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3 years ago

The first step in turning a rock into a sediment is ________. Select one:

Physics

1 answer:

Debora [2.8K]3 years ago

6 0

Answer:

The first step in turning a rock into a sediment is Compaction .

Explanation:

Lithification is a process of changing rock into sediments. There are two steps for a rock to lithify. These steps are as follows

The first step of lithification is compaction where sediments are erosed together by weight of the it. Thus, the upper layers of sediments causes compaction of lower layers.
The next process of lithification is cementation. In this fluids fill the space between the loose particles.

Hence, the first step for turning rock into sediments is compaction.

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30. A box with mass 20kg is on a cement floor. The coefficient of static friction between the box and floor is 0.25. A man is pu

Mnenie [13.5K]

Answer:

84.05

Explanation:

F=mg×0.25
F=20x9.81×0.25
f=49.05N
F=35N

F=f+F

F=49.05+35

=84.05

7 0

3 years ago

Plzzz someone help me

Margarita [4]

1) The object slows down due to kinetic friction.
2) The coefficient of kinetic friction is higher on a carpet than on the bare floor, therefore the object would slow down quicker on the carpet

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3 years ago

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How much distance does it take to stop a car going 30 m/s (67 mph) if the brakes can apply a force equal to one half the car’s w

denpristay [2]

Answer:

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2 years ago

Calculate the amount of energy transferred when a 40w light bulb is left on for 30 minutes.

Alinara [238K]

Answer:

E=72000J or 72kj

Explanation:

The formula is E=pt you need to convert your t from minutes to seconds before proceeding

3 0

3 years ago

New 5G networks utilize millimeter-wave radiation. Millimeter-wave radiation refers to electromagnetic waves with frequencies in

seraphim [82]

Answer:

It corresponds to 1mm-10 mm range.

Explanation:

Electromagnetic waves (such as the millimeter-wave radiation) travel at the speed of light, which is 3*10⁸ m/s in free space.
As in any wave, there exists a fixed relationship between speed, frequency and wavelength, as follows:

$v = \lambda * f (1)$

Replacing v= c=3*10⁸ m/s, and the extreme values of f (which are givens), in (1) and solving for λ, we can get the free-space wavelengths that correspond to the 30-300 GHz range, as follows:

$\lambda_{low} = \frac{c}{f_{high}} = \frac{3e8m/s}{300e9Hz} = 1 mm (2)$

$\lambda_{high} = \frac{c}{f_{low}} = \frac{3e8m/s}{30e9Hz} = 10 mm (3)$

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3 years ago