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zaharov [31]
3 years ago
12

Which of the following best describes a coastal plain?

Physics
2 answers:
evablogger [386]3 years ago
5 0
Area near a sea having flat land and low relief
Katarina [22]3 years ago
4 0

<em>Hello,</em>

<em>The correct answer is....</em>

<em>B</em><em>) Area near a sea having flat land and low relief.</em>

<em>Hope this helps!!!! :)</em>

<em>**(</em><em>Vanessa</em><em>)**</em>

<em>**Can you please mark me Brainliest answer!!!!! Thanks!!!!! :)</em>

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A 0.950 kg block is attached to a spring with spring constant 16.0 N/m . While the block is sitting at rest, a student hits it w
Bumek [7]

Answer:

Explanation:

Given that,

Mass attached m = 0.95kg

Spring constant k = 16N/m

Instantaneous speed v = 36cm/s = 0.36m/s

Amplitude A=?

When x = 0.7A

Using conservation of energy

∆K.E + ∆P.E = 0

K.E(final) — K.E(initial) + P.E(final) — P.E(initial) = 0

At the beginning immediately the hammer hits the mass, the potential energy is 0J, Therefore, P.E(initial) = 0J, so the speed is maximum.

Also, at the end, at maximum displacement, the speed is zero, therefore, K.E(final) = 0

So, the equation becomes

— K.E(initial) + P.E(final) = 0

K.E(initial) = P.E(final)

½mv² = ½kA²

mv² = kA²

0.95 × 0.36² = 16×A²

0.12312 = 16•A²

A² = 0.12312/16

A² = 0.007695

A = √0.007695

A = 0.088 m

A = 8.8cm

B. Speed at x = 0.7A

Using the same principle above

K.E(initial) = P.E(final)

½mv² = ½kA²

Where A = 0.7A = 0.7 × 0.088 = 0.0614m

Then,

½× 0.95 × v² = ½ × 16 × 0.0614²

0.475v² = 0.0310644

v² = 0.0310644/0.475

v² = 0.0635

v = √0.0635

v = 0.252 m/s

v = 25.2 cm/s

8 0
4 years ago
What is the name given to the initial 150 counts in 2 minutes?
Whitepunk [10]

Divide CFU of Dilution. Divide the CFU of the dilution (the number of colonies you counted) by the result from step 4. For this example, you work out 46 ÷ 1/1000, which is the same as 46 x 1,000. The result is 46,000 CFU in the original sample.

7 0
3 years ago
When you changed from low to high power, how did the change affect the working distance of the lens?
Basile [38]

The working distance gets shorter as the magnification gets bigger. In order to focus, the high-power objective lens must be significantly nearer to the specimen than the low-power lens. Magnification is negatively correlated with working distance.

Magnification change The magnification of a specimen is increased by switching from low power to high power. The magnification of an image is determined by multiplying the magnification of the objective lens by the magnification of the ocular lens, or eyepiece.

The geometry of the optical system connects the magnifying power, or how much the thing being observed seems expanded, and the field of view, or the size of the object that can be seen.

To know more about  working distance

brainly.com/question/13551539

#SPJ4

4 0
1 year ago
Which statement describes a scientific theory?
PSYCHO15rus [73]

Answer:

A theory changes based on new observations and testing.

Explanation:

A scientific theory is a product of multiple trials and repeated experiments. It usually follows after carefully conducting and testing the validity of the hypothesis.

A scientific theory provides an explanation into how something behaves.

A law just states a finding will not explain it.

Most theories are tenable and can be improved upon when new observations and testing are carried out.

4 0
3 years ago
Two polarizers A and B are aligned so that their transmission axes are vertical and horizontal, respectively. A third polarizer
Reptile [31]

Answer:

Explanation:

Let the angle between the first polariser and the second polariser axis is θ.

By using of law of Malus

(a)

Let the intensity of light coming out from the first polariser is I'

I' = I_{0}Cos^{2}\theta     .... (1)

Now the angle between the transmission axis of the second and the third polariser is 90 - θ. Let the intensity of light coming out from the third polariser is I''.

By the law of Malus

I'' = I'Cos^{2}\left ( 90-\theta \right )

So,

I'' = I_{0}Cos^{2}\theta Cos^{2}\left ( 90-\theta \right )

I'' = I_{0}Cos^{2}\theta Sin^{2}\theta

I'' = \frac{I_{0}}{4}Sin^{2}2\theta

(b)

Now differentiate with respect to θ.

I'' = \frac{I_{0}}{4}\times 2 \times 2 \times Sin2\theta \times Cos 2\theta

I'' = \frac{I_{0}}{2}\times Sin 4\theta

7 0
3 years ago
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