Gravitational force is reduced by:
B. The square of the distance..... hope that helps ;)
They protested by marching in the streets.
We have all the charges for q1, q2, and q3.
Since k = 8.988x10^2, and N=m^2/c^2
F(1) = F (2on1) + F (3on1)
F(2on1) = k |q1 q2| / r(the distance between the two)^2
k^ | 3x10^-6 x -5 x 10^-6 | / (.2m)^2
F(2on1) = 3.37 N
Since F1 is 7N,
F(1) = F (2on1) + F (3on1)
7N = 3.37 N + F (3on1)
Since it wil be going in the negative direction,
-7N = 3.37 N + F (3on1)
F(3on1) = -10.37N
F(3on1) = k |q1 q3| / r(the distance between the two)^2
r^2 x F(3on1) = k |q1 q3|
r = sqrt of k |q1 q3| / F(3on1)
= .144 m (distance between q1 and q3)
0 - .144m
So it's located in -.144m
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Answer:
11 m/s
Explanation:
Draw a free body diagram. There are two forces acting on the car:
Weigh force mg pulling down
Normal force N pushing perpendicular to the incline
Sum the forces in the +y direction:
∑F = ma
N cos θ − mg = 0
N = mg / cos θ
Sum the forces in the radial (+x) direction:
∑F = ma
N sin θ = m v² / r
Substitute and solve for v:
(mg / cos θ) sin θ = m v² / r
g tan θ = v² / r
v = √(gr tan θ)
Plug in values:
v = √(9.8 m/s² × 48 m × tan 15°)
v = 11.2 m/s
Rounded to 2 significant figures, the maximum speed is 11 m/s.
Answer:
0.54m
Explanation:
Step one:
given data
length of seesaw= 3m
mass of man m1= 85kg
weight = mg
W1= 85*10= 850N
mass of daughter m2= 35kg
W2= 35*10= 350N
distance from the center= (1.5-0.2)= 1.3m
Step two:
we know that the sum of clockwise moment equals the anticlockwise moment
let the distance the must sit to balance the system be x
taking moment about the center of the system
350*1.3=850*x
455=850x
divide both sides by 850
x=455/850
x=0.54
Hence the man must sit 0.54m from the right to balance the system
