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:
New temperature T2 = -124.7 °C
Explanation:
Given:
Old volume V1 = 135 cm³
New volume V2 = 140 cm³3
Old temperature T1 = 143 K
Find:
New temperature T2
Computation:
V1 / T1 = V2 / T2
135 / 143 = 140 / T2
New temperature T2 = 148.3 K
New temperature T2 = 148.3 - 273
New temperature T2 = -124.7 °C
Answer: Hmmmm. Let's see...
Explanation: A pebble has the same density and a boulder when it rolls down a hill. The pebble will most-likely roll down very fast which can make it dense.
<h2 /><h2>I really hope this helped! like at all...</h2><h2>Truthfully, I don't know a lot of this type of stuff.. But worth a shot, right?</h2>
I guess you could call them that. In chemistry, we call them Metalloids though.
I uploaded the answer to a file hosting. Here's link:
tinyurl.com/wpazsebu
