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4 years ago

How does energy travel in a mechanical wave?

Physics

1 answer:

VLD [36.1K]4 years ago

7 0

Answer: in a medium

Explanation:

Mechanical waves require medium to transfer energy. The medium particles vibrate about their fixed position to transfer energy. There is no transport of medium particles.

Example, sound waves are mechanical waves. The sound travels through compression and rarefaction of the medium particles.

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A 20.0 μf capacitor is charged to a potential difference of 850 v. the terminals of the charged capacitor are then connected to

liberstina [14]

(a) $Q = 1.70\times 10^{-2}\;\text{C}$ ;
(b) $V_\text{final} = 5.31\times 10^{2}\;\text{V}$ ;
(c) $E_\text{final} = 4.52\;\text{J}$ ;
(d) $\Delta E = 2.82\;\text{J}$ .

All four values are in 3 sig. fig.

<h3>Explanation</h3>

(a)

$Q = C\cdot V = 20.0\times 10^{-6} \times 850\;\text{V} = 1.70\times 10^{-2}\;\text{J}$ .

(b)

Sum of the final charge on the two capacitors should be the same as the sum of the initial charge. Voltage of the two capacitors should be the same. That is:

$C_1\cdot V_\text{final} +C_2 \cdot V_\text{final} = C_1\cdot V_\text{initial}$ ;

$(C_1+C_2)\cdot V_\text{final} = C_1\cdot V_\text{initial}$ ;

$\displaystyle V_\text{final} = \frac{C_1}{C_1+C_2}\cdot V_\text{initial}\\\phantom{V_\text{final}} = \frac{20.0\;\mu\text{F}}{20.0\;\mu\text{F} + 12.0\;\mu\text{F}} \times 850\;\text{V}\\\phantom{V_\text{final}} =531\;\text{V}$ .

(c)

$\displaystyle E = \frac{1}{2}\cdot C\cdot V^{2}$ .

$\displaystyle E_\text{final} = \frac{1}{2} (C_1 + C_2) \cdot {V_\text{final}}^{2} \\\phantom{E_\text{final}} = \frac{1}{2} \times (20.0\times 10^{-6} + 12.0\times 10^{-6}) \times 531.25\\\phantom{E_\text{final}} = 4.52\;\text{J}$ .

(d)

Initial energy of the system, which is the same as the initial energy in the $20.0\;\mu\text{F}$ capacitor:

$\displaystyle E_\text{initial} = \frac{1}{2} \times 20.0\times 10^{-6} \times 850^{2} = 7.225\;\text{J}$ .

Change in energy:

$\Delta E = 7.225\;\text{J} - 4.516\;\text{J} = 2.70\;\text{J}$ .

4 0

4 years ago

What Could be the control in an experiment to measure the effects of gas additives on fuel

alina1380 [7]

Well,

A control in an experiment would basically be the "normal" version of your test subjects.

In a drug testing experiment with people, the control group would be the people who don't take the drug.

In an experiment on the effects of salt on potatoes, the control group would be a potato without salt on it.

So in an experiment to measure the effects of gas additives on fuel, the control would be fuel without additives.

8 0

4 years ago

What new characteristics did john dalton add to the model of the atom

kicyunya [14]

Answer:

All atoms of an element are identical. The atoms of different elements vary in size and mass. Compounds are produced through different whole-number combinations of atoms. A chemical reaction results in the rearrangement of atoms in the reactant and product compounds.

HOPE THIS HELPED!!!!!!!!!!!!!!!!!!! XDDDDDDDDDDD

7 0

3 years ago

Y u y u bully me121212121

Ede4ka [16]

Answer:

wassup

Explanation:

3 0

3 years ago

Read 2 more answers

What is the force of an object with a mass of 30 kg that is free falling?

disa [49]

Answer:

F = 294.3 [N]

Explanation:

To solve this problem we must use Newton's second law which tells us that force is equal to the product of mass by acceleration. It is this particular case the acceleration is due to the gravitational acceleration since the body is in free fall.

Therefore we have:

F = m*g

where:

F = force [N]

m = mass = 30 [kg]

g = gravity acceleration = 9.81 [m/s^2]

F = 30*9.81

F = 294.3 [N]

4 0

4 years ago