light then the path difference is given as

$\Delta x = 2\mu t + \frac{\lambda}{2}$

now we know that for constructive interference of light the path difference is given as

$\Delta x = n\lambda$

so we will have

$2\mu t + \frac{\lambda}{2} = N\lambda$

so we will have

$4\mu t = \lambda$

$\lambda = 2(1.40)(120nm)$

$\lambda = 672 nm$

so this is nearly red colour light

8 0

3 years ago