We know that:
W=Fs
200J=150N*s
s=200J/150N
s=1,33m
Answer:
y₀ = 1020.3 m
Explanation:
This is a projectile launching exercise, in this case as the package is released its initial vertical velocity is zero.
y = y₀ +
t - ½ g t²
when it reaches the ground its height is zero
0 = y₀ + 0 - ½ g t²
y₀ = ½ g t²
let's calculate
y₀ = ½ 9.8 14.43²
y₀ = 1020.3 m
At a definite point, the bridge would begin oscillating to the matching rhythm as that
of the marching footsteps.
This oscillation would touch a determined peak when the bridge can
no longer tolerate its own
power and later collapses. So, soldiers are
systematic to break their steps
while passing a bridge.
The answer to the question is gonna be A
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



Now it is given that it's specific gravity is 10.8
So density of rock is

mass of the stone will be



now change in potential energy is given as

here
g = gravity on planet = 0.278 m/s^2
H = height lifted upwards = 15 cm


Now energy supplied by internal circuit of robot is given by

V = voltage supplied = 10 V
i = current = 1.83 mA
t = time = 12 s


Now efficiency is defined as the ratio of output work with given amount of energy used


so efficiency will be 23 %