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3 years ago

If the index of refraction of water is 1.33, calculate the speed of light in water.

Physics

1 answer:

astra-53 [7]3 years ago

6 0

Answer:

The correct option is;

2.26 × 10⁸

Explanation:

The index of refraction, in a medium, n = The speed of light in air/(The speed of light in the medium)

The speed of light = 299792458 m/s

Given that the index of refraction of water = 1.33, we have;

1.33 = 299792458/(The speed of light in water)

The speed of light in water = 299792458/1.33 = 225407863.58 m/s ≈ 2.26 × 10⁸ m/s

The speed of light in water ≈ 2.26 × 10⁸ m/s

Therefore;

v ≈ 2.26 × 10⁸ m/s.

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Answer:

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Explanation:

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3 years ago

An automatic dryer spins wet clothes at an angular speed of 5.2 rad/s. Starting from rest, the dryer reaches its operating speed

Dvinal [7]

<h2>Time taken by dryer to come up to speed is 1.625 seconds.</h2>

Explanation:

We have equation of motion v = u + at

Initial velocity, u = 0 rad/s

Final velocity, v = 5.2 rad/s

Time, t = ?

Acceleration, a = 3.2 rad/s²

Substituting

v = u + at

5.2 = 0 + 3.2 x t

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6 0

3 years ago

what is the electric potential at point A in the electric field created by a point charge of 5.5 • 10^-12 C? estimate k as 9.00

Hitman42 [59]

The electric potential at point A in the electric field= 0.099 x 10 ⁻¹v

<u>Explanation</u>:

Given data,

charge = 5.5 x 10¹² C

k =9.00 x 10⁹

The electric potential V of a point charge can found by,

V= kQ / r

Assuming, r=5.00×10⁻² m

V= 5.5 x 10⁻¹²C x 9.00 x 10⁹ / 5.00×10⁻² m

V= 49.5 x 10⁻³/ 5.00×10⁻²

Electric potential V= 0.099 x 10⁻¹v

3 0

3 years ago

An object, when pushed with a net force F, has an acceleration of 4 m/s . Now twice the force is applied to an object that has f

hjlf

Answer:

B. 2 m/s

B. Acceleration = 4.05 m/s² and Tension = 297.5 N.

Explanation:

A force is applied on a mass m whose acceleration is 4 m/s

Force = mass × acceleration

a = F/m = 4 m/s

4 m/s = F/m

F = 4 m/s (m)

If Force of 2F is applied on a mass of 4m ; it acceleration is as follows:

2F/4 m = F/ 2m

4m/s (m) / 2m = 2 m/s

a = 2 m/s

Given that

mass $m_1$ = 30 kg

mass $m_2$ = 50 kg

$\mu$ = 0.1

From the question; we can arrive at two cases;

That :

$m_{2} a _ \ {net} }= m_2g - T$ ----- equation (1)

$m_{1} a _ \ {net} }= T - mg sin \theta - F$ ---- equation (2)

50 a = 50 g - T

30 a = T - 30 g sin 30 - 4 × 30 g cos 30

By summation

80 a = $[ 50 - 30 * \frac{1}{2} - 0.1 *30* \frac{\sqrt{3}}{2}]$ g

80 a = 32. 4 × 10 m/s ² (using g as 10m/s²)

80 a = 324 m/s ²

a = 324/80

a = 4.05 m/s²

From equation , replace a with 4.05

50 × 4.05 = 50 × 10 - T

T = 500 -202.5

T =297.5 N

8 0

3 years ago