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3 years ago

What purpose would the sand serve?

Physics

2 answers:

maxonik [38]3 years ago

7 0

Answer:

Sand is a granular material composed of finely divided rock and mineral particles.

Explanation:

Alexus [3.1K]3 years ago

4 0

Answer:

Sand can also refer to a textural class of soil or soil type; i.e., a soil containing more than 85 percent sand-sized particles by mass.

Explanation:

Interesting fact...

Why are there differences between the cooling rates of sand and water? Sand cools off faster than water because sand is a solid. Water heats up slower and cools off slower than sand. Sand heats up faster and cools off faster than water.

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Change the following sentences into Passive Voice: (5M)

geniusboy [140]

Answer:

passive voice

many messengers all over the world was sent by emperor Ashoka to preach Buddhism.

Which pen was liked by you?

Has your passport size photo been taken for the application form?

A beautiful bicycle was given to me on my birthday by my father.

The plants is being watered by the gardener.

Paul said that he will never leave you and he will always be with you.

7 0

3 years ago

A 3kg horizontal disk of radius 0.2m rotates about its center with an angular velocity of 50rad/s. The edge of the horizontal di

Lyrx [107]

Answer:

Explanation:

From the information given:

The angular speed for the block $\omega = 50 \ rad/s$

Disk radius (r) = 0.2 m

The block Initial velocity is:

$v = r \omega \\ \\ v = (0.2 \times 50) \\ \\ v= 10 \ m/s$

Change in the block's angular speed is:

$\Delta _{\omega} = \omega - 0 \\ \\ = 50 \ rad/s$

However, on the disk, moment of inertIa is:

$I= mr^2 \\ \\ I = (3 \times 0.2^2) \\ \\ I = 0.12 \ kgm^2$

The time t = 10s

∴

Frictional torques by the wall on the disk is:

$T = I \times (\dfrac{\Delta_{\omega}}{t}) \\ \\ = 0.12 \times (\dfrac{50}{10}) \\ \\ =0.6 \ N.m$

Finally, the frictional force is calculated as:

$F = \dfrac{T}r{}$

$F= \dfrac{0.6}{0.2} \\ \\ F = 3N$

8 0

3 years ago

Two objects with masses m and 3m are connected by a compressed spring so that the energy stored in the spring is 375 joules. If

cluponka [151]

Answer:

281.25 J

Explanation:

We are told that the two objects with masses m and 3m.

Also that energy stored in the spring is 375 joules.

Now, initially the centre of mass of the system took place at rest, it means v1 = v and v2 = v/3

Thus, from principle of conservation of energy, we have;

½mv² + ½(3m)(v/3)² = 375J

(m + 3m/9)½v² = 375

(4/3)m × ½v² = 375

Multiply both sides by ¾ to get;

½mv² = 375 × ¾

½mv² = 281.25 J

Therefore, energy of lighter body is 281.25 J

7 0

3 years ago

Volcanic ash and sulfur dioxide spewed out of Mt. Pinatubo in 1991. These materials can reflect incoming solar radiation. Over t

gogolik [260]

The suspended ash made for some some spectacular sunsets! Sulfuric acid was spread worldwide, increasing acidity of rain. Ash deflected energy from the sun, causing a slight drop on global temps for a few years.

7 0

3 years ago

Read 2 more answers

A 25.0-g sample of copper at 363 K is placed in I 00.0 g of water at 293 K. The copper and water quickly come to the sa me tempe

Simora [160]

Answer:

Final temperature is 295K

Explanation:

Where the sample of copper is placed in the water, the heat transferred from the copper is equal that the heat absorbed by the water.

The heat transferred from the copper is:

C× $\frac{1mol}{63,546g}$ ×mass×ΔT

Where C is molar heat capacity of copper (24,5J/molK)

Mass is 25,0g

And ΔT is final temperature - initial temperature (X-363K)

Also, the heat absorbed by the water is:

-C× $\frac{1mol}{18,02g}$ ×mass×ΔT

Where C is molar heat capacity of water (75,2J/molK)

Mass is 100,0g

And ΔT is final temperature - initial temperature (X-293K)

As heat transferred is equal to heat absorbed:

24,5J/molK× $\frac{1mol}{63,546g}$ ×25,0g×(X-363K) = -75,2J/molK× $\frac{1mol}{18,02g}$ ×100,0g× (X-293K)

9,64X J/K - 3499J = - 417X J/K + 122273J

426,64X J/K = 125772 J

Final temperature is 295K

I hope it helps!

6 0

3 years ago