Answer:
The contact angle is the angle, conventionally measured through the liquid, where a liquid–vapor interface meets a solid surface.
Explanation:
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Answer:
2.50 m/s
Explanation:
This question can be solved using momentum conservation equation
combined mass of crow and feeder = 450+670=1120 gm
let the recoil speed of feeder be v m/s
Then applying momentum conservation we get;
1120×1.5 = 670×v
v= 2.50 m/s
the speed at which the feeder initially recoils backwards = 2.50 m/s
Answer:
Explanation:
Force = q ( v x B)
- 5.6 x 10⁻⁹ (v x - 1.25 k )
- 3.4x 10⁻⁷i + 7.4 x 10⁻⁷j
Let v = ai+bj +ck
Force = - 5.6 x 10⁻⁹ [(ai+bj +ck) x - 1.25 k )]
= - 5.6 x 10⁻⁹ ( 1.25aj - 1.25bi )
= - 7 a j + 7 b i
( 7bi - 7aj ) x 10⁻⁹
Comparing with given force
7b x 10⁻⁹ b = - 3.4 x 10⁻⁷
b = - 48.57
- 7 a x 10⁻⁹ = 7.4 x 10⁻⁷
a = - 105.7
velocity
= -105.7 i - 48.57 j + ck
b ) Component along k can not be obtained .
c ) v . F = ( -105.7 i - 48.57 j + ck ) . −(3.40×10−7N) ˆı +(7.40×10−7N) ˆȷ
= 105.7 x 3.4 x 10⁻⁷ - 48.57 x 7.4 x 10⁻⁷
= 359.38 x 10⁻⁷ - 359.38 x 10⁻⁷
=0
angle between v and F = 90 degree
We can solve the problem by using Snell's law:

where

is the refractive index of the first medium (in this case, air, so

)

is the refractive index of the second medium (in this case, water, so

)

is the angle of incidence of the light, with respect to the vertical, so


is the angle of refraction of the light inside the water, with respect to the vertical
Re-arranging the equation and using the data of the problem, we can find the the angle of refraction of the light inside the water: