Two oppositely charged particles are attracted to each other with a force of 2.7 newtons. What the new force be if only one of t
2 answers:
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Answer:
The force magnitude is 1.75 μN acting outward from the origin towards x₂
Explanation:
The given parameters, are;
The location of q₁ = The origin
The location of q₂ = 14.7 mm from q₁
The repulsive force exerted on q₂ by q₁ = 2.62 μN
The location the particle q₂ is located to = 18.0 mm from q₁
By Coulomb's law, we have;
Where;
k = Coulomb constant ≈ 8.99 × 10⁹ kg·m³/(s²·C²)
r = The distance between the particles
F = The force acting between the particles
When r = 14.7 mm F = 2.62 μN
∴ q₂ × q₁ = r² × F/k = (14.7×10^(-3))²×2.62×10^(-6)/(8.9875517923^9) ≈ 1.48 × 10⁻¹⁸
q₂ × q₁ = 1.48 × 10⁻¹⁸ C²
When the distance is increased to 18.0 mm, we have;
F = (8.9875517923^9) × 1.4796647 × 10^(-18)/((18×10^(-3))²) = 1.75 × 10⁻⁶ N
∴ F = 1.75 μN.
Therefore;
The force magnitude is 1.75 μN outward from the origin towards x₂.
Answer:
a) 24 J
b) Gravitational Force
c) 45 J
d) 0
e) 6.782m/s
Explanation:
a) m = 3kg
v = 4m/s
h = 1.5m
KE = ?
0.5 * 3 * 16 = 24J
b) Gravitational force
c) F = ma = 3 * 10 = 30N
Work done = Force * distance = 30 * 1.5 = 45J
d) Final Kinetic Energy of the ball is zero because the ball eventually stops moving
e) velocity of ball as it strikes the ground = v
where
v is the velocity as it strikes the ground
u is the initial velocity
a is acceleration
s is the distance
Now since the ball is thrown downwards, a is positive because the velocity of the ball is increasing as the gravitational force acts on it
u = 4m/s
a = 10
s = 1.5
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