Answer
is: V<span>an't
Hoff factor (i) for this solution is 1,81.
Change in freezing point from pure solvent to
solution: ΔT =i · Kf · b.
Kf - molal freezing-point depression constant for water is 1,86°C/m.
b - molality, moles of solute per
kilogram of solvent.
</span><span>b = 0,89 m.
ΔT = 3°C = 3 K.
i = </span>3°C ÷ (1,86 °C/m · 0,89 m).
i = 1,81.
Answer:

Explanation:
Given that,
The mass of a Hubble Space Telescope, 
It orbits the Earth at an altitude of 
We need to find the potential energy the telescope at this location. The formula for potential energy is given by :

Where
is the mass of Earth
Put all the values,

So, the potential energy of the telescope is
.
Answer:
Explanation:
a ) Slit separation d = .1 x 10⁻³ m
Screen distance D = 4 m
wave length of light λ = 650 x 10⁻⁹ m
Width of central fringe = λ D / d
= 
= 26 mm
b ) Distance between 1 st and 2 nd bright fringe will be equal to width of dark fringe which will also be equal to 26 mm
c ) Angular separation between the central maximum and 1 st order maximum will be equal to angular width of fringe which is equal to
λ / d
= 
= 6.5 x 10⁻³ radian.

The momentul of the system preserves:
Ok, we found the speed after the collision.
Now, because the impact is plastic, it produces heat, sound energy and who knows what other forms of energy. We denote all this wasted energy with

.
Now, we write the energy conservation law:

From the above equation, you find

, and then conclude that the sound energy can certainly not be greater than this.