work done is product of force and displacement of point of application of force
so here we have to check the product of force and displacement both
Now we will put the least to maximum work in the following order
1. -A man exerts strenuous effort in pushing a stationary wall
2. -A flea pushes a speck of dirt 1 cm
3. -A farmer pushes a 2 kg wheelbarrow 20 m
4. -A cannon launches a 3 kg cannonball a distance of 200
5. -A 2000 kg car travels 400 m down a road
6. -Space shuttle Atlantis launches from the ground into near-Earth orbit
Note: I'm not sure what do you mean by "weight 0.05 kg/L". I assume it means the mass per unit of length, so it should be "0.05 kg/m".
Solution:
The fundamental frequency in a standing wave is given by

where L is the length of the string, T the tension and m its mass. If we plug the data of the problem into the equation, we find

The wavelength of the standing wave is instead twice the length of the string:

So the speed of the wave is

And the time the pulse takes to reach the shop is the distance covered divided by the speed:
3 hours 30 min i believe because 126/36 is 3.5 and 3.5 hours is 3 hours and 30 min. Brainliest pls
Answer:
Explanation:
Calculate the volume of the lead

Now calculate the bouyant force acting on the lead


This force will act in upward direction
Gravitational force on the lead due to its mass will act in downward direction
Hence the difference of this two force

If V is the volume submerged in the water then bouyant force on the bobber is

Equate bouyant force with the tension and gravitational force

Now Total volume of bobble is

=
The answer is B. Unbalanced force