PHYSICS COLLEGE

20.0 moles, 1840 g, of a nonvolatile solute, C 3H 8O 3 is added to a flask with an unknown amount of water and stirred. The solution is allowed to reach 90.0°C . The vapor pressure of pure water at this temperature is 528.8 mm Hg. The vapor pressure of the solution is 423.0 mm Hg. How many kg of water was present?

Answers

Answer 1

Answer:

0.144 kg of water

Explanation:

From Raoult's law,

Mole fraction of solvent = vapor pressure of solution ÷ vapor pressure of solvent = 423 mmHg ÷ 528.8 mmHg = 0.8

Let the moles of solvent (water) be y

Moles of solute (C3H8O3) = 2 mole

Total moles of solution = moles of solvent + moles of solute = (y + 2) mol

Mole fraction of solvent = moles of solvent/total moles of solution

0.8 = y/(y + 2)

y = 0.8(y + 2)

y = 0.8y + 1.6

y - 0.8y = 1.6

0.2y = 1.6

y = 1.6/0.2 = 8

Moles of solvent (water) = 8 mol

Mass of water = moles of water × MW = 8 mol × 18 g/mol = 144 g = 144/1000 = 0.144 kg

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A hemispherical surface (half of a spherical surface) of radius R is located in a uniform electric field of magnitude E that is parallel to the axis of the hemisphere. What is the magnitude of the electric flux through the hemisphere surface?

A glass plate 2.95 mmmm thick, with an index of refraction of 1.60, is placed between a point source of light with wavelength 600 nmnm (in vacuum) and a screen. The distance from source to screen is 1.25 cm. How many wavelengths are there between the source and the screen?

Answers

Answer:

$N_T=2086285.67$

Explanation:

Given;

Thickness of the glass plate, $x=2.95* 10^(-3)\ m$

refractive index of the glass plate, $n=1.6$

wavelength of light source in vacuum, $\lambda=600* 10^(-9)\ m$

distance between the source and the screen, $d=1.25\ m$

Distance travelled by the light from source to screen in vacuum:

$d_v=d-x$

$d_v=1.25-0.00295$

$d_v=1.24705\ m$

So the no. of wavelengths in the vacuum:

$N=(d_v)/(\lambda)$

$N=(1.24705)/(6* 10^(-7))$

$N\approx2.0784* 10^(6)$ .......................(1)

Now we find the wavelength of the light wave in the glass:

$n=(\lambda)/(\lambda')$

where:

$\lambda'=$ wavelength of light in the medium of glass.

$1.6=(600* 10^(-9))/(\lambda')$

$\lambda'=375* 10^(-9)\ m=375\ nm$

Now the no. of wavelengths in the glass:

$N'=(x)/(\lambda')$

$N'=(2.95* 10^(-3))/(375* 10^(-9))$

$N'=7.8667* 10^(3)$ ............................(2)

From (1) & (2):

total no. of wavelengths are there between the source and the screen:

$N_T=N+N'$

$N_T=2086285.67$

A basketball player jumps straight up for a ball. To do this, he lowers his body 0.250 m and then accelerates through this distance by forcefully straightening his legs. This player leaves the floor with a vertical velocity sufficient to carry him 0.960 m above the floor. (a) Calculate his velocity (in m/s) when he leaves the floor. 4.33 Correct: Your answer is correct. m/s (b) Calculate his acceleration (in m/s2) while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.250 m.

Answers

Answer:

a. 3.73 m/s b. 27.8 m/s²

Explanation:

(a) Calculate his velocity (in m/s) when he leaves the floor.

Using the conservation of energy principles,

Potential energy gained by basketball player = kinetic energy loss of basket ball player

So, ΔU + ΔK = 0

ΔU = -ΔK

mg(h' - h) = -1/2m(v'² - v²)

g(h' - h) = -1/2(v'² - v²) where g = acceleration due to gravity = 9.8 m/s², h' = 0.960 m, h = 0.250 m, v' =0 m/s (since the basketball player momentarily stops at h' = 0.960 m) and v = velocity with which the basketball player leaves the floor

Substituting the values of the variables into the equation, we have

9.8 m/s²(0.960 m - 0.250 m) = -1/2((0 m/s)² - v²)

9.8 m/s²(0.710 m) = -1/2(-v²)

6.958 m²/s² = v²/2

v² = 2 × 6.958 m²/s²

v² = 13.916 m²/s²

v = √(13.916 m²/s²)

v = 3.73 m/s

(b) Calculate his acceleration (in m/s2) while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.250 m.

Using v² = u² + 2as where u = initial speed of basketball player before lengthening = 0 m/s, v = final speed of basketball player after lengthening = 3.73 m/s, a = acceleration during lengthening and s = distance moved during lengthening = 0.250 m

So, making, a subject of the formula, we have

a = (v² - u²)/2s

Substituting the values of the variables into the equation, we have

a = ((3.73 m/s)² - (0 m/s)²)/(2 × 0.250 m)

a = (13.913 m²/s² - 0 m²/s²)/(0.50 m)

a = 13.913 m²/s²/(0.50 m)

a = 27.83 m/s²

a ≅ 27.8 m/s²

What is the revolution ratio of two spur gears with one having 12 teeth and the other having 36 teeth

Answers

3:1 is the ratio i hope this belps

A block m1 rests on a surface. A second block m2 sits on top of the first block. A horizontal force F applied to the bottom block pulls both blocks at constant velocity. Here m1 = m2 = m.(a)
What is the normal force exerted by the surface on the bottom block? (Use the following as necessary: m and g as necessary.)

Answers

(a) The normal force exerted by the surface on the bottom block is N1 = 2mg.

Given that,

A block m1 rests on a surface.
A second block m2 sits on top of the first block.
A horizontal force F applied to the bottom block pulls both blocks at constant velocity. Here m1 = m2 = m.

Based on the above information, we can say that the N1 is 2mg.

Learn more: brainly.com/question/17429689

Answer:

N = 2mg

Explanation:

Assuming the surface is horizontal

The surface must provide enough normal force to prevent the masses from accelerating in the vertical direction.

Consider different points along one spoke of a wheel rotating with constant angular velocity. Which of the following is true regarding the centripetal acceleration at a particular instant of time?a. The magnitude of the centripetal acceleration is greater for points on the spoke closer to the hub than for points closer to the rim
b. both the magnitude and the direction of the centripetal acceleration depend on the location of the point on the spoke.
c. The magnitude of the centripetal acceleration is smaller for points on the spoke closer to the hub than for points closer to the rim but the direction of the acceleration is the same at all points on this spoke.
d. The magnitude and direction of the centripetal acceleration is the same at all points on this spoke.

Answers

Answer:

Option (a).

Explanation:

Let the angular velocity is w.

The centripetal acceleration is given by

$a = r w^2$

where, r is the distance between the axle and the spoke.

So, more is the distance more is the centripetal acceleration.

(a) For the points on the spoke closer to the hub than for points closer to the rim is larger distance, so the centripetal force is more.

The statement is true.

(b) The direction of centripetal acceleration is always towards the center, so the statement is false.

(d) It is false.

Option (a) is correct.

At a given instant, a 2.2 A current flows in the wires connected to a parallel-plate capacitor. What is the rate at which the electric field is changing between the plates if the square plates are 2.0 cm on a side? Express your answer using two significant figures

Answers

Answer:

Check attachment for better understanding

Explanation:

Given that,

Current in wire I =2.2A

Capacitor plate dimension is 2cm by 2cm

s=2cm=2/100 = 0.02m

Rate at which electric field Is changing dE/dt?

The current in the wires must also be the displacement current in the capacitor. We find the rate at which the electric field is changing from

ID = ε0•A•dE/dt

Where ε0 is a constant

ε0= 8.85×10^-12C²/Nm²

Area of the square plate is

A =s² =0.02² = 0.0004m²

Then,

Make dE/dt the subject of formula

dE/dt = ID/ε0A

dE/dt = 2.2 / (8.85×10^-12 ×4×10^-4)

dE/dt = 6.215×10^14 V/m-s

dE/dt = 6.215×10^14 N/C.s

The rate at which the electric field is changing between the plates is 6.215×10^14 N/C.s

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