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Aleks [24]
3 years ago
13

What is surface tension ​

Physics
1 answer:
harina [27]3 years ago
5 0

Answer:

Surface tension is,  the surface where the water meets the air, water molecules cling even more tightly to each other.

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A force of 100 newtons was necessary to lift a rock. a total of 150 joules of work was done. how far was the rock lifted
Makovka662 [10]

Answer:

1.5m

Explanation:

150÷100=1.5

Work done ÷force

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a string with fixed ends is made to go into standing wave patterns. at a given tension, the lowest frequency for which a certain
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700 hz

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Some minerals are evenly distributed around the planet, while other minerals are quite rare.
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it all depends on Region and how long they have been there some minerals crush faster then some others so some might only take a few thousand while others might take millions

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2 years ago
A weight lifter struggles but manages to keep a heavy barbell abovehis head. Occasionally he slips and the barbell starts to fal
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3 years ago
Read 2 more answers
A wooden object (conically shaped) has a diameter of 8cm and height of 14cm. It floats in oil with 6cm of its height above oil l
baherus [9]

Answer:

(a) The density of the object is 316/343 × the density of the oil

(b) The fraction of oil displaced after immersing the object is 0.461 of the oil volume

Explanation:

(a) The volume, V of a cone of height, h and base diameter, D = 2×r is given by the following equation;

V = \dfrac{\pi r^{2} h}{3}

The volume of the object is therefore;

\dfrac{\pi \times 4^{2} \times 14}{3} = 74\tfrac{2}{3}\pi \, cm^3

Where 6 cm is above the oil level we have;

\dfrac{\pi \times \left (6 \times \dfrac{4}{14}   \right )^{2} \times 6}{3} = 5\tfrac{43}{49}\pi \, cm^3 above the oil level

Therefore, volume of the oil displaced = 68\tfrac{116}{147}\pi cm³ = 216.11 cm³

The density of the object is thus;

\dfrac{68\tfrac{116}{147}\pi}{ 74\tfrac{2}{3}\pi} \times  Density \ of \ the \ oil = \dfrac{316}{343}  \right ) \times  Density \ of \ the \ oil

The density of the object = 316/343 × the density of the oil.

(b) The volume of the oil = 2 × Volume of the object = 2 \times 74\tfrac{2}{3}\pi \, cm^3 = 149\tfrac{1}{3}\pi \, cm^3

The fraction of the volume displaced, x, after immersing the object is given as follows;

x = \dfrac{68\tfrac{116}{147}\pi}{ 149\tfrac{1}{3}\pi} = \dfrac{158}{343} = 0.461

The fraction of oil displaced after immersing the object = 0.461 of the volume of the oil

8 0
3 years ago
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