All categories

2 years ago

A light beam travels through a diamond (n=2.42) to pure water (n=1.33). The refracted ray will:

Physics

1 answer:

Marrrta [24]2 years ago

3 0

Answer:

b) bend away from the normal

Explanation:

According to snell's law , if i be the angle of incidence and r be the angle of refraction

sin i / sin r = 1.33 / 2.42

sin i / sin r = .55

Hence sin r > sin i

r > i

In other words angle of refraction will be more than angle of incidence . So, the ray will bend away from the normal .

You might be interested in

HELPPP PLS BESTIES A pattern of stars that has been named is called a constellation. One winter evening, Jason found the star ca

Damm [24]

Answer:

D) Since the stars would move from East to West just as the Sun and Moon do.

5 0

2 years ago

If a plane has an airspeed of 40 m/s and is experiencing a crosswind of 30 m/s, what is its ground speed in m/s?

Anon25 [30]

<h2>Answer:50 $ms^{-1}$ </h2>

Explanation:

Let $v_{a}$ be the airspeed.

Let $v_{w}$ be the cross wind speed.

We know that,ground speed is the vector sum of airspeed and cross wind speed and airspeed is perpendicular to cross wind speed.

If $v_{1}$ and $v_{2}$ are two perpendicular vectors,the resultant vector has the magnitude $\sqrt{|v|_{1}^{2}+|v|_{2}^{2}}$

Given,

$v_{a}=40ms^{-1}\\v_{c}=30ms^{-1}$

So,the ground speed is $\sqrt{40^{2}+30^{2}}=\sqrt{2500}=50ms^{-1}$

6 0

3 years ago

Pressure __________ with depth to support the fluid weight above

eduard

Answer:Hydrostatic

Explanation: I think this is the answer, not sure. Sorry

5 0

2 years ago

The 4kg head of a sledge hammer is moving at 6m/s when it strikes a chisel, driving it into a log. The duration of the impact (o

LUCKY_DIMON [66]

Answer:

The average impact force is 12000 newtons.

Explanation:

By Impact Theorem we know that impact done by the sledge hammer on the chisel is equal to the change in the linear momentum of the former. The mathematical model that represents the situation is now described:

$\bar F \cdot \Delta t = m \cdot (v_{2}-v_{1})$ (1)

Where:

$\bar F$ - Average impact force, in newtons.

$\Delta t$ - Duration of the impact, in seconds.

$m$ - Mass of the sledge hammer, in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final velocity, in meters per second.

If we know that $\Delta t = 0.0020\,s$ , $m = 4\,kg$ , $v_{1} = -6\,\frac{m}{s}$ and $v_{2} = 0\,\frac{m}{s}$ , then we estimate the average impact force is: