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2 years ago

9

During a storm, the waves at this lighthouse were 5.0 meters tall top to bottom, and 10.0 m long. The waves impacted every 6.0 s

econds. What is the frequency?

1 answer:

Sonbull [250]2 years ago

7 0

Answer:

more info

Explanation:

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A combustion reaction usually gives off heat and light is it true or false

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The answer is true I think

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4 years ago

To store stacks of clean plates, a cafeteria uses a closed cart with a spring-loaded shelf inside. Customers can take plates off

ruslelena [56]

The answer for the following problem is mentioned below.

The option for the question is "A" approximately.

<u><em>Therefore the elastic potential energy of the string is 20 J.</em></u>

Explanation:

Given:

Spring constant (k) = 240 N/m

amount of the compression (x) = 0.40 m

To calculate:

Elastic potential energy (E)

We know;

<em>According to the formula;</em>

E = $\frac{1}{2}$ × k × x × x

<u>E = </u> $\frac{1}{2}$ <u> × k ×(x)²</u>

where;

E represents the elastic potential energy

K represents the spring constant

x represents amount of the compression in the string

So therefore,

Substituting the values in the above formula;

E = $\frac{1}{2}$ × 240 × (0.40)²

E = $\frac{1}{2}$ × 240 × 0.16

E = $\frac{1}{2}$ × 38.4

E = 19.2 J or approximately 20 J

<u><em>Therefore the elastic potential energy of the string is 20 J.</em></u>

5 0

3 years ago

The amplitude of a wave is dependent upon what? A. the amount of energy carried by the wave B. the maximum displacement from the

storchak [24]

Given that amplitude and energy are related, the amplitude of a wave is dependent upon A. the amount of energy carried by the wave.
These two are connected, so one affects the other and vice versa.

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3 years ago

Read 2 more answers

Prove 1/P = 1/P1 + 1/P2

Sati [7]

Answer:

Proof:

|----------(R1)-------------(R2)---------|

Let the resistances be R1 , R2 connected in series. And the potential difference be V ;

We know that :

P1 = V²/R1 ..........(1) and

P2 = V²/R2 ...........(2)

In series combination , the net resistance is the algebraic addition of individual resistance :

R = R1 + R2

=> V²/P = V²/P1 + V²/P2

=> 1/P = 1/P1 + 1/P2

Explanation:

Copied it from a different brainly question that is this question. Please try to find it on brainly instead of asking it'd make it much easier.

7 0

3 years ago

A pendulum consists of a 2.7 kg stone swinging on a 4.0 m string of negligible mass. The stone has a speed of 8.1 m/s when it pa

Brut [27]

Answer:

a). v=4.77 $\frac{m}{s}$

b). β=80.61

c). E=88.5J

Explanation:

a).

$m1=2.7 kg\\L=4m\\v_{1}=8.1 \frac{m}{s}\\\beta=63$

$\frac{m*v_{1} ^{2} }{2}+m*g*y_{1}=\frac{m*v_{2}^{2} }{2}+m*g*y_{2}\\\frac{m*v_{1} ^{2} }{2}=\frac{m*v_{2}^{2} }{2}+m*g*y_{2}\\v_{2}=\sqrt{v_{1}^{2}-2*g*y_{2}}\\y_{2} =L-L*cos(\beta)\\v_{2}=\sqrt{v_{1}^{2}-2*g*(L-L*cos(\beta))}\\v_{2}=\sqrt{8.1^{2}-2*9.8*(4(1-*cos(63))}\\v_{2}=\sqrt{8.1^{2}-42.8}\\v_{2}=\sqrt{64-42.8}\\v_{2}=\sqrt{21.19}=4.77 \frac{m}{s}$

b).

$\frac{m*v_{1}^{2}}{2} +m*y_{1} =\frac{m*v_{2}^{2}}{2} +m*y_{2}\\\frac{m*v_{1}^{2}}{2} +0 =0 +m*y_{2}\\y_{2}=\frac{v_{1}^{2}}{2*g}=\frac{(8.1^\frac{m}{s}){2}}{2*9.8\frac{m}{s^{2} }}=3.34 m\\ 3.34m=L(1-cos(\alpha)\\3.34m=4*(1-cos(\alpha ))\\cos(\alpha )=\frac{4-3.34}{4}=0.16\\(\alpha )= cos^{-1}*(0.163)\\ (\alpha )= 80.61$

c).

$E=\frac{m*v_{1} ^{2} }{2}\\ E=\frac{(2.7kg)*(8.1\frac{m}{s})^{2}}{2}=\frac{177.147\frac{kg*m^{2}}{s^{2}}}{2}\\\\E=88.5 J$

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3 years ago