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
The magnitude<span> of a </span>velocity<span> vector is </span>called<span> speed. Supposethat a wind is blowing in from the direction at a speed of 50 km/h. (This meansthat the direction from which the wind blows is west of the northerly direction.) Apilot is steering a plane in the direction at an airspeed (speed in still air) of250 km/h
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Answer:
i wish i knew your answer.
Explanation:
Answer:by book 2 do you mean the original or re, if you you mean who taught kaneki to control his kagune it was yoshimura (the manager) but if your talking about how to fight with control of his kagune its was Yomo
Explanation: if I'm wrong correct me and i'll do more research
Answer:
a) 43.20V
b) 2.71W/s
c) 40.25s
d) 7.77Nm
Explanation:
(a) The emf of a rotating coil with N turns is given by:
N: turns
B: magnitude of the magnetic field
A: area
w: angular velocity
the emf max is given by:
(b) the maximum rate of change of the magnetic flux is given by:
(c) 
(d) The torque is given by:
