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2 years ago

Use words in the box to complete the diagram, and the definitions below it. Some words will be used

Physics

2 answers:

Masteriza [31]2 years ago

7 0

The correct words to fill the given blanks would be as follows:

- Peak

- Trough

- Equilibrium Line

- Oscillation

- Frequency, Hertz

- Period, seconds

The "peak" is denoted as the highest or maximum point that the wave attains while 'Trough' is characterized as the base of the wave.
The 'equilibrium line' is characterized as the extent to which the wave flattens or compresses.
Oscillation is defined as the process in which an object moves in the "back and forth motion."
Frequency is the amount of occurrences that waves travel in a second which is denoted in the terms of Hertz.
Lastly, period is the duration required by wave to accomplish one motion is denoted in terms of seconds.
Amplitude is the breadth or magnitude of the waves.

Learn more about "Oscillation" here:

brainly.com/question/9820984

Lostsunrise [7]2 years ago

3 0

Answer:

There ya go

Explanation:

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2 years ago

Read 2 more answers

Four pairs of objects have the masses as described below, along with the distances between

lord [1]

Answer:

<h2>Mass of 1 Kg and 2 Kg, 1 meters apart.</h2>

Explanation:

The gravitational force is defined as

$F=G\frac{m_{1} m_{2} }{r^{2} }$

By definition, the gravitational force depends directly on the product of the masses and indirectly on the distance between the masses, which means the further they are, the less gravitational force would be. And, the greater the masses, the greater the gravitational force.

Among the options, the pair that would have the greatest gravitational force is Mass of 1 Kg and 2 Kg, with 1 meter between them.

Notice that the last choice includes the same masses but with a greater distance between them, that means it would be a weaker graviational force.

Therefore, the right answer is the second choice.

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3 years ago

Which of the following provides evidence that Earth's Moon is rotating and revolving at the same rate?

kkurt [141]

B. We can see only one side of the Moon from Earth.

( we only see one side of the moon because the moon rotates around the Earth)

6 0

2 years ago

All biomes don’t have the same level of biodiversity. What seems to be the optimal conditions for high biodiversity?

irinina [24]

Answer:

See the answer below

Explanation:

The optimal conditions for high biodiversity seem to be a <u>warm temperature</u> and <u>wet climates</u>.

<em>The tropical areas of the world have the highest biodiversity and are characterized by an average annual temperature of above 18 </em> $^oC$ <em> and annual precipitation of 262 cm. The areas are referred to as the world's biodiversity hotspots. </em>

Consequently, it follows logically that the optimal conditions for high biodiversity would be a warm temperature of above 18 $^oC$ and wet environment with annual precipitation of not less than 262 cm.

The variation in temperature and precipitation across biomes can thus be said to be responsible for the variation in the level of biodiversity in them.

6 0

3 years ago

A confined aquifer with a porosity of 0.15 is 30 m thick. The potentiometric surface elevation at two observation wells 1000 m a

AlekseyPX

Answer:

Part (a) The flow rate per unit width of the aquifer is 1.0875 m³/day

Part (b) The specific discharge of the flow is 0.0363 m/day

Part (c) The average linear velocity of the flow is 0.242 m/day

Part (d) The time taken for a tracer to travel the distance between the observation wells is 4132.23 days = 99173.52 hours

Explanation:

Part (a) the flow rate per unit width of the aquifer

From Darcy's law;

$q = -Kb\frac{dh}{dl}$

where;

q is the flow rate

K is the permeability or conductivity of the aquifer = 25 m/day

b is the aquifer thickness

dh is the change in th vertical hight = 50.9m - 52.35m = -1.45 m

dl is the change in the horizontal hight = 1000 m

q = -(25*30)*(-1.45/1000)

q = 1.0875 m³/day

Part (b) the specific discharge of the flow

$V = \frac{Q}{A} = \frac{q}{b} = -K\frac{dh}{dl}\\\\V = -(25 m/d).(\frac{-1.45 m}{1000 m}) = 0.0363 m/day$

V = 0.0363 m/day

Part (c) the average linear velocity of the flow assuming steady unidirectional flow

Va = V/Φ

Φ is the porosity = 0.15

Va = 0.0363 / 0.15

Va = 0.242 m/day

Part (d) the time taken for a tracer to travel the distance between the observation wells

The distance between the two wells = 1000 m

average linear velocity = 0.242 m/day

Time = distance / speed

Time = (1000 m) / (0.242 m/day)

Time = 4132.23 days

$= 4132.23 days *\frac{24 .hrs}{1.day} = 99173.52, hours$

4 0

2 years ago