When the wood block is dropped in the water and settles so that it floats and is no longer bobbing up and down, how do the buoya
1 answer:
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Answer:
570 N
Explanation:
Draw a free body diagram on the rider. There are three forces: tension force 15° below the horizontal, drag force 30° above the horizontal, and weight downwards.
The rider is moving at constant speed, so acceleration is 0.
Sum of the forces in the x direction:
∑F = ma
F cos 30° - T cos 15° = 0
F = T cos 15° / cos 30°
Sum of the forces in the y direction:
∑F = ma
F sin 30° - W - T sin 15° = 0
W = F sin 30° - T sin 15°
Substituting:
W = (T cos 15° / cos 30°) sin 30° - T sin 15°
W = T cos 15° tan 30° - T sin 15°
W = T (cos 15° tan 30° - sin 15°)
Given T = 1900 N:
W = 1900 (cos 15° tan 30° - sin 15°)
W = 570 N
The rider weighs 570 N (which is about the same as 130 lb).
Answer:
Distance traveled will be 5.6307 m
Explanation:
Time t = 3 sec
We have given force F = 25 N
We know that force is given by F = ma
So ma = 25 -----------eqn 1
Weight is given by W = 196 N
We know that weight is given by W = mg
So mg = 196 -----------------eqn 2
From equation 1 and equation 2
Initial velocity is given as 0 so u = 0 m/sec
From second equation of motion