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katovenus [111]
2 years ago
5

A common small-molecular weight (and therefore fast diffusing for an organic molecule) ingredient in perfumes is vanillin, the p

rimary component of vanilla bean extract (molecular weight = 152). the d for vanillin in air is 0.114 cm2/s. if i open a bottle of vanilla on the other side of the room 3 meters away, and the air is still so there is no convection, about how long would i have to wait before i could expect to smell the vanilla?
Physics
2 answers:
natima [27]2 years ago
7 0
What's the answer shshshsbshsbsbs sbo
mote1985 [20]2 years ago
4 0

Answer:

2.54 seconds

Explanation:

Gathering the data:

let molecular weight = 152 g/mol

diffusion coefficient = 0.114 cm²/s

Distance = 3 meters

The most important thing to know is Fick's Law of diffusion. The law states that the flux is directly proportional to the concentration gradient. In other words

where dc = change in concentration

            D = diffusion coefficient

            Dx = change in the distance, cm

solving:

j = dc/3 (-0.114)

 this gives t = 2.54 seconds

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Leanne is riding a bike. The forward force from her pedalling is 18N. There is a backward force of 6N from friction and a backwa
olchik [2.2K]
the net force acting on the bike is=
18 - 6 - 12 = 0 N.

Since the net force acting on the bike is zero, the bike is moving with constant velocity. Therefore, the speed of the bike does not change.
4 0
2 years ago
Read 3 more answers
If we double only the amplitude of a vibrating ideal mass-and-spring system, the mechanical energy of the system:
Lelechka [254]

Answer:

D. increases by a factor of 4.

Explanation:

General equation of SHM

Lets taken the general equation of the displacement given as

x = A sinω t

A=Amplitude ,t=time ,ω=natural frequency

We know that speed V

V=\dfrac{dx}{dt}

V= A ω cosωt

The mechanical energy of spring mass system

U=\dfrac{1}{2}KA^2

K=Spring constant

Now when Amplitude A become 2 times then the mechanical energy will become 4 times.

Therefore the answer is D.

8 0
2 years ago
If gravity between the Sun and Earth suddenly vanished, Earth would continue moving in
Ksenya-84 [330]

Answer:

Earth would continue moving by uniform motion, with constant velocity, in a straight line

Explanation:

The question can be answered by using Newton's first law of motion, also known as law of inertia, which states that:

"an object keeps its state of rest or of uniform motion in a straight line unless acted upon by an external net force different from zero"

This means that if there are no forces acting on an object, the object stays at rest (if it was not moving previously) or it continues moving with same velocity (if it was already moving) in a straight line.

In this problem, the Earth is initially moving around the Sun, with a certain tangential velocity v. When the Sun disappears, the force of gravity that was keeping the Earth in circular motion disappears too: therefore, there are no more forces acting on the Earth, and so by the 1st law of Newton, the Earth will continue moving with same velocity v in a straight line.

6 0
2 years ago
I pull the throttle in my racing plane at a = 12.0 m/s2. I was originally flying at v = 100. m/s. Where am I when t = 2.0s, t =
Helen [10]
Summary:
a= 12.0 m/(s^2)
v= 100m/s
t1= 2.0s => s1=?
t2=5.0s => s2=?
t3=10.0s => s3=?
——————
Solution:
• when t1=2.0 s, I have gone:
S1= v*t1 + 1/2*a*(t1^2)
=100.0 *2 + 1/2*12.0*(2.0^2)
=224 (m)

• when t2=5.0s, I have gone
S2=v*t2+ 1/2*a*(t2^2)
= 100*5.0+ 1/2*12.0*(5.0^2)
=650 (m)

•when t3= 10.0s, I have gone:
S3=v*t3+ 1/2*a*(t3^2)
=100*10.0+ 1/2*12*(10.0^2)
=1600 (m)
7 0
2 years ago
A uniform disk has a mass of 3.7 kg and a radius of 0.40 m. The disk is mounted on frictionless bearings and is used as a turnta
yuradex [85]

Answer:

1.25 kgm²/sec

Explanation:

Disk inertia, Jd =

Jd = 1/2 * 3.7 * 0.40² = 0.2960 kgm²

Disk angular speed =

ωd = 0.1047 * 30 = 3.1416 rad/sec

Hollow cylinder inertia =

Jc = 3.7 * 0.40² = 0.592 kgm²

Initial Kinetic Energy of the disk

Ekd = 1/2 * Jd * ωd²

Ekd = 0.148 * 9.87

Ekd = 1.4607 joule

Ekd = (Jc + 1/2*Jd) * ω²

Final angular speed =

ω² = Ekd/(Jc+1/2*Jd)

ω² = 1.4607/(0.592+0.148)

ω² = 1.4607/0.74

ω² = 1.974

ω = √1.974

ω = 1.405 rad/sec

Final angular momentum =

L = (Jd+Jc) * ω

L = 0.888 * 1.405

L = 1.25 kgm²/sec

5 0
2 years ago
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