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e-lub [12.9K]
3 years ago
14

Two identical beakers are both equally filled with a liquid. Beaker A contains water. Beaker B contains a liquid that is denser

than water. Which statement is true? A. The air pressure on the beaker containing dense liquid is greater. B. The weight of both liquids is the same. C. The area at the bottom of both beakers is the same. D. The pressure at the bottom of both beakers is the same.
Physics
1 answer:
Leviafan [203]3 years ago
7 0
Correct answer is C: the two beakers are identical, so the area at the bottom of the two liquids is the same.

The other statements, instead, are false. 
Statement A is false because the air pressure on top of the two liquids is the same (it is equal to the atmospheric pressure).
Statement B is false because the weight of liquid B is greater than the weight of liquid A, because liquid B is more dense, and the weight is equal to
mg= \rho V g
where \rho is the liquid density and V the volume. Since the volume of the two liquids is the same, the weight of liquid B is greater than the weight of liquid A.
Statement D is false because the pressure at the bottom of beaker B is greater. In fact, the pressure of a column of liquid is 
p= \rho g h
where \rho is the density and h is the height of the column of liquid: since the height of the two liquids is the same, but the density of liquid B is greater than the density of liquid A, then its pressure at the bottom is also greater.
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Which of the following is an example of a example of a decomposition recreation
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You blow across the open mouth of an empty test tube and producethe fundamental standing wave of the air column inside the testt
sertanlavr [38]

Answer:

(a). The frequency of this standing wave is 0.782 kHz.

(b). The frequency of the fundamental standing wave in the air is 1.563 kHz.

Explanation:

Given that,

Length of tube = 11.0 cm

(a). We need to calculate the frequency of this standing wave

Using formula of fundamental frequency

f_{1}=\dfrac{v}{4l}

Put the value into the formula

f_{1}=\dfrac{344}{4\times0.11}

f_{1}=781.81\ Hz

f_{1}=0.782\ kHz

(b). If the test tube is half filled with water

When the tube is half filled the effective length of the tube is halved

We need to calculate the frequency

Using formula of fundamental frequency of the fundamental standing wave in the air

f_{1}=\dfrac{v}{4(\dfrac{L}{2})}

Put the value into the formula

f_{1}=\dfrac{344}{4\times\dfrac{0.11}{2}}

f_{1}=1563.63\ Hz

f_{1}=1.563\ kHz

Hence, (a). The frequency of this standing wave is 0.782 kHz.

(b). The frequency of the fundamental standing wave in the air is 1.563 kHz.

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In April 1974, Steve Prefontaine completed a 10.0 km race in a time of 27 min , 43.6 s . Suppose "Pre" was at the 7.43 km mark a
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Answer:

Acceleration, a = 0.101 m/s²

Explanation:

Average speed = total distance / total time.

At the 7.43km mark, total distance = 7.43km or 7430m

Total time = 25 * 60 s = 1500s

Average speed = 7430m/1500s = 4.95m/s

He then covers (10 - 7.43)km = 2.57 km = 2570 m

in t = 27m43.6s - 25min = 2m43.6s = 163.6 s

Then he accelerates for 60 s, and maintains this velocity V, for the remaining (163.6 - 60)s = 103.6 s.

From V = u + at; V = 4.95m/s + a *60s

Distance covered while accelerating is

s = ut + ½at² = 4.95m/s * 60s + ½ a *(60s)² = 297m + a*1800s²

Distance covered while at constant velocity, v after accelerating is

D = velocity * time

Where v = 4.95m/s + a*60s

D = (4.95m/s + a*60s) * 103.6s = 512.82m + a*6216s²

Total distance covered after initial 7.43 km, S + D = 2570 m, so

2570 m = 297m + a*1800s² + 512.82m + a*6216s²

2570 = 809.82 + a*8016

a = 809.82m / 8016s² = 0.101 m/s²

8 0
3 years ago
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