a car travelling at 50km/h from rest covers a distance of 10km in 40minutes. Calculate the acceleration
1 answer:
Answer:
Explanation:
We can solve the problem by using the following suvat equation:
where
v is the final velocity
u is the initial velocity
a is the acceleration
s is the displacement
For the car in this problem:
u = 0 (it starts from rest)
is the final velocity
s = 10 km = 10 000 m is the displacement
Solving for a, we find:
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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:
The question is incomplete. However, I believe, it is asking for the acceleration of the elevator. This is 3.16 m/s².
Explanation:
By Hooke's law,
F is the force on a spring, k is the spring constant and e is the extension or compression.
From the question,
This is the force on the mass suspended on the spring. Its acceleration, a, is given by
This acceleration is more than the acceleration due to gravity, g = 9.8 m/s². Hence the elevator must be moving up with an acceleration of
12.96 - 9.8 m/s² = 3.16 m/s²