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2 years ago

What is the frequency of a wave with speed 3m/s and wavelength 6m?

Physics

1 answer:

Korolek [52]2 years ago

3 0

The frequency of the radio wave is

Explanation:

Let's use the this formula:

C represents the speed of light, which has a value of approximately

3.00

represents the wavelength and the units should be in meters.

v represents the frequency, which should have units of

−

. Hz is hertz or reciprocal seconds.

Now let's rearrange the equation to solve for v:

3.00

(

)

−

1The frequency of the radio wave is

Explanation:

Let's use the this formula:

C represents the speed of light, which has a value of approximately

3.00

represents the wavelength and the units should be in meters.

v represents the frequency, which should have units of

−

. Hz is hertz or reciprocal seconds.

Now let's rearrange the equation to solve for v:

3.00

(

)

−

1The frequency of the radio wave is

Explanation:

Let's use the this formula:

C represents the speed of light, which has a value of approximately

3.00

represents the wavelength and the units should be in meters.

v represents the frequency, which should have units of

−

. Hz is hertz or reciprocal seconds.

Now let's rearrange the equation to solve for v:

3.00

(

)

−

1The frequency of the radio wave is

Explanation:

Let's use the this formula:

C represents the speed of light, which has a value of approximately

3.00

represents the wavelength and the units should be in meters.

v represents the frequency, which should have units of

−

. Hz is hertz or reciprocal seconds.

Now let's rearrange the equation to solve for v:

3.00

(

)

−

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What is the magnitude of the electric field at the point if the electric potential in the region is given by V 2.00xyz2, where V

Hatshy [7]

Answer:

Electric field at a point ( x , y , z) is $E=-2yz^2-2xz^2-4xyz$ .

Explanation:

Given :

Electric potential in the region is , $V = 2xyz^2\ .$

We need to find the electric field .

We know , electric field , $E=-\dfrac{dV}{dr}$ { Here r is distance }

In coordinate system ,

$E=-\dfrac{dV}{\delta x }-\dfrac{dV}{\delta y }-\dfrac{dV}{\delta z }$ { $\delta$ is partial derivative }

Putting all values we get ,

$E=-\dfrac{2xyz^2}{\delta x }-\dfrac{2xyz^2}{\delta y }-\dfrac{2xyz^2}{\delta z }\\\\E=-2yz^2-2xz^2-4xyz$

Hence , this is the required solution.

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2 years ago

A sail boat moves north for a distance of 10 km when blown by a wind 30° east of south with the force of 5.00×10^4 N. how much w

erma4kov [3.2K]

<h3><u>Answer;</u></h3>

a) 5.00 x 10^8 J

<h3><u>Explanation;</u></h3>

The work done to move the sailboat is calculated through the equation;

W = F x d

where F is force and d is the distance.

Substituting the known values from the given above,

W = (5.00 x 10⁴ N)(10 km)(1000 m/ 1km)

= 5.00 x 10⁸ J

Thus, the work done is <u>5.00 x 10⁸Joules</u>

7 0

2 years ago

An athlete whose mass is 87.0 kg is performing weight-lifting exercises. Starting from the rest position, he lifts, with constan

expeople1 [14]

Answer: X = 52,314.12 N

Explanation: Let X be the force the feet of the athlete exerts on the floor.

According to newton's third law of motion the floor gives an upward reaction based on the weight of the athlete and the barbell which is known as the normal reaction ( based on the mass of the athlete and the barbell)

Mass of athlete = 87kg, mass of barbell = 600/ hence total normal reaction from the floor = 87* 61.22/ 9.8 *9.8 = 52,200N.

The athlete lifts the barbell from rest thus making it initial velocity u=0, distance covered = S = 0.65m and the time taken = 1.3s

The acceleration of the barbell is gotten by using the equation of constant acceleration motion

S= ut + 1/2at²

But u = 0

S = 1/2at²

0.65 = 1/2 *a (1.3)²

0.65 = 1.69 * a/2

0.65 * 2 = 1.69 * a

a = 0.65 * 2/ 1.69

a = 0.77m/s²

According to newton's second law of motion

Resultant force = mass * acceleration

And resultant force in this case is

X - 52,200 = (87 + 61.22) * 0.77

X - 52,200 = 148.22 * 0.77

X - 52, 200 = 114.132

X = 114.132 + 52,200

X = 52,314.12 N

6 0

3 years ago

Two long, straight parallel wires are placed 38 cm apart, one above the other. The top and bottom wires are carrying currents 4.

ElenaW [278]

Answer:

The force per unit length (N/m) on the top wire is 16.842 N/m

Explanation:

Given;

distance between the two parallel wire, d = 38 cm = 0.38 m

current in the first wire, I₁ = 4.0 kA

current in the second wire, I₂ = 8.0 kA

Force per unit length, between two parallel wires is given as;

$\frac{F}{L} = \frac{\mu_oI_1I_2 }{2\pi d }$

where;

μ₀ is constant = 4π x 10⁻⁷ T.m/A

Substitute the given values in the above equation and calculate the force per unit length

$\frac{F}{L} = \frac{\mu_oI_1I_2 }{2\pi d } = \frac{4\pi *10^{-7}*4000*8000 }{2\pi *0.38} = 16.842 \ N/m$

Therefore, the force per unit length (N/m) on the top wire is 16.842 N/m

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3 years ago

An object moves from point A to point C along the rectangle shown in the figure below.

Naily [24]

Answer:

Hello friend where is the figure of the question

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2 years ago