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Furkat [3]
3 years ago
15

Determine the position in which a solid cylindrical block of wood of diameter 0.3 m and length 0.4 m will float in water. Take s

pecific gravity of wood as 0.5
Physics
1 answer:
yan [13]3 years ago
4 0

Answer:

0.2 m

Explanation:

Diameter = 0.3 m

radius, r = 0.15 m

Length, H = 0.4 m

density of wood, d = 0.5 g/cm^3 = 500 kg/m^3

density of water, d = 1000 kg/m^3

Let h be the depth of cylinder immersed in water.

By the principle of floatation.

Buoyant force = Weight of cylinder

Volume immeresed x density of water x g = Volume of cylinder x density of wood x g

A x h x 1000 x g = A x H x 500 x g

1000 h = 500 x 0.4

h = 0.2 m

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A ray of light incident in air strikes a rectangular glass block of refractive index 1.50, at an angle of incidence of 45°. Calc
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Answer:

Approximately 28^{\circ}.

Explanation:

The refractive index of the air n_{\text{air}} is approximately 1.00.

Let n_\text{glass} denote the refractive index of the glass block, and let \theta _{\text{glass}} denote the angle of refraction in the glass. Let \theta_\text{air} denote the angle at which the light enters the glass block from the air.

By Snell's Law:

n_{\text{glass}} \, \sin(\theta_{\text{glass}}) = n_{\text{air}} \, \sin(\theta_{\text{air}}).

Rearrange the Snell's Law equation to obtain:

\begin{aligned} \sin(\theta_{\text{glass}}) &= \frac{n_{\text{air}} \, \sin(\theta_{\text{air}})}{n_{\text{glass}}} \\ &= \frac{(1.00)\, (\sin(45^{\circ}))}{1.50} \\ &\approx 0.471\end{aligned}.

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\begin{aligned} \theta_{\text{glass}} &= \arcsin (0.471) \approx 28^{\circ}\end{aligned}.

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