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3 years ago

a waves amplitude is 0.5 meters. If the amplitude is increased to 1 metro, how does its energy change

Physics

1 answer:

marysya [2.9K]3 years ago

5 0

Answer:

The energy becomes 4 times greater.

Explanation:

We know that the energy of a wave is proportional to the square of its amplitude

E ∝ Amplitude^2

Since the original amplitude = 0.5 m

and the new amplitude becomes = 1 m

We are doubling the amplitude. This means that the new energy will be affected by a factor of 4

E_new ∝ (2*Amplitude)^2 =

E_new ∝ 4*(Amplitude)^2

E_new = 4*E

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Which principle(s) of the cell theory shows that all organisms are somehow similar?

Stolb23 [73]

Explanation:

The unified cell theory states that: all living things are composed of one or more cells; the cell is the basic unit of life; and new cells arise from existing cells.

hope it helps

plz mark as brainliest

7 0

3 years ago

Use the graph below to answer the following question: if average acceleration is calculated using the equation, “ change in velo

sergiy2304 [10]

Answer:

$a=9\ cm/s^2$

Explanation:

Average Acceleration

Acceleration is a physical magnitude defined as the change of velocity over time. When we have experimental data, we can compute it by calculating the slope of the line in velocity vs time graph.

Note: We cannot see if the time axis is numbered in increments of 1 second, and we'll assume that.

When $t_2=4\ sec$ , the graph shows a value of $v_2=36\ cm/s$

When $t_1=0\ sec$ , the object is at rest, $v_1=0$

We compute the average acceleration as

$\displaystyle a=\frac{v_2-v_1}{t_2-t_1}$

$\displaystyle a=\frac{36\ cm/s-0\ cm/s}{4\ sec-0\ sec}$

$\displaystyle a=\frac{36\ cm/s}{4\ s}$

$\boxed{a=9\ cm/s^2}$

6 0

3 years ago

A ball is thrown upward at a 45° angle. Inthe absence of air resistance, the ballfollows aA. tangential curve.B. sine curve.C. p

Evgesh-ka [11]

As ball is projected up in air at an angle of 45 degree without any air resistance

Let the initial speed will be v

now we will have

In x direction

$v_x = v cos45$

in y direction

$v_y = vsin45$

now displacement in x direction

$x = (vcos45)t + 0$

displacement in y direction

$y = (vsin45)t - \frac{1}{2}gt^2$

now from above two equations we have

$y = (vsin45)\frac{x}{vcos45} - \frac{1}{2}g(\frac{x}{vcos45})^2$

$y = xtan45 - \frac{1}{2v^2cos^245}gx^2$

so above equation is a quadratic equation and hence it will be a parabolic curve

so correct answer will be

C. parabolic curve.

8 0

3 years ago

You need to build a prototype with machined parts that withstand a saline corrosive environment and temperatures above 200 degre

kondor19780726 [428]

Answer:

Stainless steel

Explanation:

I will try to order the solutions from the least correct to the most correct.

Since a temperature greater than 200 ° F is required, that is to say approximately 93 ° c, Polycaprolactone is the least indicated. Its melting point is approximately 60 ° C, so it would not serve the required application.

On the other hand we have Untreated aluminum, which although it has a melting point higher than the required one, without a zinc and magnesium treatment it will easily oxidize in a salty environment, so it cannot be used in this choice either.

We have to compare the two steels.

The Mild Steel has a better corrosion resistance than the previous ones, but in a long-term cycle it will end up full of corrosion and therefore its properties will be highly affected.

Finally, we have stainless steel, which, as the name implies, contains in some of its variations chromium, zinc or magnesium in its alloys, which makes it highly resistant to corrosion.

In addition its melting point is above 1500 ° c.

The best choice is stainless steel.

5 0

3 years ago

At what distance will the weight of a body be halved , Earth's radius=6.4×10^6

Sholpan [36]

A body of mass m has weight

F = GMm/r²

on the surface of the Earth, where G is the universal gravitational constant, M is the mass of the Earth, and r is it's radius.

If the weight is to be halved, then we have

1/2 F = 1/2 GMm/r² = (1/√2)² GMm/r² = GMm/(√2 r²)

so the distance between the body and the planet's center needs to be

√2 × 6.4 × 10⁶ m ≈ 9.1 × 10⁶ m

5 0

3 years ago