The magnitude of the Normal Force on a
1 answer:
Answer:
5. Is greater than mg, always
Explanation:
If the cone has an inclination of angle β, the sum of forces will be:
x-axis (centripetal axis):
N*sin β = m*ax where ax is the centripetal acceleration
y-axis:
N*cos β - m*g = m*ay where ay is the vertical acceleration. If the block starts falling down, ay will be negative. If the block starts sliding up, ay will be positive. If the block does not move up nor down, ay=0.
Solving for N:
If ay is positive or zero, N will be greater than mg. If ay is negative, N will be less than mg.
If the block is sliding along a horizontal circular path (not up, nor down), ay = 0, so N will always be greater than mg.
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So the answer is option b.
Answer:
x = 0.775m
Explanation:
Conceptual analysis
In the attached figure we see the locations of the charges. We place the charge q₃ at a distance x from the origin. The forces F₂₃ and F₁₃ are attractive forces because the charges have an opposite sign, and these forces must be equal so that the net force on the charge q₃ is zero.
We apply Coulomb's law to calculate the electrical forces on q₃:
(Electric force of q₂ over q₃)
(Electric force of q₁ over q₃)
Known data
q₁ = 15 μC = 15*10⁻⁶ C
q₂ = 6 μC = 6*10⁻⁶ C
Problem development
F₂₃ = F₁₃
(We cancel k and q₃)
q₂(2-x)² = q₁x²
6×10⁻⁶(2-x)² = 15×10⁻⁶(x)² (We cancel 10⁻⁶)
6(2-x)² = 15(x)²
6(4-4x+x²) = 15x²
24 - 24x + 6x² = 15x²
9x² + 24x - 24 = 0
The solution of the quadratic equation is:
x₁ = 0.775m
x₂ = -3.44m
x₁ meets the conditions for the forces to cancel in q₃
x₂ does not meet the conditions because the forces would remain in the same direction and would not cancel
The negative charge q₃ must be placed on x = 0.775 so that the net force is equal to zero.
Answer:
80%
Explanation:
800 / 1000 = 0.8
Efficiency = 0.8 *100 = 80%