The watch hand covers an angular displacement of 2π radians in 60 seconds.
ω = 2π/60
ω = 0.1 rad/s
v = ωr
v = 0.1 x 0.08
v = 8 x 10⁻³ m/s
Answer:
a= 92. 13 m/s²
Explanation:
Given that
Amplitude ,A= 0.165 m
The maximum speed ,V(max) = 3.9 m/s
We know that maximum velocity in the SHM given as
V(max) = ω A
ω=Angular speed
A=Amplitude

ω=23.63 rad/s
The maximum acceleration given as
a = ω² A
a= (23.63)² x 0.165 m/s²
a= 92. 13 m/s²
Therefore the maximum magnitude of the acceleration will be 92. 13 m/s².
Answer:
Diagram C
Explanation:
We are given that Sulfur with atomic number 16.
We have to find that which diagram shows the electronic configuration of sulfur.
S=16
Its Diagram C
A. krypton (atomic mass 83.8 amu)
B. argon (atomic mass 39.95 amu)
C. xenon (atomic mass 131.3 amu)
D. radon (atomic mass 222 amu)
hope this helps!
Answer:
Solution given:
frequency[f]=60,500,000Hz
velocity[V]=300,000,000m/s
wave length=?
we have
wave length=
=
=
=4.96 m
Option A.4.96m