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alexira [117]
3 years ago
9

If two waves pass a point every second what is the frequency of the waves

Physics
1 answer:
marishachu [46]3 years ago
7 0
[two waves] pass a point [every second]... The answer is in the question (B)
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An ideal parallel - plate capacitor consists of two parallel plates of area A separated by a distance d. This capacitor is conne
Svetradugi [14.3K]

Answer:

The capacitance is cut in half.

Explanation:

The capacitance of a plate capacitor is directly proportional to the area A of the plates and inversely proportional to the distance between the plates d. So if the distance was doubled we should expect that the capacitance would be cut in half. That can be verified by the following equation that is used to compute the capacitance in such cases:

C = (\epsilon)*(A/d)

Where \epsilon is a constant that represents the characteristics for the insulator between the plates. A is the area of the plates and d is the distance between them. When we double d we have a new capacitance, given by:

C_new = (\epsilon)*(A/2d)

C_new = (1/2)*[(\epsilon)*(A/d)]

Since C = (\epsilon)*(A/d)] we have:

C_new = (1/2)*C

4 0
3 years ago
Which object will be the easiest for a magnet to pull? a 1-gram piece of paper, a 2-gram eraser, a 3-gram steel paper clip, a 4-
Vlad1618 [11]
Steel paper clip because it can be moved by the magnet and it is lighter than the iron nail
6 0
3 years ago
WHY ARE ALL GIRLS THE SAME, don't even say their not cause if you say that then I guess you don't have a life!!!!!​
Zinaida [17]
Because they are. it’s just how life works
6 0
2 years ago
The moon has a diameter of 3.48 x 106 m and is a distance of 3.85 x 108 m from the earth. The sun has a diameter of 1.39 x 109 m
Mrrafil [7]

Answer:

0.00903 rad

0.00926 rad

6.268\times 10^{-6}

Explanation:

s = Diameter of the object

r = Distance between the Earth and the object

Angle subtended is given by

\theta=\frac{s}{r}

For the Moon

\theta_m=\dfrac{3.48\times 10^6}{3.85\times 10^8}\\\Rightarrow \theta_m=0.00903\ rad

The angle subtended by the Moon is 0.00903 rad

For the Sun

\theta_s=\dfrac{1.39\times 10^9}{1.5\times 10^{11}}\\\Rightarrow \theta_s=0.00926\ rad

The angle subtended by the Sun is 0.00926 rad

Area ratio is given by

\frac{A_m}{A_s}=\dfrac{\pi r_m^2}{\pi r_s^2}\\\Rightarrow \frac{A_m}{A_s}=\dfrac{d_m^2}{d_s^2}\\\Rightarrow \frac{A_m}{A_s}=\dfrac{(3.48\times 10^{6})^2}{(1.39\times 10^9)^2}\\\Rightarrow \frac{A_m}{A_s}=6.268\times 10^{-6}

The area ratio is 6.268\times 10^{-6}

3 0
3 years ago
The emf induced in a coil that is rotating in a magnetic field will be at a maximum at which moment?
adelina 88 [10]
TLDR: It will reach a maximum when the angle between the area vector and the magnetic field vector are perpendicular to one another.

This is an example that requires you to investigate the properties that occur in electric generators; for example, hydroelectric dams produce electricity by forcing a coil to rotate in the presence of a magnetic field, generating a current.

To solve this, we need to understand the principles of electromotive forces and Lenz’ Law; changing the magnetic field conditions around anything with this potential causes an induced current in the wire that resists this change. This principle is known as Lenz’ Law, and can be described using equations that are specific to certain situations. For this, we need the two that are useful here:

e = -N•dI/dt; dI = ABcos(theta)

where “e” describes the electromotive force, “N” describes the number of loops in the coil, “dI” describes the change in magnetic flux, “dt” describes the change in time, “A” describes the area vector of the coil (this points perpendicular to the loops, intersecting it in open space), “B” describes the magnetic field vector, and theta describes the angle between the area and mag vectors.

Because the number of loops remains constant and the speed of the coils rotation isn’t up for us to decide, the only thing that can increase or decrease the emf is the change in magnetic flux, represented by ABcos(theta). The magnetic field and the size of the loop are also constant, so all we can control is the angle between the two. To generate the largest emf, we need cos(theta) to be as large as possible. To do this, we can search a graph of cos(theta) for the highest point. This occurs when theta equals 90 degrees, or a right angle. Therefore, the electromotive potential will reach a maximum when the angle between the area vector and the magnetic field vector are perpendicular to one another.

Hope this helps!
6 0
3 years ago
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