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galina1969 [7]
3 years ago
11

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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If the distance between two charges is halved, the electrical force between them increases by a factor 4.

In fact, the magnitude of the electric force between two charges is given by:
F= k \frac{q_1 q_2}{r^2}
where
k is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges

We see that the magnitude of the force F is inversely proportional to the square of the distance r. Therefore, if the radius is halved:
r'= \frac{r}{2}
the magnitude of the force changes as follows:
F'=k \frac{q_1 q_2}{r'^2}=k \frac{q_1 q_2}{( \frac{r}{2})^2 }=k \frac{q_1 q_2}{ \frac{r^2}{4} } =4k  \frac{q_1 q_2}{r^2}=4 F
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3 years ago
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Many web sites describe how to add wires to your clothing to keep you warm while riding your motorcycle. The wires are added to
Tomtit [17]

Answer:

P=42.075W

Explanation:

The power provided by a resistor (wire in this case) is given by:

P=\frac{V^2}{R}.

The resistance of a wire is given by:

R=\frac{\rho L}{A}

Where for the resistivity the one of the copper should be used: \rho=1.68\times10^{-8}\Omega m.

The area A is that of a circle, which written in terms of its diameter is:

A=\pi r^2=\pi (d/2)^2=\frac{\pi d^2}{4}

Putting all together:

P=\frac{AV^2}{\rho L}=\frac{\pi d^2V^2}{4\rho L}

Which for our values is:

P=\frac{\pi (0.00025m)^2(12V)^2}{4(1.68\times10^{-8}\Omega m)(10m)}=42.075W

7 0
3 years ago
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Air is being blown into a spherical balloon at the rate of 1.68 in.3/s. Determine the rate at which the radius of the balloon is
NISA [10]

Answer: 0.006in/s

Explanation:

Let the rate at which air is being blown into a spherical balloon be dV/dt which is 1.68in³/s

Also let the rate at which the radius of the balloon is increasing be dr/dt

Given r = 4.7in and Π = 3.14

Applying the chain rule method

dV/dt = dV/dr × dr/dt

If the volume of the sphere is 4/3Πr³

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dV/dr = 4Πr²

If r = 4.7in

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dV/dr = 277.45in²

Therefore;

1.68 = 277.45 × dr/dt

dr/dt = 1.68/277.45

dr/dt = 0.006in/s

7 0
3 years ago
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