To see how deep a well is you drop a rock in. The rock hits the water 10 seconds later.
1 answer:
Answer:
490m
Explanation:
Given parameters:
Time taken = 10s
Unknown:
Depth of the well = ?
Solution:
To find the depth of the well, we apply the appropriate motion equation:
h = ut + gt²
h is the depth
u is the initial velocity
g is the acceleration due to gravity = 9.8m/s²
t is the time taken
Input the parameters and solve;
The ball was dropped from rest; u = 0
h = gt²
h = x 9.8 x 10² = 490m
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Answer:
x = 41.2 m
Explanation:
The electric force is a vector magnitude, so it must be added as vectors, remember that the force for charges of the same sign is repulsive and for charges of different sign it is negative.
In this case the fixed charges (q₁ and q₂) are positive and separated by a distance (d = 100m), the charge (q₃ = -1.0 10⁻³ C)) is negative so the forces are attractive, such as loads q₃ must be placed between the other two forces subtract
F = F₁₃ - F₂₃
let's write the expression for each force, let's set a reference frame on the charge q1
F₁₃ =
F₂₃ =
they ask us that the net force be zero
F = 0
0 = F₁₃ - F₂₃
F₁₃ = F₂₃
k \frac{q_1 q_3}{x^2} =k \frac{q_2 q_3}{(d-x)^2}
q1 / x2 = q2 / (d-x) 2
(d-x)² = x²
we substitute
(100 - x)² = 2/1 x²
100- x = √2 x
100 = 2.41 x
x = 41.2 m