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soldi70 [24.7K]
3 years ago
12

Which of the following statements is true regarding electromagnetic waves traveling through a vacuum?

Physics
1 answer:
kap26 [50]3 years ago
6 0

Answer:

C. All waves have the same speed.

Explanation:

Wave equation is given as;

V = fλ

where;

V is the speed of the wave

f is the frequency of the wave

λ is the wavelength

The speed of the wave depends on both wavelength and frequency

The speed of the electromagnetic waves in a vacuum is 3 x 10⁸ m/s, this also the speed of light which is constant for all electromagnetic waves.

Therefore, the correct option is "C"

C. All waves have the same speed.

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Line segment gj is a diameter of circle l. angle k measures (4x 6)°. circle l is inscribed with triangle g j k. line segment g j
choli [55]

The value of x in the given right triangle in a semicircle is determined as 21.

<h3>What is the measure of a triangle in a semicircle?</h3>

The triangle in a semicircle is always a right angle triangle.

From the figure shown, we can say that the triangle  G J K is right triangle and m<K = 90degrees.

Given that m<K = 4x + 6, we will can use the following equation to find the value of x as shown:

4x + 6 = 90

4x = 90 - 6

4x = 84

x = 21

Thus, the value of x in the given right triangle in a semicircle is determined as 21.

Learn more about right angle here: brainly.com/question/64787

#SPJ4

8 0
2 years ago
A transformer is intended to decrease the value of the alternating current from 500 amperes to 25 amperes. The primary coil cont
EastWind [94]

Answer:

The number of turns in secondary coil is 4000

Explanation:

Given:

Current in primary coil I_{P} = 500 A

Current in secondary coil I_{S} = 25 A

Number of turns in primary coil N_{P} = 200

In case of transformer the relation between current and number of turns is given by,

     \frac{N_{S} }{N_{P}  } = \frac{I_{P} }{I_{S} }

For finding number of turns in secondary coil,

     N_{S} = \frac{I_{P} }{I_{S} }  N_{P}

     N_{S} = \frac{500}{25} \times 200

     N_{S} = 4000

Therefore, the number of turns in secondary coil is 4000

5 0
3 years ago
Two children are balanced on a seesaw that has a mass of 18.0 kg. The first child has a mass of 26.0 kg and sits 1.60 m from the
Mashcka [7]

Answer:

1.28 m

Explanation:

As shown in the diagram attached,

According to the principle of moment,

For a body at equilibrium,

Sum of clockwise moment = sum of anticlockwise moment.

Taking moment about the pivot,

W₁(1.6)+W(0.133) = W₂(x)............... Equation 1

Where W₁ = Weight of the first child, Wₓ = Weight of the seesaw, W₂ = weight of the second child, x = distance of the second child from the pivot.

But,

W = mg

Where g = 9.8 m/s², m = mass of the body

Therefore,

W₁ = 26×9.8 = 254.8 N,

Wₓ = 18×9.8 = 176.4 N

W₂ = 34.4×9.8 = 337.12 N

Substitute these values into equation 1

(254.8×1.6)+(176.4×0.133) = 337.12(x)

407.68+23.4612 = 337.12x

337.12x = 431.1412

x = 431.1412/337.12

x = 1.2789

x ≈ 1.28 m

7 0
2 years ago
An amoeba has 1.00 x 1016 protons and a net charge of 0.300 pC. Assuming there are 1.88 x 106 fewer electrons than protons, If y
Xelga [282]

Answer:

The fraction of the protons would have no electrons =1.88\times 10^{-10}

Explanation:

We are given that

Amoeba has total number of protons=1.00\times 10^{16}

Net charge, Q=0.300pC

Electrons are fewer than protons=1.88\times 10^6

We have to find the fraction of protons would have no electrons.

The fraction of the protons would have no electrons

=\frac{Fewer\;electrons}{Total\;protons}

The fraction of the protons would have no electrons

=\frac{1.88\times 10^{6}}{1.00\times 10^{16}}

=1.88\times 10^{-10}

Hence, the fraction of the protons would have no electrons =1.88\times 10^{-10}

6 0
3 years ago
sest yourse 1. A pencil lies on the dashboard of a car. a) i) What happens to the pencil when the car suddenly stops? suddenly a
julia-pushkina [17]

Answer:

1. the pencil would have the momentum and would keep going until it hits the windshield. 2. when the car suddenly accelerates, the pencil would be inert and it would move toward the back of the car until a constant speed from the car is reached.

8 0
2 years ago
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