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2 years ago

How are the amplitude and energy of a wave related

Physics

2 answers:

vovangra [49]2 years ago

4 0

B is most likely the answer (Wave amplitude and energy are not related)

In-s [12.5K]2 years ago

3 0

I would go with the last two because they are the same please give me brainiest if i am right

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What is the gravitational potential energy of a 2.5kg object that is 300m above the surface of the earth? g=10m/s

Alexus [3.1K]

Answer:

7350 J

Explanation:

Gravitational Potential Energy: This is defined as the energy possessed by a body due to it's position in the gravitational field. The S.I unit is Joules(J).

Applying,

E.p = mgh..................... Equation 1

Where E.p = Gravitational potential Energy, m = mass of the object, h = height of the object above the surface of the earth, g = acceleration due to gravity.

Given: m = 2.5 kg, h = 300 m

Constant: g = 9.8 m/s²

Substitute these values into equation 1

E.p = 2.5(300)(9.8)

E.p = 7350 J.

4 0

3 years ago

on a recent adventure trip, Anita Break went rock climbing. Anita was able to perform 1370 J of work in 100 seconds. Determine A

Zanzabum

      Power = (energy) / (time)

   =    (1370 joules) / (100 seconds)

   = 13.7 joules/second

   = 13.7 watts .

That's not an awful lot of power, especially for a strenuous activity like
rock-climbing. Shoot ! Even I could probably perform at that level.

Compare 13.7 watts to the light power coming out of a 20-watt night light.

   13.7 watts = 0.018 horsepower. (rounded)

5 0

3 years ago

Name at least 4 other gases in the atmosphere besides oxygen and nitrogen

Irina18 [472]

Carbon dioxide
Helium
Argon
Hydrogen

3 0

3 years ago

Read 2 more answers

Four weightlifters (A-D) enter a competition. The mass, distance, and time of their lifts are shown in the table.

siniylev [52]

Let Pa, Pb, Pc, and Pd be the powers delivered by weightlifters A, B, C, and D, respectively.

Use this equation to determine each power value:

P = W÷Δt

P is the power, W is the work done by the weightlifter, and Δt is the elapsed time.

A) Determining Pa:

Pa = W÷Δt

The weightlifter does work to lift the weight up a certain distance. Therefore the work done is equal to the weight's gain in gravitational potential energy. The equation for gravitational PE is

PE = mgh

PE is the potential energy, m is the mass of the weight, g is the acceleration of objects due to earth's gravity, and h is the distance the weight was lifted.

We can equate W = PE = mgh, therefore we can make the following substitution:

Pa = mgh÷Δt

Given values:

m = 100.0kg

g = 9.81m/s²

h = 2.25m

Δt = 0.151s

Plug in the values and solve for Pa

Pa = 100.0×9.81×2.25÷0.151

<u>Pa = 14600W</u> (watt is the SI derived unit of power)

B) Determining Pb:

Let us use our new equation derived in part A to solve for Pb:

Pb = mgh÷Δt

Given values:

m = 150.0kg

g = 9.81m/s²

h = 1.76m

Δt = 0.052s

Plug in the values and solve for Pb

Pb = 150.0×9.81×1.76÷0.052

C) Determining Pc:

Pc = mgh÷Δt

Given values:

m = 200.0kg

g = 9.81m/s²

h = 1.50m

Δt = 0.217s

Plug in the values and solve for Pc

Pc = 200.0×9.81×1.50÷0.217

D) Determining Pd:

Pd = mgh÷Δt

Given values:

m = 250.0kg

g = 9.81m/s²

h = 1.25m

Δt = 0.206s

Plug in the values and solve for Pd

Pd = 250.0×9.81×1.25÷0.206

Compare the following power values:

Pa = 14600W, Pb = 49800W, Pc = 13600W, Pd = 14900W

Pc is the lowest value.

Therefore, weightlifter C delivers the least power.

7 0

2 years ago

Consider three identical metal spheres, A, B, and C. Sphere A carries a charge of +6q. Sphere B caries a charge of-2q. Sphere C

miskamm [114]

<h2>20. How much charge is on sphere B after A and B touch and are separated?</h2><h3>Answer:</h3>

$\boxed{q_{B}=+2q}$

<h3>Explanation:</h3>

We'll solve this problem by using the concept of electric potential or simply called potential $V$ , which is <em>the energy per unit charge, </em>so the potential $V$ at any point in an electric field with a test charge $q_{0}$ at that point is:

$V=\frac{U}{q_{0}}$

The potential $V$ due to a single point charge q is:

$V=k\frac{q}{r}$

Where $k$ is an electric constant, $q$ is value of point charge and $r$ is the distance from point charge to where potential is measured. Since, the three spheres A, B and C are identical, they have the same radius $r$ . Before the sphere A and B touches we have:

$V_{A}=k\frac{q_{A}}{r_{A}} \\ \\ V_{B}=k\frac{q_{B}}{r_{A}} \\ \\ But: \\ \\ \ r_{A}=r_{B}=r$

When they touches each other the potential is the same, so:

$V_{A}= V_{B} \\ \\ k\frac{q_{A}}{r}=k\frac{q_{B}}{r} \\ \\ \boxed{q_{A}=q_{B}}$

From the principle of conservation of charge <em>the algebraic sum of all the electric charges in any closed system is constant. </em>So:

$q_{A}+q_{B}=q \\ \\ q_{A}=+6q \ and \ q_{B}=-2q \\ \\ So: \\ \\ \boxed{q_{A}+q_{B}=+4q}$

Therefore:

$(1) \ q_{A}=q_{B} \\ \\ (2) \ q_{A}+q_{B}=+4q \\ \\ (1) \ into \ (2): \\ \\ q_{A}+q_{A}=+4q \therefore 2q_{A}=+4q \therefore \boxed{q_{A}=q_{B}=+2q}$

So after A and B touch and are separated the charge on sphere B is:

$\boxed{q_{B}=+2q}$

<h2>21. How much charge ends up on sphere C?</h2><h3>Answer:</h3>

$\boxed{q_{C}=+1.5q}$

<h3>Explanation:</h3>

First: A and B touches and are separated, so the charges are:

$q_{A}=q_{B}=+2q$

Second: C is then touched to sphere A and separated from it.

Third: C is to sphere B and separated from it

So we need to calculate the charge that ends up on sphere C at the third step, so we also need to calculate step second. Therefore, from the second step:

Here $q_{A}=+2q$ and C carries no net charge or $q_{C}=0$ . Also, $r_{A}=r_{C}=r$

$V_{A}=k\frac{q_{A}}{r} \\ \\ V_{C}=k\frac{q_{C}}{r}$

Applying the same concept as the previous problem when sphere touches we have:

$k\frac{q_{A}}{r} =k\frac{q_{C}}{r} \\ \\ q_{A}=q_{C}$

For the principle of conservation of charge:

$q_{A}+q_{C}=+2q \\ \\ q_{A}=q_{C}=+q$

Finally, from the third step:

Here $q_{B}=+2q \ and \ q_{C}=+q$ . Also, $r_{B}=r_{C}=r$

$V_{B}=k\frac{q_{B}}{r} \\ \\ V_{C}=k\frac{q_{C}}{r}$

When sphere touches we have:

$k\frac{q_{B}}{r} =k\frac{q_{C}}{r} \\ \\ q_{B}=q_{C}$

For the principle of conservation of charge:

$q_{B}+q_{C}=+3q \\ \\ q_{A}=q_{C}=+1.5q$

So the charge that ends up on sphere C is:

$q_{C}=+1.5q$

<h2>22. What is the total charge on the three spheres before they are allowed to touch each other.</h2><h3>Answer:</h3>

$+4q$

<h3>Explanation:</h3>

Before they are allowed to touch each other we have that:

$q_{A}=+6q \\ \\ q_{B}=-2q \\ \\ q_{C}=0$

Therefore, for the principle of conservation of charge <em>the algebraic sum of all the electric charges in any closed system is constant, </em>then this can be expressed as:

$q_{A}+q_{B}+q_{C}=+6q -2q +0 \\ \\ \therefore q_{A}+q_{B}+q_{C}=+4q$

Lastly, the total charge on the three spheres before they are allowed to touch each other is:

$+4q$

8 0

3 years ago