A 2.0-kg block is on a perfectly smooth (frictionless) ramp that makes an angle of 30^\circ30 ∘ with the horizontal. What is
the force of the ramp on the block?
1 answer:
Answer:
Explanation:
Since the surface is frictionless therefore there will be no friction force on block but there will be weight of block which we can divide in to two components i.e. mgcosθ &mgsinθ which is perpendicular and parallel to the surface respectively.
In response to mgcosθ ramp will apply a normal force to the block which will be of equal magnitude to that of mgcosθ.
Therefore Ramp will apply a Force of mgcosθ on block where m is the mass of block.
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Answer:
The third charged particle must be placed at x = 0.458 m = 45.8 cm
Explanation:
To solve this problem we apply Coulomb's law:
Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:
Formula (1)
F: Electric force in Newtons (N)
K : Coulomb constant in N*m²/C²
q₁, q₂: Charges in Coulombs (C)
d: distance between the charges in meters (m)
Equivalence
1μC= 10⁻⁶C
1m = 100 cm
Data
K = 8.99 * 10⁹ N*m²/C²
q₁ = +5.00 μC = +5.00 * 10⁻⁶ C
q₂= +7.00 μC = +7.00 * 10⁻⁶ C
d₁ = x (m)
d₂ = 1-x (m)
Problem development
Look at the attached graphic.
We assume a positive charge q₃ so F₁₃ and F₂₃ are repulsive forces and must be equal so that the net force is zero:
We use formula (1) to calculate the forces F₁₃ and F₂₃
F₁₃ = F₂₃
We eliminate k and q₃ on both sides
We eliminate 10⁻⁶ on both sides
We solve the quadratic equation:
In the option x₂, F₁₃ and F₂₃ will go in the same direction and will not be canceled, therefore we take x₁ as the correct option since at that point the forces are in opposite way .
x = 0.458m = 45.8cm
