I’LL MARK YOU BRAINIEST IF U ANSWER IT
1 answer:
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Answer:
1) d = h- L - mg / k , 2) k = 2 mg/h (-1 +2 L / h)
Explanation:
1) Let's use the translational equilibrium equation
-W = 0
F_{e} = W
k x = mg
x = mg / k
from the statement of the exercise the height of the bridge, for the reference system in the river, is
h = L + x + d
x = h - L - d
we substitute
h - L -d = mg / k
d = h- L - mg / k
2) They ask us for the spring constant. For this part we can use energy conservation
Starting point. At the point before jumping
Em₀ = U = m g h
Final Point. When it's hanging
Em_f = + U = ½ k x² + mg d
Em₀ = Em_f
mg h = 1 / k x² + m g d
in the exercise they indicate that Kate touches the surface of the river, so the distance d = 0
mg h = ½ k x²
k = 2 mg h / x²
we substitute the value of x
k = 2mg h / (h -L)²
k = 2mg h / (- L + h)²
we simplify the expression
k = 2mg h / [h² (1- L / h)²]
k = 2m /h (1- L / h)⁻²
In these jumps the bridge height is always greater than the length of the rope L / h <1, so we can expand the last expression
(1- L / h)⁻² = 1 - 2 (1 -L / h) + 2 3/2! (1 -L / h)² + ...
for simplicity let's keep up to the linear term, we substitute in the solution
k = 2 mg/h [1 - 2 (1- L / h)]
k = 2 mg/h (-1 +2 L / h)
Answer:
40 ms¯².
Explanation:
To solve this problem, we shall illustrate the question with a diagram.
The attached photo gives a better understanding of the question.
From the attached photo:
Velocity (v) = 160 ms¯¹
Time (t) = 4 secs.
Acceleration (a) =?
Acceleration (a) = Velocity (v) /time (t)
a = v/t
a = 160/4
a = 40 ms¯²
Therefore, the initial acceleration of the rocket is 40 ms¯².
