Dave throws a 10 kg bowling ball straight up in the air. At the very tippy-top of its path, what is it's momentum?
1 answer:
You might be interested in
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:
(a) 7 m
(b) 1 m
Explanation:
Given:
The magnitude of displacement vector 'a' is 3 m
The magnitude of displacement vector 'b' is 4 m.
The vector 'c' is the vector sum of vectors 'a' and 'b'.
(a)
Now, when the angle between the vectors is 0°, it means that the vectors are in the same direction. When vectors are in the same direction, then their resultant magnitude is simply the sum of their magnitudes.
So, magnitude of 'c' when 'a' and 'b' are in same direction is given as:
Therefore, the magnitude of vector 'c' is 7 m when angle between 'a' and 'b' is 0°.
(b)
When the angle between the vectors is 180°, it means that the vectors are exactly in the opposite direction. When the vectors are in opposite direction, then their resultant magnitude is the subtraction of their magnitudes.
So, magnitude of 'c' when 'a' and 'b' are in opposite direction is:
Therefore, the magnitude of vector 'c' is 1 m when angle between 'a' and 'b' is 180°.