That looks like something I have right down
Answer:
10%
Explanation:
Efficiency = work done / energy used
e = (10 m × 100 N) / (10,000 J)
e = 0.1
The efficiency is 0.1, or 10%.
Answer:
1 ) Distribution of mass within the ball
2 ) Height of the ramp
Explanation:
Acceleration of a rolling body down an inclined plane is given by the following formula
a = g sinθ / ( 1 + k² / R² )
k is radius of gyration , R is radius of the spherical object ,
when acceleration is more , velocity will also be more .
for objects in which masses are lying in the periphery like in hollow sphere , the value of k²/R² will be high so denominator of the expression will be high so acceleration will be less , hence velocity on reaching the bottom will be less.
On mass of the ball , velocity will not depend .
If height is increased , ball will have acceleration for greater time so velocity will be high.
On radius it will not depend because , radius r and k increases proportionately.
Answer:
The entropy of the gas increases
Explanation:
The entropy of an isolated system tends to increase. but this gas gas is not isolated. It has a piston perfoming work on it and it can exchange heat with it's surroundings.
While the system's entropy decreases, the surroundings' entropy increases, and the entropy change of the universe, which is the sum of the entropy changes of the system and the surroundings, will be equal or greater than zero.
Answer:
![\omega=\sqrt{\frac{4g}{3R}}](https://tex.z-dn.net/?f=%5Comega%3D%5Csqrt%7B%5Cfrac%7B4g%7D%7B3R%7D%7D)
Explanation:
According to the law of the conservation of energy, the gravitational potential energy of the physical system will be converted into rotational kinetic energy. So, we have:
![U=K_{rot}\\mgR=\frac{I_s\omega^2}{2}](https://tex.z-dn.net/?f=U%3DK_%7Brot%7D%5C%5CmgR%3D%5Cfrac%7BI_s%5Comega%5E2%7D%7B2%7D)
Is the moment of inertia of the system, that is the sum of the moments of inertia of the disk and the small object.
![mgR=\frac{(I_d+I_o)\omega^2}{2}\\mgR=\frac{(\frac{mR^2}{2}+mR^2)\omega^2}{2}\\mgR=\frac{(\frac{3}{2}mR^2)\omega^2}{2}\\gR=\frac{3}{4}R^2 \omega^2\\\omega^2=\frac{4g}{3R}\\\omega=\sqrt{\frac{4g}{3R}}](https://tex.z-dn.net/?f=mgR%3D%5Cfrac%7B%28I_d%2BI_o%29%5Comega%5E2%7D%7B2%7D%5C%5CmgR%3D%5Cfrac%7B%28%5Cfrac%7BmR%5E2%7D%7B2%7D%2BmR%5E2%29%5Comega%5E2%7D%7B2%7D%5C%5CmgR%3D%5Cfrac%7B%28%5Cfrac%7B3%7D%7B2%7DmR%5E2%29%5Comega%5E2%7D%7B2%7D%5C%5CgR%3D%5Cfrac%7B3%7D%7B4%7DR%5E2%20%5Comega%5E2%5C%5C%5Comega%5E2%3D%5Cfrac%7B4g%7D%7B3R%7D%5C%5C%5Comega%3D%5Csqrt%7B%5Cfrac%7B4g%7D%7B3R%7D%7D)