Two particles with charges are initially very far apart (effectively an infinite distance apart). They are then fixed at positio
ns that are apart. What is which is the change in the electric potential energy?
1 answer:
Answer:
potential energy increases.
Explanation:
The potential energy between the two charged particles is given by
U = k Q q / r
If they are very far apart then r tends to infinity and the potential energy is zero.
If they come closer then the potential energy between the two charged particles increases.
Thus, the potential energy increases.
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Answer:
It's the third one in the picture ...
r = 3x - 2y
Explanation:
The ball rises for v/g seconds; which equals 14.7/9.8=1.5 seconds . After this time, it’s height will be:
h(t)=g/2(1.5)²+14.7(1.5)
=-4.9 x 2.25 + 22.05
=11.025m
The ball then falls for 49+11.025=60.025m, which takes:
g/2t²=60.025
t²=12.25
t=3.5 secs
Total time: 1.5+3.5=5 seconds
Answer: 117.8 nm
Explanation:
Given,
Nonreflective coating refractive index : n = 1.21
Index of refraction: 
= 1.52
Wave length of light = λ = 570 nm = 
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Answer:
We are given x= bt +ct²
So
A. bxt= m
Because m/s*s= m
So b= m/s and c= m/s²
B.
x= bt-ct²
So at x=0 t=0
x=0 t= 2
We have
bt = ct² so t = b/c at x= 0
So b-2ct= 0
B. To find velocity we use
dx / dt = b - 2 Ct
C. At rest wen V= 0
We have t= b/2c
D. To find acceleration we use
dv / dt = - 2C
<span>They are emitted by the unstable nuclei of certain atoms.
That's all I could find out; Sorry I couldn't be more of an help.</span>
