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3 years ago

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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An unknown number of identical light bulbs are connected to a 15 V battery in parallel. The current through the battery is 2 A.

zaharov [31]

Answer:

An unknown number of identical light bulbs are connected to a 15 V battery in parallel. The current through the battery is 2 A. If the light bulbs are connected to the battery in series, the current through the battery is 5 mA. How many bulbs are there?

3 bulbs are there from the analogy given above

Explanation:

7 0

3 years ago

A research vessel is mapping the bottom of the ocean using sonar. It emits a short sound pulse called "ping" downward. The frequ

Svetradugi [14.3K]

Answer:

d = 4180.3m

wavelengt of sound is 0.251m

Explanation:

Given that

frequency of the sound is 5920 Hz

v=1485m/s

t=5.63s

let d represent distance from the vessel to the ocean bottom.

an echo travels a distance equivalent to 2d, that is to and fro after it reflects from the obstacle.

$velocity=\frac{distance}{time}\\\\ v=\frac{2d}{t} \\\\vt=2d\\\\d=\frac{vt}{2}$

$d=\frac{1485*5.63}{2}\\d= 4180.3m$

wavelengt of sound is $\lambda$ = v/f

= (1485)/(5920)

= 0.251 m

7 0

3 years ago

Two wheel gears are connected by a chain. The larger gear has a radius of 8 centimeters and the smaller gear has a radius of 3 c

dezoksy [38]

Answer:

B.The linear velocity of the gears is the same. The linear velocity is 432π centimeters per minute.

Explanation:

As we know that the small gear completes 24 revolutions in 20 seconds

so the angular speed of the smaller gear is given as

$\omega = 2\pi\frac{24}{20}$

$\omega = 2.4\pi rad/s$

Now we know that the tangential speed of the chain is given as

$v = r \omega$

so we have

$v = (3 cm)(2.4\pi)$

$v = 7.2 \pi cm/s$

$v = 432\pi cm/min$

Since both gears are connected by same chain so both have same linear speed and hence correct answer will be

B.The linear velocity of the gears is the same. The linear velocity is 432π centimeters per minute.

8 0

4 years ago

Unpolarized light is incident upon two polarization filters that do not have their transmission axes aligned. If 21% of the ligh

Debora [2.8K]

Answer:

49.6°

Explanation:

$I_0$ = Unpolarized light

$I_2$ = Light after passing though second filter = $0.21I_0$

Polarized light passing through first filter

$I_1=\frac{I_0}{2}$

Polarized light passing through second filter

$I_2=\frac{I_0}{2}cos^2\theta\\\Rightarrow 0.21I_0=\frac{I_0}{2}cos^2\theta\\\Rightarrow cos^2\theta=\frac{0.21I_0}{\frac{I_0}{2}}\\\Rightarrow cos\theta=\sqrt{\frac{0.21I_0}{\frac{I_0}{2}}}\\\Rightarrow \theta=cos^{-1}\sqrt{\frac{0.21I_0}{\frac{I_0}{2}}}\\\Rightarrow \theta=cos^{-1}\sqrt{0.21\times 2}\\\Rightarrow \theta=cos^{-1}\sqrt{0.42}\\\Rightarrow \theta=49.6^{\circ}$

The angle between the two filters is 49.6°

5 0

4 years ago

A small racquet ball launcher is set up 5 meters from a 12 meter tall wall. It launches a racket ball at 30 m/s at an angle of 4

Strike441 [17]

Answer:20.82 m

Explanation:

Given

distance between wall and launcher=5 m

initial velocity=30 m/s

Launching angle $=45^{\circ}$

Height of wall=12 m

maximum height by ball

$h_{max}=\frac{u^2sin^2\theta }{2g}$

$h_{max}=\frac{30^2sin^{2}45}{2\times 9.8}$

h=22.95 m

$y=xtan\theta -\frac{gx^2}{2u^2cos^\theta }$

y=4.72 m

so ball will strike with wall

and for perfect restitution final velocity of ball will be same as the horizontal velocity before its impact, only direction will be opposite and vertical velocity will be zero

thus it seems as if someone throw the ball with horizontal velocity of 30cos45 from a height of 4.72

Time required to cover 4.72 m

$t=\sqrt{\frac{2h}{g}}$