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:
Explanation:
GIVEN
diameter = 15 fm =m
we use here energy conservation
there will be some initial kinetic energy but after collision kinetic energy will zero
on solving these equations we get kinetic energy initial
J ..............(i)
That is, the alpha particle must be fired with 35.33 MeV of kinetic energy. An alpha particle with charge q = 2 e
and gains kinetic energy K =e∆V ..........(ii)
by accelerating through a potential difference ∆V
Thus the alpha particle will
just reach the nucleus after being accelerated through a potential difference ∆V
equating (i) and second equation we get
e∆V = 35.33 Me V
The P value for the given data set is 25127. For finding P value, we have to must find the Z value.
<h3>How to get the z scores?</h3>If we've got a normal distribution, then we can convert it to standard normal distribution and its values will give us the z score.
The Z value is calculated as;
Z = (X - μ) / σ
Z = (4.007 - 3.6) / 0.607
Z = 0.67051
The P value for the given data set is 25127.
Learn more about z-score here:
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