Answer: Angle 59 degree
Explanation: Given that the
n1 = 1.0
n2 = 1.5
Øi = 35 degree
From Snell law, which says that
n1/n2 = sinØ1/ sinØ2
Substitute all the parameters into the formula
1/1.5 = sin 35/sinØ2
Cross multiply
Sin Ø2 = 1.5 sin35
SinØ2 = 1.5 × 0.573 = 0.860
Ø2 = sin^-1(0.860)
Ø2 = 59.36 degree
Ø2 = 59 degree ( approximately)
It has angle 59 degree when passing from air to glass
Answer:
Explanation:
The unknown charge can not remain in between the charge given because force on the middle charge will act in the same direction due to both the remaining charges.
So the unknown charge is somewhere on negative side of x axis . Its charge will be negative . Let it be - Q and let it be at distance - x on x axis.
force on it due to rest of the charges will be equal and opposite so
k3q Q / x² =k 8q Q / (L+x)²
8x² = 3 (L+x)²
2√2 x = √3 (L+x)
2√2 x - √3 x = √3 L
x(2√2 - √3 ) = √3 L
x = √3 L / (2√2 - √3 )
Let us consider the balancing force on 3q
force on it due to -Q and -8q will be equal
kQ . 3q / x² = k3q 8q / L²
Q = 8q (x² / L²)
so charge required = - 8q (x² / L²)
and its distance from x on negative x side = √3 L / (2√2 - √3 )
Answer:
a. 240 N due east
b. 540 N due west
Explanation:
Let east be the reference direction
(a) if the resultant force has a magnitude of 390 N and points east, and the 1st force is 150N due East, then the additional force would also due east and has a magnitude of
390 - 150 = 240 N
(b) if the resultant force has a magnitude of 390 N and points west, it would be -390N is eastern reference, and the 1st force is 150N due East, then the additional force would also due east and has a magnitude of
-390 - 150 = -540 N
This force would point west