
Actually Welcome to the concept of Efficiency.
Here we can see that, the Input work is given as 2.2 x 10^7 J and the efficiency is given as 22%
The efficiency is => 22% => 22/100.
so we get as,
E = W(output) /W(input)
hence, W(output) = E x W(input)
so we get as,
W(output) = (22/100) x 2.2 x 10^7
=> W(output) = 0.22 x 2.2 x 10^7 => 0.484 x 10^7
hence, W(output) = 4.84 x 10^6 J
The useful work done on the mass is 4.84 x 10^6 J
Answer:
= 1,386 m / s
Explanation:
Rocket propulsion is a moment process that described by the expression
- v₀ =
ln (M₀ / Mf)
Where v are the velocities, final, initial and relative and M the masses
The data they give are the relative velocity (see = 2000 m / s) and the initial mass the mass of the loaded rocket (M₀ = 5Mf)
We consider that the rocket starts from rest (v₀ = 0)
At the time of burning half of the fuel the mass ratio is that the current mass is
M = 2.5 Mf
- 0 = 2000 ln (5Mf / 2.5 Mf) = 2000 ln 2
= 1,386 m / s
10,000 units of momentum.
p=mv
20,000=m(2v)
10,000=mv
Answer:

Explanation:
Given:
- volume of oil in the cylinder,

- volume of the oil level when the ice is immersed,

- the volume level of oil when the ice melted,

<u>Now, therefore the volume of ice:</u>



<u>Now the volume of water:</u>



As we know that the relative density is the ratio of density of the substance to the density of water.
<u>So, the relative density of ice:</u>
.....................(1)
as we know that density is given as:

now eq. (1)

where, m = mass of the water or the ice which remains constant in any phase


