Answer:
Explanation:
Given
mass of archer ![m=0.3\ kg](https://tex.z-dn.net/?f=m%3D0.3%5C%20kg)
Average force ![F_{avg}=201\ N](https://tex.z-dn.net/?f=F_%7Bavg%7D%3D201%5C%20N)
extension in arrow ![x=1.3\ m](https://tex.z-dn.net/?f=x%3D1.3%5C%20m)
Work done to stretch the bow with arrow
![W=F\cdot x](https://tex.z-dn.net/?f=W%3DF%5Ccdot%20x)
![W=201\times 1.3=261.3\ m](https://tex.z-dn.net/?f=W%3D201%5Ctimes%201.3%3D261.3%5C%20m)
This work done is converted into kinetic Energy of arrow
![W=\frac{1}{2}mv^2](https://tex.z-dn.net/?f=W%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
where v= velocity of arrow
![261.3=\frac{1}{2}\times 0.3\times v^2](https://tex.z-dn.net/?f=261.3%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%200.3%5Ctimes%20v%5E2)
![v=\sqrt{1742}](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B1742%7D)
![v=41.73\ m/s](https://tex.z-dn.net/?f=v%3D41.73%5C%20m%2Fs)
(b)if arrow is thrown vertically upward then this energy is converted to Potential energy
![W=mgh](https://tex.z-dn.net/?f=W%3Dmgh)
![261.3=0.3\times 9.8\times h](https://tex.z-dn.net/?f=261.3%3D0.3%5Ctimes%209.8%5Ctimes%20h)
![h=\frac{261.3}{0.3\times 9.8}](https://tex.z-dn.net/?f=h%3D%5Cfrac%7B261.3%7D%7B0.3%5Ctimes%209.8%7D)
![h=88.87\ m](https://tex.z-dn.net/?f=h%3D88.87%5C%20m)
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 %
Answer:
C: equal to mg
Explanation:
in free-fall, gravity is always the net force on an object
Your lungs aren’t the ones that make the sound