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
Answer:
Given:
Fundamental frequency: 470Hz
T1:310k,T2:315k
Calculating velocity
Recall v=(331m/s)✓[T1/273k)
V=331✓(310/273)
V1=331*(1.0656)=352.72m/s
V2=331✓(315/273)=355.5m/s
Fundamental frequency=4L
F2=F1(V2/V1)
F2=470(355.5/352.72)=474.4Hz
Beat=[F2-F1]=474.4-470=4.4Hz
Explanation:
Isn’t it a light box , mirror and / or angle measurer?
Answer: 14.28 m/s
Explanation:
Assuming the girl is spinning with <u>uniform circular motion</u>, her centripetal acceleration is given by the following equation:
(1)
Where:
is the <u>centripetal acceleration</u>
is the<u> tangential speed</u>
is the <u>radius</u> of the circle
Isolating from (1):
(2)
<u />
Finally:
This is the girl's tangential speed
You could put ice, put it on wheels!