Answer:
The students displacement is 3·√3 blocks, 45° North of West
Explanation:
From the question, we have;
The distance North in which the student walks = 3.0 blocks
The next walking distance West by the student = 3.0 blocks
The displacement of the student = The shortest distance between the student start and final point
By representing the motion of the student in vector format, we have;
Therefore the magnitude of d is given as follows;
The direction of the student = tan⁻¹(3/3) = 45° West of North
The students displacement = 3·√3 blocks, 45° North of West.
Answer:
43.16°
Explanation:
λ = Wavelength = 1.4×10⁻¹⁰ m
θ₁ = 20°
n can be any integer
d = distance between the two slits
Since for the first bright fringe, n₁ = 1
n₂ = 2 for second order line
The relation between the distance of the slits and the angle through which it is passed is:
dsinθ=nλ
As d and λ are constant
∴ Angle by which the second order line appear is 43.16°
Answer:
a) -1.25 rev/s² and 23.3 rev
b) 2.67s
Explanation:
a) ω = (500 rev/min)(1min/ 60s) => 8.333 rev/s
ω= (200 rev/min)(1min/ 60s) => 3.333rev/s
time 't'= 4 s
angular acceleration 'α'=?
constant angular acceleration equation is given by,
ω= ω
+ α
t
α= (ω
- ω
)/t => (3.333-8.333)/4
α= -1.25 rev/s²
θ-θ = ω
t + 1/2α
t²
=(8.333)(4) + 1/2 (-1.25)(4)²
=23.3 rev
b) ω=0 (comes to rest)
ω = 3.333 rev/s
α= -1.25 rev/s²
ω= ω
+ α
t
t= (ω - ω
)/α
=> (0- 3.333)/-1.25
t= 2.67s