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Oliga [24]
3 years ago
13

Two capillary tubes, made of the same substance, are lowered into a water bath. The radius of the larger tube is twice that of t

he smaller tube. If water rises a height 8.8 cm above the surface of the water bath in the smaller tube, how high (in cm) will it rise in the larger tube?
Physics
1 answer:
Alinara [238K]3 years ago
8 0

Answer: 4.4 cm.

Explanation:

Rise of water in the smaller tube= h1=8.8 centimeter(cm)

Radius of the smaller tube= r

Rise of water in the larger tube= h2 (in centimeter).

The radius of the larger tube is twice that of the smaller tube means that;

The radius in the larger tube is 2r ( 2 multiply by the radius r, of the smaller tube)

Using Jurin's law;

height or rise of liquid is inversely proportional to its radius, r.

That is; hr= constant.

Therefore, we have;

h1 × r1 = h2 × r2.

Rise in smaller tube × radius of the smaller tube = height of the larger tube × radius of the larger tube.

8.8 cm × r = h2 × 2r

= (8.8cm)r = (h2) 2r

Divide both sides by 2r, we then have;

8.8cm r/ 2r = h2

h2= 4.4cm

Therefore, the height or rise in large tube is half of that of the smaller tube.

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This question is incomplete the complete question is

A diver bounces straight up from a diving board, avoiding the diving board on the way down, and falls feet first into a pool. She starts with a velocity of 4.00 m/s and her takeoff point is 1.80 m above the pool. (a) What is her highest point above the board? (b) How long a time are her feet in the air? (c) What is her velocity when her feet hit the water?

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Explanation:

(a) To reach maximum distance

g=-9.8m/s^{2}\\ Vf=0\\v_{b}^{2}=v_{a}^{2}+2gx_{s} \\  x_{s}=\frac{0-(3^{2} )}{-2*9.8}\\ x_{s}=0.459m

(b) For Time

To find t we must find t1 and t2

as

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t_{1}=(Vb-Va)/g \\t_{1}=(0-3)/9.8\\t_{1}=0.306s

For T2

x_{l}=Vbt+(1/2)gt_{2}^{2}\\   as\\x_{l}=x_{1}+x_{s}\\x_{l}=1.8+0.459\\x_{l}=2.259\\so\\t_{2}=\frac{2.259*2}{9.8} \\t_{2}=0.6789s

For Total Time

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t=0.306+0.6789

t=0.984s

(c) To find Vc

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Vc=(0)+(9.8)(0.6789)

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