Suppose you increase your walking speed from 4 m/s to 12 m/s in a period of 2 s. What is your acceleration?
2 answers:
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Answer:
120 m
Explanation:
Given:
wavelength 'λ' = 2.4m
pulse width 'τ'= 100T ('T' is the time of one oscillation)
The below inequality express the range of distances to an object that radar can detect
τc/2 < x < Tc/2 ---->eq(1)
Where, τc/2 is the shortest distance
First we'll calculate Frequency 'f' in order to determine time of one oscillation 'T'
f = c/λ (c= speed of light i.e 3 x m/s)
f= 3 x / 2.4
f=1.25 x hz.
As, T= 1/f
time of one oscillation T= 1/1.25 x
T= 8 x s
It was given that pulse width 'τ'= 100T
τ= 100 x 8 x => 800 x
s
From eq(1), we can conclude that the shortest distance to an object that this radar can detect:
= τc/2 => (800 x
x 3 x
)/2
=120m
Answer:
Δ L = 2.57 x 10⁻⁵ m
Explanation:
given,
cross sectional area = 1.6 m²
Mass of column = 26600 Kg
Elastic modulus, E = 5 x 10¹⁰ N/m²
height = 7.9 m
Weight of the column = 26600 x 9.8
= 260680 N
we know,
Young's modulus=
stress =
=
= 162925
strain =
now,
Δ L = 2.57 x 10⁻⁵ m
The column is shortened by Δ L = 2.57 x 10⁻⁵ m