Answer:
Explanation:
According to given:
- molecular mass of glycerin,
- molecular mass of water,
- ∵Density of water is
- ∴mass of water in 316 mL,
- mass of glycerin,
- pressure of mixture,
- temperature of mixture,
<em>Upon the formation of solution the vapour pressure will be reduced since we have one component of solution as non-volatile.</em>
<u>moles of water in the given quantity:</u>
<u>moles of glycerin in the given quantity:</u>
<u>Now the mole fraction of water:</u>
<em>Since glycerin is non-volatile in nature so the vapor pressure of the resulting solution will be due to water only.</em>
I'll tell you how I look at this, although I may be missing something important.
Position = x(t) = 0.5 sin(pt + p/3)
Speed = position' = x'(t) = 0.5 p cos(pt + p/3)
Acceleration = speed' = position ' ' = x ' '(t) = -0.5 p² sin(pt + p/3)
At (t = 1.0),
x ' '(t) = -0.5 p² sin( 4/3 p )
In order to evaluate this, don't I still have to know what 'p' is ? ?
I don't think it can be evaluated with the information given in the question.
I think the correct answer from the choices listed above is option D. One advantage of using electromagnets in devices would be that electromagnets can <span>easily be turned on and off. Hope this answers the question. Have a nice day.</span>
Answer:
S = 16 m
Explanation:
Given that
The frequency of the water waves, f = 4 Hz
The wavelength of the water waves, λ = 2 m
The time the waves reached the shore, t = 2 s
The relation between the velocity, wavelength, and the frequency of the wave is given by the relation,
v = f λ m/s
Substituting the given values in the above equation,
v = 4 x 2
= 8 m/s
The velocity of the water waves is v = 8 m/s
The distance between the shore and boat is given by
s = v x t
= 8 x 2
= 16 m
Hence, the distance between the boat and the shore is, s = 16 m
Answer
Pressure, P = 1 atm
air density, ρ = 1.3 kg/m³
a) height of the atmosphere when the density is constant
Pressure at sea level = 1 atm = 101300 Pa
we know
P = ρ g h
h = 7951.33 m
height of the atmosphere will be equal to 7951.33 m
b) when air density decreased linearly to zero.
at x = 0 air density = 0
at x= h ρ_l = ρ_sl
assuming density is zero at x - distance
now, Pressure at depth x
integrating both side
now,
h = 15902.67 m
height of the atmosphere is equal to 15902.67 m.
