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3 years ago

What is the strongest interparticle force is each substance ch3cl?

Physics

1 answer:

Veronika [31]3 years ago

8 0

Hydrogen bonding hope this helps

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A very powerful vacuum cleaner which has a hose of circular cross section can lift a brick of mass 12 kg when the hose is placed

valkas [14]

Answer:

A). 1.9 cm

Explanation:

m = Mass of brick = 12 kg

g = Acceleration due to gravity = 9.81 m/s²

r = Radius of hose

A = Area = $\pi r^2$

F = Force = $mg$

Let us assume that the pressure required to lift the brick would be atmospheric pressure

$P=\dfrac{F}{A}\\\Rightarrow P=\dfrac{F}{\pi r^2}\\\Rightarrow r=\sqrt{\dfrac{F}{\pi P}}\\\Rightarrow r=\sqrt{\dfrac{12\times 9.81}{\pi\times 101325}}\\\Rightarrow r=0.01923\ m=1.9\ cm$

The radius of the hose should be 1.9 cm

6 0

3 years ago

A wave has a frequency of 46 Hz and a wavelength of 1.7 meters. What is the speed of the wave?

olga55 [171]

Answer:0.588..

Explanation:

7 0

3 years ago

A 6.00-μf parallel-plate capacitor has charges of 40.0 μc on its plates. how much potential energy is stored in this capacitor?

dedylja [7]

Thew energy stored in a capacitor of capacitance $C$ and voltage between the plates $V$ is

$E=\frac{1}{2} CV^2=\frac{1}{2C} Q^2$ .

Substituting numerical value

$E=\frac{1}{2*6*10^{-6}} (40*10^{-6})^2\\ E=133.33\; \mu J$

7 0

3 years ago

A furlong is an old british unit of length equal to 0.125 mi, derived from the length of a furrow in an acre of ploughed land. a

castortr0y [4]

The speed of light is: c = 3x10^8 m/s <span>

or

c = 186,000,000 miles/sec = 1.86x10^8 mi/s

1 furlong = 0.125 mile

1 fortnight = 2 weeks(7d/wk)(24h/d)(3600s/h)
= 1209600s = 1.2096x10^6 s

Therefore,

c =1.86x10^8 mi/s(1furl/0.125mi)(1.2096x10^6s/fort)

<span>c = 18x10^14 furlong/fortnight = 18x10^8 Mfurlong/fortnight</span></span>

4 0

3 years ago

A 6.0 kg mass is placed on a 20º incline which has a coefficient of friction of 0.15. What is the acceleration of the mass down

Leona [35]

Answer:

Explanation:

The form of Newton's 2nd Law that we use for this is:

F - f = ma where F is the Force pulling the mass down the ramp forward, f is the friction trying to keep it from moving forward, m is the mass and a is the acceleration (and our unknown).

We know mass and we can find f, but we don't have F. But we can solve for that by rewriting our main equation to reflect F:

$wsin\theta-\mu F_n=ma$ That's everything we need.

w is weight: 6.0(9.8). Filling in:

6.0(9.8)sin20 - .15(6.0)(9.8) = 6.0a and

2.0 × 10¹ - 8.8 = 6.0a and

11 = 6.0a so

a = 1.8 m/s/s

6 0

3 years ago