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4 years ago

What type of wave requires a medium to transfer energy

Physics

2 answers:

Digiron [165]4 years ago

8 0

Answer:

Mechanical waves.

Explanation:

A mechanical wave is a perturbation on the mechanical properties of the material medium (position, velocity, pressure, etc. ) in which it propagates.

Mechanical waves need a source that creates the perturbation and, a physical medium in which perturbation propagates, the medium can be a gas, a liquid or, a solid.

Some examples are sound, earthquakes and, waves on a guitar string.

vivado [14]4 years ago

6 0

A mechanical wave
glad to help out!

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kirza4 [7]

Answer:

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3 years ago

How do you find displacement when all you have is time with no velocity, and then in order to find velocity you have to have dis

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4 years ago

A proton has a speed of 3.50 Ã 105 m/s when at a point where the potential is +100 V. Later, itâs at a point where the potential

ruslelena [56]

Answer:

(a). The change in the protons electric potential is 0.639 kV.

(b). The change in the potential energy of the proton is $1.022\times10^{-16}\ J$

(c). The work done on the proton is $-8\times10^{-18}\ J$ .

Explanation:

Given that,

Speed $v= 3.50\times10^{5}\ m/s$

Initial potential V=100 V

Final potential = 150 V

(a). We need to calculate the change in the protons electric potential

Potential energy of the proton is

$U=qV=eV$

Using conservation of energy

$K_{i}+U_{i}=K_{f}+U_{f}$

$\dfrac{1}{2}mv_{i}^2+eV_{i}=\dfrac{1}{2}mv_{f}^2+eV_{f}$

$]\dfrac{1}{2}mv_{i}^2-]\dfrac{1}{2}mv_{f}^2=e(V_{f}-V_{i})$

$\dfrac{1}{2}mv_{i}^2-]\dfrac{1}{2}mv_{f}^2=e\Delta V$

$\Delta V=\dfrac{m(v_{i}^2-v_{f}^2)}{2e}$

Put the value into the formula

$\Delta V=\dfrac{1.67\times10^{-27}(3.50\times10^{5}-0)^2}{2\times1.6\times10^{-19}}$

$\Delta V=639.2=0.639\ kV$

(b). We need to calculate the change in the potential energy of the proton

Using formula of potential energy

$\Delta U=q\Delta V$

Put the value into the formula

$\Delta U=1.6\times10^{-19}\times639.2$

$\Delta U=1.022\times10^{-16}\ J$

(c). We need to calculate the work done on the proton

Using formula of work done

$\Delta U=-W$

$W=q(V_{2}-V_{1})$

$W=-1.6\times10^{-19}(150-100)$

$W=-8\times10^{-18}\ J$

Hence, (a). The change in the protons electric potential is 0.639 kV.

(b). The change in the potential energy of the proton is $1.022\times10^{-16}\ J$

(c). The work done on the proton is $-8\times10^{-18}\ J$ .

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