Answer:
10.8 N
Explanation:
The question requires the force between them, hence, we only need the magnitude of the force without considering what direction it's acting.
Parameters given:
Q1 = 15 * 10^(-6) C
Q2 = 10 * 10^(-6) C
The diagram explains better.
The electrostatic force BETWEEN Q1 and Q2 is:
F = (k * Q1 * Q2)/r²
Using Pythagoras theorem:
r² = 0.25² + 0.25² = 0.0625 + 0.0625
r² = 0.125
=> F = [9 * 10^9 * 15 * 10^(-6) * 10 * 10^(-6)]/0.125
F = 1.35/0.125
F = 10.8 N
Answer:
ω = 10.75 rad/s
Explanation:
The stone leaves the circular path with a horizontal speed
v₀ = vt = R*ω
So
ω = v₀ / R
we are given that R = x / 25.7 so ω = 25.7*v₀ / x
Kinematics gives x = v₀*t
With this substitution for x the expression for ω becomes ω = 25.7 / t
Kinematics also gives for the vertical displacement y that
y = v₀y*t + 0.5*ay*t²
we know that v₀y = 0 m/s
since the stone is launched horizontally, so that
y = 0.5*ay*t² ⇒ t = √(2*y / ay)
Using this result for t in the expression for ω and assuming that upward is positive, we get
ω = 25.7 / √(2*y / ay)
⇒ ω = 25.7 / √(2*(-28) / (-9.8))
⇒ ω = 10.75 rad/s
The answer is C) Solar eclipse.
Hope this helps!
Answer:
0.619 cm
Explanation:
V = 2.17 m/s, diameter internal, D = 1.60 cm
v = 14.5 m/s, inside diameter, d = ?
According to the equation of continuity,
A V = a v
3.14 x (1.60/2)^2 x 2.17 = 3.14 x (d/2 )^2 x 14.5
5.5552 / 4 = d^2 x 3.625
d = 0.619 cm
