There can be an experiment run using two or more cleansers at the same time as the manufacturers' one
Answer:
Δt'/ T% = 90.3%
Explanation:
Simple harmonic movement is described by the expression
x = A cos (wt)
we find the time for the two points of motion
x = - 0.3 A
-0.3 A = A cos (w t₁)
w t₁ = cos -1 (-0.3)
remember that angles are in radians
w t₁ = 1.875 rad
x = 0.3 A
0.3 A = A cos w t₂
w t₂ = cos -1 (0.3)
w t₂ = 1,266 rad
Now let's calculate the time of a complete period
x= -A
w t₃ = cos⁻¹ (-1)
w t₃ = π rad
this angle for the forward movement and the same time for the return movement in the oscillation to the same point, which is the definition of period
T = 2 t₃
T = 2π / w s
now we can calculate the fraction of time in the given time interval
Δt / T = (t₁ -t₂) / T
Δt / T = (1,875 - 1,266) / 2pi
Δt / T = 0.0969
This is the fraction for when the mass is from 0 to 0.3, for regions of oscillation of greater amplitude the fraction is
Δt'/ T = 1 - 0.0969
Δt '/ T = 0.903
Δt'/ T% = 90.3%
Answer:
They move outwards.
They don't intersect each other at any point.
They show the electric field.
Explanation:
Answer:
<h3> b. 1.18</h3>
Explanation:
The fundamental frequency in string is expressed as;
F1 = 1/2L√T/m .... 1
L is the length of the string
T is the tension
m is the mass per unit length
If the tension is increased by 40%, the new tension will be;
T2 = T + 40%T
T2 = T + 0.4T
T2 = 1.4T
The new fundamental frequency will be;
F2 = 1/2L√1.4T/m ..... 2
Divide 1 by 2;
F2/F = (1/2L√1.4T/m)/1/2L√T/m)+
F2/F = √1.4T/m ÷ √T/m
F2/F = √1.4T/√m ×√m/√T
F2/F = √1.4T/√T
F2/F = 1.18√T/√T
F2/F = 1.18
F2 = 1.18F
Hence the fundamental frequency of vibration changes by a factor of 1.18
40 seconds I’m pretty sure sorry if I’m wrong
