Answer:
Explanation:
For entry of light into tube of unknown refractive index
sin ( 90 - 25 ) / sinr = μ , μ is the refractive index of the tube , r is angle of refraction in the medium of tube
r = 90 - C where C is critical angle between μ and body medium in which tube will be inserted.
sin ( 90 - 25 ) / sin( 90 - C) = μ
sin65 / cos C = μ
sinC = 1.33 / μ , where 1.33 is the refractive index of body liquid.
From these equations
sin65 / cos C = 1.33 / sinC
TanC = 1.33 / sin65
TanC = 1.33 / .9063
TanC = 1.4675
C= 56°
sinC = 1.33 / μ
μ = 1.33 / sinC
= 1.33 / sin56
= 1.33 / .829
μ = 1.6 Ans
They are said to be directly related.
a) directly related.
This is Charles' Law.
Answer:
91.84 m/s²
Explanation:
velocity, v = 600 m/s
acceleration, a = 4 g = 4 x 9.8 = 39.2 m/s^2
Let the radius of the loop is r.
he experiences a centripetal force.
centripetal acceleration,
a = v² / r
39.2 x r = 600 x 600
r = 3600 / 39.2
r = 91.84 m/s²
Thus, the radius of the loop is 91.84 m/s².
Answer and explanation:
When you are changing a car tire, the most important thing is to keep the total diameter as equal as possible.
The total car tire diameter can be calculated as:

The profile of this tire is 75 (the higher/taller relation), therefore a 5 percent lower profile would be:
pr=0.95·75=71.25
The problem is that the profiles are normalized and the nearest profile available is 70.
If we take a theorical tire with a profile of 71.25:

The theorical tire size should be 205/71 R15.
If we look for a real tire size, we should look for a tire with a diameter nearest to 26.5'' and a profile of 70.
The best option for real tire size is: Tire 225/70 R14 (wheel diameter of 26.4'') or 205/70 R15 (wheel diameter of 26.3'').
Answer:
51.96 m/s^-1
Explanation:
a) see the attachment
b) As we know the velocity of the projectile has two component, horizontal velocity v_ox. and vertical velocity v_oy as shown in the figure. At the highest point of the trajectory, the projectile has only horizontal velocity and vertical velocity is zero. Therefore at the highest point of the trajectory, the velocity of the projectile will be
v_ox=v_o*cosФ
=60*cos (30)
= 51.96 m/s^-1