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3 years ago

Which type of wave results from a moderately sloping coastal region?

Physics

2 answers:

Naddik [55]3 years ago

8 0

Hey there!

Your answer: Spilling breaker

Spilling breaker usually occurs when a beach or ocean is flat, and as the waves of the wind continues to happen, slowly the region would eventually become a slope.

It's almost like play-dough. Let's say that we set a perfect flat surface of play-dough on the table. As we continue slide our hands one direction, doesn't the play dough have more on one side than the other? It eventually contains a slope if you add enough from the first place.

Your answer: Spilling breaker

Assoli18 [71]3 years ago

4 0

The answer would be "Spilling breaker"

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Which conversion is the function of a photovoltaic cell? A) sunlight to mechanical energy B) electricity to heat C) sunlight to

irakobra [83]

Answer:

C. Sunlight to electricity

Explanation:

Photo voltaic cells covert sunlight to the electricity. It absorbs the light from the sun and use that sunlight to create electric current. The number of photo voltaic cells are connected together in a solar panel to generate enough electricity to power up a home.

3 0

2 years ago

The uniform 100-kg beam is freely hinged about its upper end A and is initially at rest in the vertical position with theta = 0.

max2010maxim [7]

Answer:

ac = 2.86 m / s²

Explanation:

Image can detail the system to determine the force in the FA to understand the system into the applicated force

m = 100 kg , L = 3 m

∑ F = 0 ⇒ Ay - 100 kg + P * cos (45) = 0

Ay = 768.86 N

∑ Mₐ = α * I ₐ

I ₐ = m * L² / 3 ⇒ I ₐ = 100 kg * 4² m / 3

Replacing

P * sin (45) * 3 = α * 100 kg * 4² m / 3

α = 1.193 rad / s²

ac = α *2 ⇒ ac = 1.193 rad / s² * 2

ac = 2.86 m / s²

4 0

3 years ago

Two parallel conducting plates are separated by 10.0 cm, and one of them is taken to be at zero volts. (a) what is the electric

Alex_Xolod [135]

The expression for the magnitude of the electric field between two uniform conducting plates is

$E=\frac{V}{d}$

Here, V is potential difference between plates and d is separation between plates.

As the potential 6.00 cm from the zero volt plate (and 4.00 cm from the other) is 420 V.

Therefore,

$E = \frac{420 \ V}{6 \times 10^{-2} m } = 70 \times 10^2 \ V/m$

Thus, the electric field strength between the plates is 7000 V/ m

8 0

3 years ago

Transform boundaries are classified under which type of fault?

cluponka [151]

Answer:

Strike-slip fault

Explanation:

Transform boundaries play the role of connecting the other plate boundary segments.

When the plates are rubbed against each other, they result in enormous amount of stresses which leads to the breaking of the part of a rock causing earthquakes. Places of occurrence of these breaks are termed as faults.

Strike slip faults results from compression which takes place horizontally, but but in this the rock displacement releases energy and takes place in a horizontal direction which is parallel to the force of compression.

8 0

3 years ago

A thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 10° F. Af

vekshin1

Answer:

$T=51.64^\circ F$

$t=180.10s$

Explanation:

The Newton's law in this case is:

$T(t)=T_m+Ce^{kt}$

Here, $T_m$ is the air temperture, C and k are constants.

We have

$70^\circ F$ in $t=0$

So:

$T(0)=70^\circ F\\T(0)=10^\circ F+Ce^{k(0)}\\70^\circ F=10^\circ F+C\\C=70^\circ F-10^\circ F=60^\circ F$

And we have $60^\circ F$ in $t=30 s$ , So:

$T(30)=60^\circ F\\T(30)=10^\circ F+(60^\circ F)e^{k(30)}\\60^\circ F=10^\circ F+(60^\circ F)e^{k(30)}\\50^\circ F=(60^\circ F)e^{k(30)}\\e^{k(30)}=\frac{50^\circ F}{60^\circ F}\\(30)k=ln(\frac{50}{60})\\k=\frac{ln(\frac{50}{60})}{30}=-0.0061$

Now, we have:

$T=10^\circ F+(60^\circ F)e^{-0.0061t}(1)$

Applying (1) for $t=1 min=60s$ :

$T=10^\circ F+(60^\circ F)e^{-0.0061*60}\\T=10^\circ F+(60^\circ F)0.694\\T=10^\circ F+41.64^\circ F\\T=51.64^\circ F$

Applying (1) for $T=30^\circ F$ :

$30^\circ F=10^\circ F+(60^\circ F)e^{-0.0061t}\\30^\circ F-10^\circ F=(60^\circ F)e^{-0.0061t}\\-0.0061t=ln(\frac{20}{60})\\t=\frac{ln(\frac{20}{60})}{-0.0061}=180.10s$

8 0

3 years ago