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Semmy [17]
3 years ago
11

The index of refraction of silicate flint glass for red light is 1.620 and for violet light is 1.660 . A beam of white light in

this glass strikes the glass–air interface at a 23.90° angle of incidence and refracts out into the air. What is the angular separation between the red and violet components of the spectrum that emerges from the glass?
Physics
1 answer:
chubhunter [2.5K]3 years ago
6 0

Answer:

The angular separation equals 0.35^{o}

Explanation:

We have according to Snell's law

n_{1}sin(\theta _{1})=n_{2}sin(\theta _{2})

\therefore \theta _{2}=sin^{-1}(\frac{n_{1}sin(\theta _{1})}{n_{2}})

Using this equation for both the colors separately we have

\theta _{red}=sin^{-1}(\frac{sin(23.90^{o})}{1.62})\\\\\theta _{red}=14.48^{o}

Similarly for violet light we have

\theta _{violet}=sin^{-1}(\frac{sin(23.90^{o})}{1.660})\\\\\theta _{violet}=14.13^{o}

Thus the angular separation becomes

\Delta \theta =14.48-14.13=0.35^{o}

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Milk with a density of 1020 kg/m3 is transported on a level road in a 9-m-long, 3-m-diameter cylindrical tanker. The tanker is c
jeka57 [31]

Solution :

Given data is :

Density of the milk in the tank, $\rho = 1020 \ kg/m^3$

Length of the tank, x = 9 m

Height of the tank, z = 3 m

Acceleration of the tank, $a_x = 2.5 \ m/s^2$

Therefore, the pressure difference between the two points is given by :

$P_2-P_1 = -\rho a_x x - \rho(g+a)z$

Since the tank is completely filled with milk, the vertical acceleration is $a_z = 0$

$P_2-P_1 = -\rho a_x x- \rho g z$

Therefore substituting, we get

$P_2-P_1=-(1020 \times 2.5 \times 7) - (1020 \times 9.81 \times 3)$

           $=-17850 - 30018.6$

           $=-47868.6 \ Pa$

           $=-47.868 \ kPa$

Therefore the maximum pressure difference in the tank is Δp = 47.87 kPa and is located at the bottom of the tank.

         

4 0
2 years ago
A hawk is flying horizontally at 18.0 m/s in a straight line, 230 m above the ground. A mouse it has been carrying struggles fre
Lisa [10]

Answer:

a) vd = 47.88 m/s

b) θ = 80.9°

c) t = 6.8 s

Explanation:

In the situation of the problem, you can assume that the trajectory of the hawk and the trajectory of the mouse form a rectangle triangle.

One side of the triangle is the horizontal trajectory of the hawk after 2.00s of flight, the other side of the triangle is the distance traveled by the mouse when it is falling down. And the hypotenuse is the trajectory of the hawk when it is trying to recover the mouse.

(a) In order to calculate the diving speed of the hawk, you first calculate the hypotenuse of the triangle.

One side of the triangle is c1 = (18.0m/s)(2.0s) = 36m

The other side of the triangle is c2 = 230m - 3m = 227 m

Then, the hypotenuse is:

h=\sqrt{(36m)^2+(227m)^2}=229.83m    (1)

Next, it is necessary to calculate the falling down time of the mouse, this can be done by using the following formula:

y=y_o+v_ot+\frac{1}{2}gt^2    (2)

yo: initial height = 230m

vo: initial vertical speed of the mouse = 0m/s

g: gravitational acceleration = -9.8m/s^2

y: final height of the mouse = 3 m

You replace the values of the parameters in (2) and solve for t:

3=230-4.9t^2\\\\t=\sqrt{\frac{227}{4.9}}=6.8s

The hawk traveled during 2.00 second in the horizontal trajectory, hence, the hawk needed 6.8s - 2.0s = 4.8 s to travel the distance equivalent to the hypotenuse to catch the mouse.

You use the value of h and 4.8s to find the diving speed of the hawk:

v_d=\frac{229.83m}{4.8s}=47.88\frac{m}{s}

The diving speed of the Hawk is 47.88m/s

(b) The angle is given by:

\theta=cos^{-1}(\frac{c_1}{h})=cos^{-1}(\frac{36m}{229.83m})=80.9 \°

Then angle between the horizontal and the trajectory of the Hawk when it is descending is 80.9°

(c) The mouse is falling down during 6.8 s

4 0
3 years ago
Help, anyone?? please:/
NeTakaya
A,E,C,B,D is the order
5 0
3 years ago
By the term universe astronomers mean
OverLord2011 [107]

'Universe' means 'Everything'. That is, all matter, all space, all time.

6 0
3 years ago
An airplane starts from rest and reaches a takeoff velocity of 60.0m/s due north in 4.0 s. How far has it traveled before takeof
Blizzard [7]
240 meters that's it
7 0
3 years ago
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