First do 1.6 m (how far he jumps) 9.8 m/s (what gravity is measured at) then times 2
= 31.36
Sq root = 5.6
Answer:
270 m/s²
Explanation:
Given:
α = 150 rad/s²
ω = 12.0 rad/s
r = 1.30 m
Find:
a
The acceleration will have two components: a radial component and a tangential component.
The tangential component is:
at = αr
at = (150 rad/s²)(1.30 m)
at = 195 m/s²
The radial component is:
ar = v² / r
ar = ω² r
ar = (12.0 rad/s)² (1.30 m)
ar = 187.2 m/s²
So the magnitude of the total acceleration is:
a² = at² + ar²
a² = (195 m/s²)² + (187.2 m/s²)²
a = 270 m/s²
Answer:
How far will the electron travel beforehitting a plate is 248.125mm
Explanation:
Applying Gauss' law:
Electric Field E = Charge density/epsilon nought
Where charge density=1.0 x 10^-6C/m2 & epsilon nought= 8.85× 10^-12
Therefore E = 1.0 x 10^-6/8.85× 10^-12
E= 1.13×10^5N/C
Force on electron F=qE
Where q=charge of electron=1.6×10^-19C
Therefore F=1.6×10^-19×1.13×10^5
F=1.808×10^-14N
Acceleration on electron a = Force/Mass
Where Mass of electron = 9.10938356 × 10^-31
Therefore a= 1.808×10^-14 /9.11 × 10-31
a= 1.985×10^16m/s^2
Time spent between plate = Distance/Speed
From the question: Distance=1cm=0.01m and speed = 2×10^6m/s^2
Therefore Time = 0.01/2×10^6
Time =5×10^-9s
How far the electron would travel S =ut+ at^2/2 where u=0
S= 1.985×10^16×(5×10^-9)^2/2
S=24.8125×10^-2m
S=248.125mm
Answer of your question is in this photo
Answer:
B.) by interfering with sound waves
Explanation:
As we know that the interference of sound waves is of two types
1). constructive interference
2). destructive interference
now we know that constructive interference means the resultant intensity will be more than the intensity of interfering waves as here two waves are in same phase.
In destructive interference the resultant of two waves is given by the minimum resultant of the intensity as here the phase of two waves are opposite to each other.
So we will say that

here in case of noise cancelling headphones we know that the phase of noise is always made in opposite phase with the sound which is used to cancelled the noise.
This will reduce the noise and we will get a clear sound