Your kinetic energy goes down and your potential energy rises. This happens till you reach the top or start falling, in which the opposite happens. Hope this helps!
"Asteroid" is the name we give to the huge number of small bodies
that orbit the sun, here in the inner solar system.
Their orbits are scattered all over the place. Most of them spend
most of the time between the orbits of Mars and Jupiter, but there are
many asteroids that sometimes come very close to Earth.
no question please so what's the problem
In order to find the efficiency first we will find the Change in Potential energy of the small stone that robot picked up
First we will find the mass of the stone
As it is given that stone is spherical in shape so first we will find its volume
![V = \frac{4}{3}\pi r^3](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B4%7D%7B3%7D%5Cpi%20r%5E3)
![V = \frac{4}{3}\pi *(\frac{0.06}{2})^3](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B4%7D%7B3%7D%5Cpi%20%2A%28%5Cfrac%7B0.06%7D%7B2%7D%29%5E3)
![V = 1.13 * 10^{-4} m^3](https://tex.z-dn.net/?f=V%20%3D%201.13%20%2A%2010%5E%7B-4%7D%20m%5E3)
Now it is given that it's specific gravity is 10.8
So density of rock is
![\rho = 10.8 * 10^3 kg/m^3](https://tex.z-dn.net/?f=%5Crho%20%3D%2010.8%20%2A%2010%5E3%20kg%2Fm%5E3)
mass of the stone will be
![m = \rho V](https://tex.z-dn.net/?f=m%20%3D%20%5Crho%20V)
![m = 10.8* 10^3 * 1.13 * 10^{-4}](https://tex.z-dn.net/?f=m%20%3D%2010.8%2A%2010%5E3%20%2A%201.13%20%2A%2010%5E%7B-4%7D)
![m = 1.22 kg](https://tex.z-dn.net/?f=m%20%3D%201.22%20kg)
now change in potential energy is given as
![\Delta U = mgH](https://tex.z-dn.net/?f=%5CDelta%20U%20%3D%20mgH)
here
g = gravity on planet = 0.278 m/s^2
H = height lifted upwards = 15 cm
![\Delta U = 1.22* 0.278 * 0.15](https://tex.z-dn.net/?f=%5CDelta%20U%20%3D%201.22%2A%200.278%20%2A%200.15)
![\Delta U = 0.051 J](https://tex.z-dn.net/?f=%5CDelta%20U%20%3D%200.051%20J)
Now energy supplied by internal circuit of robot is given by
![E = Vit](https://tex.z-dn.net/?f=E%20%3D%20Vit)
V = voltage supplied = 10 V
i = current = 1.83 mA
t = time = 12 s
![E = 10* 1.83 * 10^{-3} * 12](https://tex.z-dn.net/?f=E%20%3D%2010%2A%201.83%20%2A%2010%5E%7B-3%7D%20%2A%2012)
![E = 0.22 J](https://tex.z-dn.net/?f=E%20%3D%200.22%20J)
Now efficiency is defined as the ratio of output work with given amount of energy used
![\eta = \frac{\Delta U}{E}*100](https://tex.z-dn.net/?f=%20%5Ceta%20%3D%20%5Cfrac%7B%5CDelta%20U%7D%7BE%7D%2A100)
![\eta = \frac{0.051}{0.22} = 0.23](https://tex.z-dn.net/?f=%5Ceta%20%3D%20%5Cfrac%7B0.051%7D%7B0.22%7D%20%3D%200.23)
so efficiency will be 23 %
Vi = As * h = 1000 * 30 = 30,000 cm^3 = Vol. of the ice.
Vb = (Di/Dw) * Vi = (0.9/1.0) * 30,000 = 27,000 cm^3 = Vol. below surface - Vol. of water displaced.
27,000cm^3 * 1g/cm^3 = 27,000 grams = 27 kg = Mass of water displaced.