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3 years ago

Why are some substances dissolve both polar and nonpolar substances?

Physics

1 answer:

dlinn [17]3 years ago

8 0

Answer:

Because water is polar and oil is nonpolar, their molecules are not attracted to each other. polar solvents dissolve polar solutes, and nonpolar solvents dissolve nonpolar solutes.

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What is the relationship between atmospheric pressure and the density of gas particles in an area of increasing pressure

mezya [45]

Answer:

this is a no brainer

Explanation:

As air pressure in an area increases, the density of the gas particles in that area increases.

8 0

3 years ago

Two objects are 5 kg and 10 kg respectively, and they're 10 m apart. if the distance between them is increased to 20 m, what hap

Sati [7]

I think it's OPTION A, IT DECREASES.

6 0

3 years ago

g In a certain binary-star system, each star has the same mass which is 8.2 times of that of the Sun, and they revolve about the

Mademuasel [1]

To solve this problem it is necessary to apply the concepts related to the Third Law of Kepler.

Kepler's third law tells us that the period is defined as

$T^2 = \frac{4\pi^2 d^3}{2GM}$

The given data are given with respect to known constants, for example the mass of the sun is

$m_s = 1.989*10^{30}$

The radius between the earth and the sun is given by

$r = 149.6*10^9m$

From the mentioned star it is known that this is 8.2 time mass of sun and it is 6.2 times the distance between earth and the sun

Therefore:

$m = 8.2*1.989*10^{30}$

$d = 6.2*149.6*10^6$

Substituting in Kepler's third law:

$T^2 = \frac{4\pi^2 d^3}{2}$

$T^2 = \frac{4\pi^2(6.2*149.6*10^9)^3}{2(6.674*10^{-11} )(8.2*1.989*10^30 )}$

$T=\sqrt{\frac{4\pi^2(6.2*149.6*10^9)^3}{2(6.674*10^{-11} )(8.2*1.989*10^30)}}$

$T = 120290789.7s$

$T = 120290789.7s(\frac{1year}{31536000s})$

$T \approx 3.8143 years$

Therefore the period of this star is 3.8years

7 0

3 years ago

The resistance created by waves on a 120-m-long ship is tested in a channel using a model that is 4 m long Y Part A If the ship

DanielleElmas [232]

Answer:

$V_m = 12.78 km/hr$

Explanation:

given,

length of the ship = 120 m

length of model of the ship = 4 m

Speed at which the ship travels = 70 km/h

speed of model = ?

by using froude's law

$F_r = \dfrac{V}{\sqrt{L g}}$

for dynamic similarities

$(\dfrac{V}{\sqrt{L g}})_P = (\dfrac{V}{\sqrt{L g}})_{model}$

$(\dfrac{V_p}{\sqrt{L_p}}) = (\dfrac{V_m}{\sqrt{L_m}})$

$(\dfrac{70}{\sqrt{120}}) = (\dfrac{V_m}{\sqrt{4}})$

$V_m = 12.78 km/hr$

hence, the velocity of model will be 12.78 km/h

6 0

3 years ago

A girl is sitting in a sled sliding horizontally along some snow (there is friction present). The mass of the girl is 29.8 kg an

jonny [76]

Answer:

28.81 m

Explanation:

Ff = -123

m * a = -123

(29.8+10.3) * a = -123

a = -123/40.1 = -3.07

We know,

v^2 = u^2 + 2as

0^2 = 13.3^2 + 2*(-3.07)*s

s = 176.89/6.14 = 28.81

[ If there's a problem with the solution, pleaase let me know ]

6 0

3 years ago