The solution would be like
this for this specific problem:
<span>
The force on m is:</span>
<span>
GMm / x^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2] ->
1
The force on 2m is:</span>
<span>
GM(2m) / (L - x)^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2]
-> 2
From (1), you’ll get M = 2mx^2 / L^2 and from
(2) you get M = m(L - x)^2 / L^2
Since the Ms are the same, then
2mx^2 / L^2 = m(L - x)^2 / L^2
2x^2 = (L - x)^2
xsqrt2 = L - x
x(1 + sqrt2) = L
x = L / (sqrt2 + 1) From here, we rationalize.
x = L(sqrt2 - 1) / (sqrt2 + 1)(sqrt2 - 1)
x = L(sqrt2 - 1) / (2 - 1)
x = L(sqrt2 - 1) </span>
= 0.414L
<span>Therefore, the third particle should be located the 0.414L x
axis so that the magnitude of the gravitational force on both particle 1 and
particle 2 doubles.</span>
Answer:
Explanation:
Given two mass on an incline code 
and 
and an angle of inclination 
. 
. Assume that is the weight being pulled up and the hanging weight.
-The equations of motion from Newton's Second Law are:
where a is the acceleration.
#Substituting for 
(tension) gives:
#and solving for 
which is the system's acceleration.
Answer:
You could try finding a familiar peer to join the activity with your child. Or ask your child who their friends are at school, or what they look for in a friend at school.
It’s c because it’s not Control so that means that it would be broken and non fix able
Answer:
E = 1.50 × 
V/m
Explanation:
given data
B = 0.50 T
solution
we know that energy density by the magnetic field is express as
...............1
and
energy density due to electric filed is
...............2
and here 
so that
E = 
...................3
put here value and we get
E = 3 × × 0.50
E = 1.50 × V/m
