the force between the electron and the proton.
a) Use F = k * q1 * q2 / d²
where k = 8.99e9 N·m²/C²
and q1 = -1.602e-19 C (electron)
and q2 = 1.602e-19 C (proton)
and d = distance between point charges = 0.53e-10 m
The negative result indicates "attraction".
the radial acceleration of the electron.
b) Here, just use F = ma
where F was found above, and
m = mass of electron = 9.11e-31kg, if memory serves
a = radial acceleration
the speed of the electron.
c) Now use a = v² / r
where a was found above
and r was given
<span> the period of the circular motion.</span>
d) period T = 2π / ω = 2πr / v
where v was found above
and r was given
Answer:
Pentan_1,5_di-al
Explanation:
OHC-CH₂-CH₂-CH₂-CHO
This is Pentan_1,5_di-al
If we break this compound, we will observe that there is presence aldehyde group and hence the functional group "al". This aldehyde is bonded to carbon 1 and carbon 5 respectively.
Also the pentan is due to presence of 5 carbon atoms.
Therefore, the IUPAC name of this compound (OHC-CH₂-CH₂-CH₂-CHO) is Pentan_1,5_di-al
A metallic bond since both are pure metals but are not ionic
<span>A pulse with an amplitude of 3+ would be considered as increased.
Peripheral Pulse Assessment Grading System is measured in 0 - 3 Scale.
0 = absent
1+ = Weak/thready pulse
2+ Normal Pulse
3+ = Full, firm pulse.
from the above scale we can find that the 3+ reading shows that the pulse is increased.</span>