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Ivenika [448]
3 years ago
6

3. The honeycomb-like appearance of this sandstone is a result of.

Physics
2 answers:
sladkih [1.3K]3 years ago
6 0

Answer:

The answer is "Option C".

Explanation:

Wedging Freeze is generated by repeated freezing. Freeze wedging occurs, whenever the water is turned into ice as a result of the 9 percent expansion. When it freezes, cracks full of water are forced to separate further. and other options are incorrect, that can be described as follows:

  • In option A, It is a state where the element converts into a liquid, that's why it is not correct.
  • In option B, It is a reaction in which the bonds of water is divided into particular substance, that's why it is not correct.
  • In option D, It is a form of mechanical or physical rock weathering, that's it is not correct.

LenaWriter [7]3 years ago
4 0

Answer

C. frost wedging.

Explanation:

  • The honey-comb-like structure formed on the sand stone is the result of stresses due to frost wedging in the sandstone.
  • As we know that the microscopic structure of snow is hexagonal and similar is the pattern being discussed here for a honeycomb.

Frost wedging is the process of wearing of rocks by the repeated cycles of freezing and melting of water in the porous cavities of the sandstone.

Due to anomalous expansion of water on freezing it exerts stress on the inner surfaces of the space it occupies in the porosity.

  • As we know that the structure of the molecular bonds is arranged in a cage-like structure of ice which wears out the particles of the rock leaving a honeycomb structured impression due to the stresses.

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A horizontal desk surface measures 1.7 m by 1.0 m. If the Earth's magnetic field has magnitude 0.42 mT and is directed 68° below
AlexFokin [52]

Answer:

The magnetic flux through the desk surface is 6.6\times10^{-4}\ T-m^2.

Explanation:

Given that,

Magnetic field B = 0.42 T

Angle =68°

We need to calculate the magnetic flux

\phi=BA\costheta

Where, B = magnetic field

A = area

Put the value into the formula

\phi=0.42\times10^{-3}\times1.7\times1.0\cos22^{\circ}

\phi=0.42\times10^{-3}\times1.7\times1.0\times0.927

\phi=6.6\times10^{-4}\ T-m^2

Hence, The magnetic flux through the desk surface is 6.6\times10^{-4}\ T-m^2.

3 0
3 years ago
A rocket in deep space has an empty mass of 220 kg and exhausts the hot gases of burned fuel at 2500 m/s. What mass of fuel is n
Scilla [17]

Answer:

Explanation:

Let fuel is released at the rate of dm / dt where m is mass of the fuel

thrust created on rocket

= d ( mv ) / dt

= v dm / dt

this is equal to force created on the rocket

= 220 dv / dt

so applying newton's law

v dm / dt = 220 dv / dt

v dm = 220 dv

dv / v = dm / 220

integrating on both sides

∫ dv / v    = ∫ dm / 220

lnv =  ( m₂ - m₁ ) / 220

ln4000 - ln 2500 = ( m₂ - m₁ ) / 220

( m₂ - m₁ ) = 220 x ( ln4000 - ln 2500 )

( m₂ - m₁ ) = 220 x ( 8.29  - 7.82 )

= 103.4 kg .

8 0
3 years ago
Write an essay on basketball history and how the game has changed over time
jarptica [38.1K]

Basket ball has gone through some changes and hence got into shape as what we play or watch nowadays. Overall rules have the same fundamental principles as set in 1891 by the founder of this game.

December 21st, in 1891, the introductory basketball game was played in Springfield, Massachusetts. It was first brought into shape by a Canadian-by birth, Dr. James Naismith. The basic idea of the new game was to keep the sports loving students in shape during the winters or in between the outdoor game seasons.

'The basket ball' which had a set of thirteen basic rules set up a strong foundation to the game and still played almost in the same style with few modifications.In the very year of 1891, after mixing and changing theme of several other played games of those days, basket ball was born.

Initially, basket ball with the 13 rules was played with 2 peach baskets setup as goals, which now a days are the baskets on a high poles although modified but with same basic idea. In the very first game played in Springfield, the set of players were able to score a single point only, in the whole game.

By comparing Naismith’s basketball and basketball of today we see a characteristic of the original game which had no dribbling, instead a player had to pass the ball to a another player of his team ,from his the very spot where he caught it. The second thing which has now changed is the fouls, in the original game if any team made 3 consecutive foul play ,the oponent received a goal in reward. Although this scoring technique doesn't exist in the modern basketball. Rather nowadays, if any team makes five fouls in one quarter, the offending team is in the penalty ,and shoot free throws are awarded against them.

All other rules are quiet intact in the modern play as well. As the game today,has extended to over 2 hundered countries and enjoyed by many on television screens, with modern umpiring and technologies being used in the game, but the original idea of the founder is still endured.

5 0
3 years ago
All but one statement applies to electromagnetic radiation
lina2011 [118]
<span>D) Electromagnetic radiation travels in the form of longitudinal waves.</span>
3 0
3 years ago
An electron is moving east in a uniform electric field of 1.55 N/C directed to the west. At point A, the velocity of the electro
valkas [14]

Answer:

Final velocity of electron, v=6.45\times 10^5\ m/s    

Explanation:

It is given that,

Electric field, E = 1.55 N/C

Initial velocity at point A, u=4.52\times 10^5\ m/s

We need to find the speed of the electron when it reaches point B which is a distance of 0.395 m east of point A. It can be calculated using third equation of motion as :

v^2=u^2+2as........(1)

a is the acceleration, a=\dfrac{F}{m}

We know that electric force, F = qE

a=\dfrac{qE}{m}

Use above equation in equation (1) as:

v^2=u^2+\dfrac{2qEs}{m}

v^2=(4.52\times 10^5\ m/s)^2+2\times \dfrac{1.6\times 10^{-19}\ C\times 1.55\ N/C}{9.1\times 10^{-31}\ kg}\times 0.395\ m

v = 647302.09 m/s

or

v=6.45\times 10^5\ m/s

So, the final velocity of the electron when it reaches point B is 6.45\times 10^5\ m/s. Hence, this is the required solution.

3 0
3 years ago
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