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3 years ago

Waves break on a beach due to:

Physics

2 answers:

Anarel [89]3 years ago

6 0

Answer:

increasing wavelength near beach

Explanation:

When wavelength increases, frequency of these waves decreases and the waves suddenly stop.

This is also called damped oscillation.

Liula [17]3 years ago

5 0

The answer is “increasing wavelength near beach”

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How to use heliocentric in your own sentence?

Alexus [3.1K]

According to the heliocentric theory, the sun is the center of everything in the universe.

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3 years ago

What is the electric potential energy of an electron at the negative end of the cable, relative to the positive end of the cable

VashaNatasha [74]

Answer:

Electric potential energy at the negative terminal: $1.92\cdot 10^{-18}J$

Explanation:

When a particle with charge $q$ travels across a potential difference $\Delta V$ , then its change in electric potential energy is

$\Delta U = q \Delta V$

In this problem, we know that:

The particle is an electron, so its charge is

$q=-1.60\cdot 10^{-19}C$

We also know that the positive terminal is at potential

$V_+=0V$

While the negative terminal is at potential

$V_-=-12 V$

Therefore, the potential difference (final minus initial) is

$\Delta V = -12-0 = -12 V$

So, the change in potential energy of the electron is

$\Delta U = (-1.6\cdot 10^{-19})(-12)=1.92\cdot 10^{-18}J$

This means that the electron when it is at the negative terminal has $1.92\cdot 10^{-18}J$ of energy more than when it is at the positive terminal.

Since the potential at the positive terminal is 0, this means that the electric potential energy of the electron at the negative end is

$1.92\cdot 10^{-18}J$

3 0

3 years ago

A small rock is thrown straight up with initial speed v0 from the edge of the roof of a building with height H. The rock travels

Crank

Answer:

$v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} }$

Explanation:

The average velocity is total displacement divided by time:

$v_{avg} =\dfrac{D_{tot}}{t}$

And in the case of vertical $v_{avg}$

$v_{avg}=\dfrac{y_{tot}}{t}$

where $y_{tot}$ is the total vertical displacement of the rock.

The vertical displacement of the rock when it is thrown straight up from height $H$ with initial velocity $v_0$ is given by:

$y=H+v_0t-\dfrac{1}{2} gt^2$

The time it takes for the rock to reach maximum height is when $y'(t)=0$ , and it is

$t=\frac{v_0}{g}$

The vertical distance it would have traveled in that time is

$y=H+v_0(\dfrac{v_0}{g} )-\dfrac{1}{2} g(\dfrac{v_0}{g} )^2$

$y_{max}=\dfrac{2gH+v_0^2}{2g}$

This is the maximum height the rock reaches, and after it has reached this height the rock the starts moving downwards and eventually reaches the ground. The distance it would have traveled then would be:

$y_{down}=\dfrac{2gH+v_0^2}{2g}+H$

Therefore, the total displacement throughout the rock's journey is

$y_{tot}=y_{max}+y_{down}$

$y_{tot} =\dfrac{2gH+v_0^2}{2g}+\dfrac{2gH+v_0^2}{2g}+H$

$\boxed{y_{tot} =\dfrac{2gH+v_0^2}{g}+H}$

Now wee need to figure out the time of the journey.

We already know that the rock reaches the maximum height at

$t=\dfrac{v_0}{g},$

and it should take the rock the same amount of time to return to the roof, and it takes another $t_0$ to go from the roof of the building to the ground; therefore,