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zavuch27 [327]
2 years ago
6

The temperature of the liquid in a container decreases as the liquid evaporates. Use kinetic theory to explain why.

Physics
1 answer:
melisa1 [442]2 years ago
6 0

Answer:

Temperature decreases because the number of collision of the molecules decreases as they escape or evaporate. Molecules are in constant motion. Increase in temperature leads to increase in average kinetic energy of the molecules.

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The hydraulic cylinder imparts a constant upward velocity vA = 0.23 m/s to corner A of the rectangular container during an inter
s344n2d4d5 [400]

Answer:

vB = 0.5418 m/s (→)

aB = - (0.3189/L)  m/s²

ωcd = (0.2117/L)  rad/s

Explanation:

a) Given:

vA = 0.23 m/s (↑) (constant value)

If

tan θ = vA/vB

For the instant when θ = 23° we have

vB = vA/ tan θ

⇒ vB = 0.23 m/s/tan 23°

⇒ vB = 0.5418 m/s (→)

b) If tan θ = vA/vB   ⇒   vA = vB*tan θ

⇒  d(vA)/dt = d(vB*tan θ)/dt

⇒  0 = tan θ*d(vB)/dt + vB*Sec²θ*dθ/dt

Knowing that  

aB = d(vB)/dt

ωcd = dθ/dt

we have

⇒  0 = tan θ*aB + vB*Sec²θ*ωcd

ωcd = - Sin (2θ)*aB/(2*vB)

If

v = ωcd*L

where v = vA*Cos θ   ⇒  ωcd = v/L = vA*Cos θ/L

⇒ vA*Cos θ/L = - Sin (2θ)*aB/(2*vB)

⇒ aB = - vA*vB/((Sin θ)*L)

We plug the known values into the equation

aB = - (0.23 m/s)*(0.5418 m/s)/(L*Sin 23°)

⇒ aB = - (0.3189/L)  m/s²

Finally we obtain the angular velocity of CD as follows

ωcd = vA*Cos θ/L

⇒ ωcd = 0.23 m/s*Cos 23°/L

⇒ ωcd = (0.2117/L)  rad/s

5 0
3 years ago
Two wires are parallel and one is directly above the other. Each has a length of 50.0 m and a mass per unit length of 0.020 kg/m
ivann1987 [24]

Answer: The time required for the impluse passing through each other is approximately 0.18seconds

Explanation:

Given:

Length,L = 50m

M/L = 0.020kg/m

FA = 5.7×10^2N

FB = 2.5×10^2N

The sum of distance travelled by each pulse must be 50m since each pulse started from opposite ends.

Ca(t) + CB(t) = 50

Where CA and CB are the velocities of the wire A and B

t = 50/ (CA + CB)

But C = Sqrt(FL/M)

Substituting gives:

t = 50/ (Sqrt( FAL/M) + Sqrt(FBL/M))

t = 50/(Sqrt 5.7×10^2/0.02) + (Sqrt(2.5×10^2/0.02))

t = 50 / (168.62 + 111.83)

t = 50/280.15

t = 0.18 seconds

4 0
3 years ago
Read 2 more answers
Guys I just need this question, cmon help me
slavikrds [6]
There can be an experiment run using two or more cleansers at the same time as the manufacturers' one
3 0
3 years ago
A radioactive isotope of the element potassium decays to produce argon. If the ratio of argon to potassium is found to be 31:1,
iris [78.8K]

Answer:

Explanation:

Argon to potassium ratio after 1 half life = 1:1

After 2 half lives = 75/25= 3:1

After 3 half lives = 87.5/12.5= 7:1

After 4 half lives = 93.75/6.25 = 15:1

After 5 half lives = 96.875/3.125 = 31/1

8 0
3 years ago
A woman on a bridge 82.2 m high sees a raft floating at a constant speed on the river below. She drops a stone from rest in an a
solong [7]

Answer:

0.71 m/s

Explanation:

We find the time it takes the stone to hit the water.

Using y = ut - 1/2gt² where y = height of bridge, u = initial speed of stone = 0 m/s, g = acceleration due to gravity = -9.8 m/s² (negative since it is directed downwards)and t = time it takes the stone to hit the water surface.

So, substituting the values of the variables into the equation, we have

y = ut - 1/2gt²

82.2 m = (0m/s)t - 1/2( -9.8 m/s²)t²

82.2 m = 0 + (4.9 m/s²)t²

82.2 m =  (4.9 m/s²)t²

t²  = 82.2 m/4.9 m/s²

t² = 16.78 s²

t = √16.78 s²

t = 4.1 s

This is also the time it takes the raft to move from 5.04 m before the bridge to 2.13 m before the bridge. So, the distance moved by the raft in time t = 4.1 s is 5.04 m - 2.13 m = 2.91 m.

Since speed = distance/time, the raft's speed v = 2.91 m/4.1 s = 0.71 m/s

5 0
3 years ago
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