The formula that is applicable here is E = kQ/r^2 in which the energy of attraction is proportional to the charges and inversely proportional to the square of the distance. In this case,
kQ1/(r1)^2 = kQ2/(r2)^2 r1=l/3, r2=2l/3solve Q1/Q2
kQ1/(l/3)^2 = kQ2/(2l/3)^2 kQ1/(l^2/9) = kQ2/(4l^2/9)Q1/Q2 = 1/4
Answer:
The force on one side of the plate is 3093529.3 N.
Explanation:
Given that,
Side of square plate = 9 m
Angle = 60°
Water weight density = 9800 N/m³
Length of small strip is
The area of strip is
We need to calculate the force on one side of the plate
Using formula of pressure
On integrating
Hence, The force on one side of the plate is 3093529.3 N.
First, you find the velocity at each component. The general equation is:
a = (v2 - v1)/t
a,x = (v2,x - v1,x)/t
-0.105 = (v2,x - 8.57)/6.67
v2,x = 7.87 m/s
a,y = (v2,y - v1,y)/t
0.101 = (v2,y - -2.61)/6.67
v2,y = -1.94 m/s
To find the final speed, find the resultant velocity by taking the hypotenuse.
v^2 = (v2,x)^2 + (v2,y)^2
v^2 = (7.87)^2 + (-1.94)^2
v = 8.1 m/s
The farthest position the mouse reaches inside the tunnel is 4 meters into the tunnel.
From the graph,
The positions reached after,
5 s = 4 m
10 s = 2 m
20 s = 2 m
35 s = 3 m
40 s = 0 m
So the farthest position here is 4 m into the tunnel.
The rate of change of positions is displacement. So displacement will be change in initial and final positions divided by change in time.
s = Δx / Δt
Therefore, the farthest position the mouse reaches inside the tunnel is 4 meters into the tunnel.
To knw more about displacement
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