Centripetal force is equal to (mv^2)/r
The way I use to answer these question is to set every variable to 1
m=1
v=1
r=1
so centripetal force =1
then change the variable we're looking at
and since we're find when it's half we could either change it to 1/2 or 2, but 2 is easier to use
m=1
v=2
r=1
((1)×(2)^2)/1=4
So the velocity in the 1st part is half the velocity in the 2nd part and the centripetal force is 4× less
The answer is the centripetal force is 1/4 as big the second time around
Refer to the diagram shown below.
The given data is
mass, kg Coordinates. m
------------- -----------------
2 (0, 0)
2 (2, 0)
4 (2, 1)
Total mass, M = 2+2+4 = 8kg
Let (x,y) be the coordinates of M.
Then, taking moments about the origin, we obtain
8x = 2*0 + 2*2 + 4*2 = 12
x = 1.5 m
8y = 2*0 + 2*0 + 4*1 = 4
y = 0.5 m
Answer: (1.5, 0.5) m
Answer:
5.619×10⁶ N
Explanation:
Applying,
F = kqq'/r²................... Equation 1
Where F = electrostatic force between the charges, k = coulomb's constant, q = first charge, q' = second charge, r = distance btween the charges
From the questiion,
Given: q = 2.5 C, q' = 2.5 C, r = 100 m
Constant: 8.99×10⁹ Nm²/C²
Substitute these values into equation 1
F = (2.5×2.5×8.99×10⁹)/100²
F = 56.19×10⁵
F = 5.619×10⁶ N