Answer:
0.62 m/s² at 68° S of E
Explanation:
Net force north = 12 - 70 = -58 N
Net force east = 33 - 10 = 23 N
Net force = √(-58² + 23²) = 62.3939... N
acceleration = F/m = 62.3939/100 = 0.623939... ≈ 0.62 m/s²
θ = arctan(-58/23) = -68.3691... ≈ 68° S of E
F=mv^2/R
----> V^2=FR/m=(350x0.9)/2.5=126
----- V=11.22 m/s
<span>Active Galactic Nuclei.</span>
Answer:
143 °
Explanation:
a ) If d be the distance between slits , λ be wavelength of light used and at angle θ nth dark fringe is formed then
d sinθ = ( 2n+1) λ/2
for first dark fringe
d sinθ = λ/2
d /λ = 1/ 2 sinθ
1 / 2 sin15
= 1.93
b )
For intensity of fringe at angle θ, the relation is
I = I₀ cos²θ
I / I₀ = cos²θ/2
Given I / I₀ =0. 1
0.1 = cos²θ/2
θ/2 = 71.5
θ = 143 °
