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2 years ago

If the humidity in a room of volume 450 m3 at 30 ∘C is 75%, what mass of water can still evaporate from an open pan?

1 answer:

5 0

From tables,

SVP at 30°C = 4.24 kPa

From ideal gas expressions;
n = PV/RT = (4.24*1000*450)/(8.314*303) = 757.4 moles

Now, 75% of 757.4 moles will evaporate leaving 20%. Then, 25% of 757.5 moles...
25% of 757.4 moles = 25/100*757.4 = 189.35 moles
Mass of 189.35 moles = 189.35 moles*18 g/mol = 3408.3 g ≈ 3.4 kg

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Two open organ pipes, sounding together, produce a beat frequency of 8.0 Hz . The shorter one is 2.08 m long. How long is the ot

s344n2d4d5 [400]

Answer:

The length of the longer pipe is L = 2.30 m

Explanation:

Given that:

Two open organ pipes, sounding together, produce a beat frequency of 8.0 Hz . The shorter one is 2.08 m long.

How long is the other pipe?

From above;

The formula for the frequency of open ended pipes can be expressed as:

$f = \dfrac{nv}{2L}$

where n = 1 ( since half wavelength exist between those two pipes)

v = 343 m/s and L = 2.08 m

Thus, the shorter pipe produces a frequency of :

$f = \dfrac{1*343}{2*2.08}$

$f = \dfrac{343}{4.16}$

$f =82.45 \ Hz$

Also; we know that the beat frequency was given as 8.0 Hz

Then,

The lower frequency of the longer pipe = ( 82.45 - 8.0 )Hz

The lower frequency of the longer pipe = 74.45 Hz

Finally;

From the above equation; make Length L the subject of the formula. Then,

The length of the longer pipe is L = $\dfrac{nv}{2f}$

The length of the longer pipe is L = $\dfrac{1*343}{2*74.45}$

The length of the longer pipe is L = $\dfrac{343}{148.9}$

The length of the longer pipe is L = 2.30 m

6 0

2 years ago

Calculate a pendulum's frequency of oscillation (in Hz) if the pendulum completes one cycle in 0.5 s.

Marina86 [1]

Time taken to complete one oscillation for a pendulum is Time Period, T = 0.5 s
Frequency of the pendulum oscillation = 1 / Time Period => f = 1 / T = 1 / 0.5
Frequency f = 2 Hz

3 0

2 years ago

Two friends of different masses are on the playground. They are playing on the seesaw and are able to balance it even though the

Westkost [7]

Answer:

They are able to balance torques due to gravity.

$F_1 L_1 = F_2L_2$

Explanation:

When two friends of different masses will balance themselves on see saw then at equilibrium position the see saw will remain horizontal

This condition will be torque equilibrium position where the see saw will not rotate

Here we can say

$F_1 L_1 = F_2L_2$

here we know that force is due to weight of two friends

and their positions are different with respect to the lever about which see saw is rotating

since both friends are of different weight so they will balance themselves are different positions as per above equation

5 0

2 years ago

The burning of fossil fuels contributes to the addition of greenhouse gases to the atmosphere. These gases trap thermal energy i

denpristay [2]

Answer:

B. there would be a global rise in temperatures

Hence, global warming

Explanation:

hope this helped! :D

5 0

2 years ago

Read 2 more answers

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a un

Kamila [148]

Answer:

a) 6738.27 J

b) 61.908 J

c) $\frac{4492.18}{v_{car} ^{2} }$

Explanation:

The complete question is

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.

Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?

Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?

Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar. Enter an expression for the mass of the car, in terms of the quantities defined here.

moment of inertia is given as

$I$ = $\frac{1}{2}$ $mr^{2}$

where m is the mass of the flywheel,

and r is the radius of the flywheel

for the flywheel with radius 1.1 m

and mass 11 kg

moment of inertia will be

$I$ = $\frac{1}{2}$ $*11*1.1^{2}$ = 6.655 kg-m^2

The maximum speed of the flywheel = 35 m/s

we know that v = ωr

where v is the linear speed = 35 m/s

ω = angular speed

r = radius

therefore,

ω = v/r = 35/1.1 = 31.82 rad/s

maximum rotational energy of the flywheel will be

E = $Iw^{2}$ = 6.655 x $31.82^{2}$ = 6738.27 J

b) second flywheel has

radius = 2.8 m

mass = 16 kg

moment of inertia is

$I$ = $\frac{1}{2}$ $mr^{2}$ = $\frac{1}{2}$ $*16*2.8^{2}$ = 62.72 kg-m^2

According to conservation of angular momentum, the total initial angular momentum of the first flywheel, must be equal to the total final angular momentum of the combination two flywheels

for the first flywheel, rotational momentum = $Iw$ = 6.655 x 31.82 = 211.76 kg-m^2-rad/s