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notsponge [240]
2 years ago
13

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

1

×

10

8

H

z

Explanation:

Let's use the this formula:

C

=

λ

×

v

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

3.00

×

10

8

m

s

λ

represents the wavelength and the units should be in meters.

v represents the frequency, which should have units of

H

z

or

s

−

1

. Hz is hertz or reciprocal seconds.

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

C

λ

=

v

v

=

3.00

×

10

8

m

s

/

(

3

m

)

=

1

×

10

8

s

−

1The frequency of the radio wave is

1

×

10

8

H

z

Explanation:

Let's use the this formula:

C

=

λ

×

v

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

3.00

×

10

8

m

s

λ

represents the wavelength and the units should be in meters.

v represents the frequency, which should have units of

H

z

or

s

−

1

. Hz is hertz or reciprocal seconds.

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

C

λ

=

v

v

=

3.00

×

10

8

m

s

/

(

3

m

)

=

1

×

10

8

s

−

1The frequency of the radio wave is

1

×

10

8

H

z

Explanation:

Let's use the this formula:

C

=

λ

×

v

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

3.00

×

10

8

m

s

λ

represents the wavelength and the units should be in meters.

v represents the frequency, which should have units of

H

z

or

s

−

1

. Hz is hertz or reciprocal seconds.

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

C

λ

=

v

v

=

3.00

×

10

8

m

s

/

(

3

m

)

=

1

×

10

8

s

−

1The frequency of the radio wave is

1

×

10

8

H

z

Explanation:

Let's use the this formula:

C

=

λ

×

v

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

3.00

×

10

8

m

s

λ

represents the wavelength and the units should be in meters.

v represents the frequency, which should have units of

H

z

or

s

−

1

. Hz is hertz or reciprocal seconds.

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

C

λ

=

v

v

=

3.00

×

10

8

m

s

/

(

3

m

)

=

1

×

10

8

s

−

1

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The force needed to keep the space shuttle moving at constant speed is 0.

The given parameters;

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The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 77.0 nC. The plates are in va
WINSTONCH [101]

Answer:

Part A: 7500 V

Part B: 2.899×10⁻³ m²

Part C: 10.27 pF or 10.27×10⁻¹² F

Explanation:

Part A:

Applying,

E = V/d................ Equation 1

Where E = electric field intensity between the plates, V = potential difference between the plates, d = distance of separation between the plates

make V the subject of the equation above,

V = Ed............. Equation 2

Given: E = 3.0×10⁶ V/m, d = 2.5 mm = 2.5×10⁻³ m

Substitute into equation 2

V =  3.0×10⁶ (2.5×10⁻³ )

V = 7.5×10³ V

V = 7500 V

Part B:

Using,

E = Q/(e₀A).................... Equation 3

Where Q = Charge on each plate of the capacitor, A = Area of each plate, e₀ = constant = dielectric = permitivity of free space

make A the subject of the equation,

A = Q/(e₀E).............. Equation 4

Given: Q = 77 nC = 77×10⁻⁹ C, E = 3.0×10⁶ V/m

Constant: e₀ = 8.854×10⁻¹² F/m

Substitute into equation 4

A = 77×10⁻⁹/(8.854×10⁻¹²× 3.0×10⁶)

A = 77×10⁻⁹/(26.562×10⁻⁶)

A = 2.899×10⁻³ m²

A = 2.899×10⁻³ m².

Part C:

Using,

Q = CV.................. Equation 5

Where C = Capacitance of the capacitor

make C the subject of the equation

C = Q/V.............. Equation 6

Given: Q = 77 nC = 77×10⁻⁹ C, V = 7500 V

Substitute into equation 6

C = 77×10⁻⁹/7500

C = 10.27×10⁻¹² F

C = 10.27 pF

5 0
3 years ago
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