The adjustments that your eyes make as they look from objects near to objects far away or from objects far away to objects close
1 answer:
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Answer:
time of collision is
t = 0.395 s
so they will collide at height of 5.63 m from ground
Explanation:
initial speed of the ball when it is dropped down is
similarly initial speed of the object which is projected by spring is given as
now relative velocity of object with respect to ball
now since we know that both are moving under gravity so their relative acceleration is ZERO and the relative distance between them is 6.4 m
Now the height attained by the object in the same time is given as
so they will collide at height of 5.63 m from ground
The value of the force, F₀, at equilibrium is equal to the horizontal
component of the tension in string 2.
Response:
- The value of F₀ so that string 1 remains vertical is approximately <u>0.377·M·g</u>
Given:
The weight of the rod = The sum of the vertical forces in the strings
Therefore;
M·g = T₂·cos(37°) + T₁
The weight of the rod is at the middle.
Taking moment about point (2) gives;
M·g × L = T₁ × 2·L
Therefore;
Which gives;
F₀ = T₂·sin(37°)
Which gives;
- F₀ ≈ <u>0.377·M·g</u>
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Learn more about equilibrium of forces here: