Answer:
Let the second medium be air (n₁=1)
The refractive index n₂ of the medium where first medium is air is found (a)
(a) n₂ = 2
Explanation:
Critical angle can be defined as the angle of incidence that provides the angle of refraction of 90°.
Refractive index of a medium can be defined as a number that describes that how fast a light will travel through that medium.
Critical angle and Refractive index are related by:
![\theta_{critical}= sin^{-1}(\frac{n_1}{n_2})](https://tex.z-dn.net/?f=%5Ctheta_%7Bcritical%7D%3D%20sin%5E%7B-1%7D%28%5Cfrac%7Bn_1%7D%7Bn_2%7D%29)
![sin \theta_{critical}=\frac{n_1}{n_2}](https://tex.z-dn.net/?f=sin%20%5Ctheta_%7Bcritical%7D%3D%5Cfrac%7Bn_1%7D%7Bn_2%7D)
To find refractive index of medium with respect to air, substitute n₁=1 (Refractive index of air is 1)
Also θ(critical)=30°
Find n₂ :
![sin30= \frac{1}{n_2}\\0.5=\frac{1}{n_2}\\n_2=\frac{1}{0.5}\\n_2=2](https://tex.z-dn.net/?f=sin30%3D%20%5Cfrac%7B1%7D%7Bn_2%7D%5C%5C0.5%3D%5Cfrac%7B1%7D%7Bn_2%7D%5C%5Cn_2%3D%5Cfrac%7B1%7D%7B0.5%7D%5C%5Cn_2%3D2)
Hey there!
The answer would be B. The sound moves from air to water.
Sound travels through different mediums. It goes fastest in solids, a little slower in liquids, and slowest in air. Sound is a very fast wave, but remember that mediums can differ that. In a vacuum space, there is no sound at all. (ex. outer space)
Hope this helps !
Answer:
t = 5.19 s
Explanation:
We have,
Height of the cliff is 132 m
It is required to find the time taken by the ball to fall to the ground. Let t is the time taken. So, using equation of kinematics as :
![y=ut+\dfrac{1}{2}gt^2\\\\\text{since}\ u=0\\\\y=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2y}{g}}\\\\t=\sqrt{\dfrac{2\times 132}{9.8}}\\\\t=5.19\ s](https://tex.z-dn.net/?f=y%3Dut%2B%5Cdfrac%7B1%7D%7B2%7Dgt%5E2%5C%5C%5C%5C%5Ctext%7Bsince%7D%5C%20u%3D0%5C%5C%5C%5Cy%3D%5Cdfrac%7B1%7D%7B2%7Dgt%5E2%5C%5C%5C%5Ct%3D%5Csqrt%7B%5Cdfrac%7B2y%7D%7Bg%7D%7D%5C%5C%5C%5Ct%3D%5Csqrt%7B%5Cdfrac%7B2%5Ctimes%20132%7D%7B9.8%7D%7D%5C%5C%5C%5Ct%3D5.19%5C%20s)
So, it will take 5.19 seconds to fall to the ground.
Answer:
The arrow is at a height of 500 feet at time t = 2.35 seconds.
Explanation:
It is given that,
An arrow is shot vertically upward at a rate of 250 ft/s, v₀ = 250 ft/s
The projectile formula is given by :
![h=-16t^2+v_ot](https://tex.z-dn.net/?f=h%3D-16t%5E2%2Bv_ot)
We need to find the time(s), in seconds, the arrow is at a height of 500 ft. So,
![-16t^2+250t=500](https://tex.z-dn.net/?f=-16t%5E2%2B250t%3D500)
On solving the above quadratic equation, we get the value of t as, t = 2.35 seconds
So, the arrow is at a height of 500 feet at time t = 2.35 seconds. Hence, this is the required solution.
The correct answer to the question is : 9375 N.
CALCULATION:
As per the question, the mass of the car m = 1500 Kg.
The diametre of the circular track D = 200 m.
Hence, the radius of the circular path R = ![\frac{D}{2}](https://tex.z-dn.net/?f=%5Cfrac%7BD%7D%7B2%7D)
= ![\frac{200}{2}\ m](https://tex.z-dn.net/?f=%5Cfrac%7B200%7D%7B2%7D%5C%20m)
= 100 m.
The velocity of the truck v = 25 m/s.
When a body moves in a circular path, the body needs a centripetal force which helps the body stick to the orbit. It acts along the radius and towards the centre.
Hence, the force acting on the car is centripetal force.
The magnitude of the centripetal force is calculated as -
Force F = ![\frac{mv^2}{R}](https://tex.z-dn.net/?f=%5Cfrac%7Bmv%5E2%7D%7BR%7D)
= ![\frac{1500\times (25)^2}{100}\ N](https://tex.z-dn.net/?f=%5Cfrac%7B1500%5Ctimes%20%2825%29%5E2%7D%7B100%7D%5C%20N)
= 9375 N. [ANS}
The centripetal force is provided to the car in two ways. It is the friction which provides the necessary centripetal force. Sometimes friction is not sufficient. At that time, the road is banked to some extent which provides the necessary centripetal force.