Answer:
A. 91 meters north
Explanation:
Take +y to be north.
Given:
v₀ = 13 m/s
a = 0 m/s²
t = 7 s
Find: Δy
Δy = v₀ t + ½ at²
Δy = (13 m/s) (7 s) + ½ (0 m/s²) (7 s)²
Δy = 91 m
The displacement is 91 m north.
Answer:
34.6 m/s
Explanation:
From conservation of momentum, the sum of initial and final momentum are equal. Momentum is a product of mass and velocity. Initial mass will be 42.8+31.5+25.9=100.2 kg
Final mass will be 31.5+25.9=57.4 kg
From formula of momentum
M1v1=m2v2
Making v2 the subject of the formula then
Substitute 100.2 kg for M1, 19.8 m/s fkr v1 and 57.4 kg for m2 then
Answer:
15 m/s or 1500 cm/s
Explanation:
Given that
Speed of the shoulder, v(h) = 75 cm/s = 0.75 m/s
Distance moved during the hook, d(h) = 5 cm = 0.05 m
Distance moved by the fist, d(f) = 100 cm = 1 m
Average speed of the fist during the hook, v(f) = ? cm/s = m/s
This can be solved by a very simple relation.
d(f) / d(h) = v(f) / v(h)
v(f) = [d(f) * v(h)] / d(h)
v(f) = (1 * 0.75) / 0.05
v(f) = 0.75 / 0.05
v(f) = 15 m/s
Therefore, the average speed of the fist during the hook is 15 m/s or 1500 cm/s
