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2 years ago

How does energy affecf a wave?

1 answer:

4 0

The energy transported by a wave is directly proportional to the square of the amplitude. So whatever change occurs in the amplitude, the square of that effect impacts the energy. This means that a doubling of the amplitude results in a quadrupling of the energy.

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Which of the circuit offers greater resistance to the flow of current 1A or 2A ?

Dafna1 [17]

1 A offers greater resistance

7 0

3 years ago

Read 2 more answers

Why is the ozone layer important?

adoni [48]

Answer:

Ozone is a gas in the atmosphere that protects everything living on the Earth from harmful ultraviolet (UV) rays from the Sun. Without the layer of ozone in the atmosphere, it would be very difficult for anything to survive on the surface.

Explanation:

Ozone layer, also called ozonosphere, region of the upper atmosphere, between roughly 15 and 35 km (9 and 22 miles) above Earth’s surface, containing relatively high concentrations of ozone molecules (O3). Approximately 90 percent of the atmosphere’s ozone occurs in the stratosphere, the region extending from 10–18 km (6–11 miles) to approximately 50 km.

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3 years ago

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A bowling ball is launched from the top of a building at an angle of 35° above the horizontal with an initial speed of 15 m/s. T

Mamont248 [21]

Let $y_0$ be the height of the building and thus the initial height of the ball. The ball's altitude at time $t$ is given by

$y=y_0+\left(15\dfrac{\rm m}{\rm s}\right)\sin35^\circ\,t-\dfrac g2t^2$

where $g=9.80\frac{\rm m}{\mathrm s^2}$ is the acceleration due to gravity.

The ball reaches the ground when $y=0$ after $t=2.9\,\mathrm s$ . Solve for $y_0$ :

$0=y_0+\left(15\dfrac{\rm m}{\rm s}\right)\sin35^\circ(2.9\,\mathrm s)-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(2.9\,\mathrm s)^2$

$\implies y_0\approx16.258\,\mathrm m$

so the building is about 16 m tall (keeping track of significant digits).

3 0

3 years ago

Consider three identical metal spheres, A, B, and C. Sphere A carries a charge of +6q. Sphere B caries a charge of-2q. Sphere C

miskamm [114]

<h2>20. How much charge is on sphere B after A and B touch and are separated?</h2><h3>Answer:</h3>

$\boxed{q_{B}=+2q}$

<h3>Explanation:</h3>

We'll solve this problem by using the concept of electric potential or simply called potential $V$ , which is <em>the energy per unit charge, </em>so the potential $V$ at any point in an electric field with a test charge $q_{0}$ at that point is:

$V=\frac{U}{q_{0}}$

The potential $V$ due to a single point charge q is:

$V=k\frac{q}{r}$

Where $k$ is an electric constant, $q$ is value of point charge and $r$ is the distance from point charge to where potential is measured. Since, the three spheres A, B and C are identical, they have the same radius $r$ . Before the sphere A and B touches we have:

$V_{A}=k\frac{q_{A}}{r_{A}} \\ \\ V_{B}=k\frac{q_{B}}{r_{A}} \\ \\ But: \\ \\ \ r_{A}=r_{B}=r$

When they touches each other the potential is the same, so:

$V_{A}= V_{B} \\ \\ k\frac{q_{A}}{r}=k\frac{q_{B}}{r} \\ \\ \boxed{q_{A}=q_{B}}$

From the principle of conservation of charge <em>the algebraic sum of all the electric charges in any closed system is constant. </em>So:

$q_{A}+q_{B}=q \\ \\ q_{A}=+6q \ and \ q_{B}=-2q \\ \\ So: \\ \\ \boxed{q_{A}+q_{B}=+4q}$

Therefore:

$(1) \ q_{A}=q_{B} \\ \\ (2) \ q_{A}+q_{B}=+4q \\ \\ (1) \ into \ (2): \\ \\ q_{A}+q_{A}=+4q \therefore 2q_{A}=+4q \therefore \boxed{q_{A}=q_{B}=+2q}$

So after A and B touch and are separated the charge on sphere B is:

$\boxed{q_{B}=+2q}$

<h2>21. How much charge ends up on sphere C?</h2><h3>Answer:</h3>

$\boxed{q_{C}=+1.5q}$

<h3>Explanation:</h3>

First: A and B touches and are separated, so the charges are:

$q_{A}=q_{B}=+2q$

Second: C is then touched to sphere A and separated from it.

Third: C is to sphere B and separated from it

So we need to calculate the charge that ends up on sphere C at the third step, so we also need to calculate step second. Therefore, from the second step:

Here $q_{A}=+2q$ and C carries no net charge or $q_{C}=0$ . Also, $r_{A}=r_{C}=r$

$V_{A}=k\frac{q_{A}}{r} \\ \\ V_{C}=k\frac{q_{C}}{r}$

Applying the same concept as the previous problem when sphere touches we have:

$k\frac{q_{A}}{r} =k\frac{q_{C}}{r} \\ \\ q_{A}=q_{C}$

For the principle of conservation of charge:

$q_{A}+q_{C}=+2q \\ \\ q_{A}=q_{C}=+q$

Finally, from the third step:

Here $q_{B}=+2q \ and \ q_{C}=+q$ . Also, $r_{B}=r_{C}=r$

$V_{B}=k\frac{q_{B}}{r} \\ \\ V_{C}=k\frac{q_{C}}{r}$

When sphere touches we have:

$k\frac{q_{B}}{r} =k\frac{q_{C}}{r} \\ \\ q_{B}=q_{C}$

For the principle of conservation of charge:

$q_{B}+q_{C}=+3q \\ \\ q_{A}=q_{C}=+1.5q$

So the charge that ends up on sphere C is:

$q_{C}=+1.5q$

<h2>22. What is the total charge on the three spheres before they are allowed to touch each other.</h2><h3>Answer:</h3>

$+4q$

<h3>Explanation:</h3>

Before they are allowed to touch each other we have that:

$q_{A}=+6q \\ \\ q_{B}=-2q \\ \\ q_{C}=0$

Therefore, for the principle of conservation of charge <em>the algebraic sum of all the electric charges in any closed system is constant, </em>then this can be expressed as:

$q_{A}+q_{B}+q_{C}=+6q -2q +0 \\ \\ \therefore q_{A}+q_{B}+q_{C}=+4q$

Lastly, the total charge on the three spheres before they are allowed to touch each other is:

$+4q$

8 0

4 years ago

A student is performing a double-slit experiment to determine the wavelength of a light source. She has measured the distance be

Harman [31]

Answer:

lights need elewctricity

Explanation:

4 0

3 years ago