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.
earths precession so its B
Answer: barometer
Explanation: An instrument that measures air pressure is called a barometer. One of the first barometers was developed in the 1600s. The original instrument had mercury in the small basin, with an upside down glass tube placed in the mercury.
The volume of the cylinder is:
V= pi* (r^2) * h.
So the volume is = pi * (5.7/2)^2 * 12 = 306.2 in^2.
Since it is scaled up by the factor of 1.5, so we have to multiply each dimension with 1.5.
That means the diameter will be 8.55 and the height will be 18. so the scaled up volume will be
=pi * ((5.7*1.5)/2)^2 * (12*1.5)
The answer then would have to be 1033.5
Answer:
100 m
Explanation:
Arthur and Betty should be walking the same amount of time if they start walking at the same time and stop when they meet = t
Speed of Arthur = 3 m/s
Speed of Betty = 2 m/s
Distance = Speed × time
Distance covered by Arthur = 3t
Distance covered by Betty = 2t
The distance covered by both of them will be 100 m
The speed of dog is 5 m/s
Spot is running back and forth at 5 m/s for 20 seconds
In 20 seconds the distance covered by the dog is
