Answer:
1/3 the distance from the fulcrum
Explanation:
On a balanced seesaw, the torques around the fulcrum calculated on one side and on another side must be equal. This means that:
where
W1 is the weight of the boy
d1 is its distance from the fulcrum
W2 is the weight of his partner
d2 is the distance of the partner from the fulcrum
In this problem, we know that the boy is three times as heavy as his partner, so
If we substitute this into the equation, we find:
and by simplifying:
which means that the boy sits at 1/3 the distance from the fulcrum.
satellite originally moves in a circular orbit of radius R around the Earth. Suppose it is moved into a circular orbit of radius 4R.
(i) What does the force exerted on the satellite then become?
eight times larger<span>four times larger </span>one-half as largeone-eighth as largeone-sixteenth as large(ii) What happens to the satellite's speed?<span>eight times larger<span>four times larger </span>one-half as largeone-eighth as largeone-sixteenth as large(iii) What happens to its period?<span>eight times larger<span>four times larger </span>one-half as largeone-eighth as largeone-sixteenth as large</span></span>
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Answer:
the force exerted by the seat on the pilot is 10766.7 N
Explanation:
The computation of the force exerted by the seat on the pilot is as follows:
Hence, the force exerted by the seat on the pilot is 10766.7 N
Answer:
Work done gets doubled.
Explanation:
The work done by a force is given by :
W = Fd
Where
F is force and d is distance move
If the force is doubled and the distance moved remain the same, it would mean that the work done becomes double of the initial work done.
Answer:
Angular velocity is same as frequency of oscillation in this case.
ω = 
x ![[\frac{L^{2}}{mK}]^{3/14}](https://tex.z-dn.net/?f=%5B%5Cfrac%7BL%5E%7B2%7D%7D%7BmK%7D%5D%5E%7B3%2F14%7D)
Explanation:
- write the equation F(r) = -K
with angular momentum <em>L</em>
- Get the necessary centripetal acceleration with radius r₀ and make r₀ the subject.
- Write the energy of the orbit in relative to r = 0, and solve for "E".
- Find the second derivative of effective potential to calculate the frequency of small radial oscillations. This is the effective spring constant.
- Solve for effective potential
- ω = x
