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3 years ago

A wave with a frequency of 500 Hz is traveling at a speed of 200 m/s. What is the wavelength ?

Physics

1 answer:

s344n2d4d5 [400]3 years ago

5 0

The wavelength of the wave is 0.40 m.

Frequency f of a wave is defined as the number of waves traveling through a point in one second. Wavelength λ is the distance traveled by the wave in one complete time period. The speed of the wave v is related to frequency and wavelength by the expression,

$v=f \lambda$

Therefore,

$\lambda =\frac{v}{f}$

Substitute 200 m/s for v and 500 Hz for f.

$\lambda =\frac{v}{f}\\ =\frac{200 m/s}{500 Hz} \\ =0.40m$

Thus, the wavelength of the wave is 0.40 m

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Two 10-cm-diameter charged rings face each other, 25.0cm apart. Both rings are charged to +20.0nC. What is the electric field st

Black_prince [1.1K]

Answer:

Part A:

$E_{midpoint}=0$

Part B:

$E_{center}=2711.7558 N/C$

Explanation:

Part A:

Formula of Electric Field Strength:

$E=\frac{1}{4\pi\epsilon}\frac{xQ}{(x^2+R^2)^{3/2}}$

Where:

x is the distance from the ring

R is the radius of the ring

$\epsilon$ is constant permittivity of free space=8.854*10^-12 farads/meter

Q is the charge

For right Ring E at the midpoint can be calculated as:

x for right plate=25/2=12.5 cm=0.125 m

Radius=R=10/2=5 cm=0.05 m

$E_{right}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.125)*(20*10^{-19})}{((0.125)^2+(0.05)^2)^{3/2}}\\E_{right}=9208.1758 N/C$

For Left Ring E at the midpoint can be calculated as:

Since charge on both plates is +ve and same in magnitude, the electric field will be same for both plates.

$E_{left}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.125)*(20*10^{-19})}{((0.125)^2+(0.05)^2)^{3/2}}\\E_{left}=9208.1758 N/C$

Electric Field at midpoint:

Both rings have same magnitude but the direction of fields will be opposite as they have same charge on them.

$E_{midpoint}=E_{left}-E_{right}\\E_{midpoint}=9208.1758-9208.1758\\E_{midpoint}=0$

Part B:

At center of left ring:

Due to left ring Electric field at center is zero because x=0.

$E_{left}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0)*(20*10^{-19})}{((0)^2+(0.05)^2)^{3/2}}\\E_{left}=0 N/C$

Due to right ring Electric field at center of left ring:

Now: x=25 cm= o.25 m (To the center of left ring)

$E_{right}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.25)*(20*10^{-19})}{((0.25)^2+(0.05)^2)^{3/2}}\\E_{right}=2711.7558 N/C$

Electric Field Strength at center of left ring is same as that of right ring.

$E_{center}=2711.7558 N/C$

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3 years ago

A 2-kg object is moving 6 m/s in a horizontal direction.

german

Answer: A) O N

Explanation:

An object in motion will maintain its state of motion. The presence of an unbalanced force changes the velocity of the object.

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2 years ago

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Answer:

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Answer:

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AysviL [449]

Answer: Provided in the explanation

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3 years ago