Answer:
25.71 kgm/s
Explanation:
Let K₁ and K₂ be the initial and final kinetic energies of object A and v₁ and v₂ its initial and final speeds.
Given that K₂ = 0.7K₁
1/2mv₂² = 0.7(1/2mv₁²)
v₂ = √0.7v₁ = √0.7 × 20 m/s = ±16.73 m/s
Since A rebounds, its velocity = -16.73 m/s and its momentum change, p₂ = mΔv = m(v₂ - v₁) = 0.7 kg (-16.73 - 20) m/s = 0.7( -36.73) = -25.71 kgm/s.
Th magnitude of object A's momentum change is thus 25.71 kgm/s
Answer:
Rise in level of fluid is 0.11 m
Rise in level of fluid in case of mercury is 0.728 cm or 7.28 mm
Solution:
As per the question:
Density of oil, 
Change in Pressure in the tank, 
Density of the mercury, 
Now,
To calculate the rise in the level of fluid inside the manometer:
We know that:
1 mmHg = 133.332 Pa
Thus

Also,

where
g = acceleration due to gravity
h = height of the fluid level

h = 0.11 m
Now, if mercury is used:


h = 0.00728 m = 7.28 mm
When the iron and the water reach thermal equilibrium, they have same temperature,

.
We can consider this as an isolated system, so the heat released by the water is equal to the heat absorbed by the iron.
The hear released by the water is:

where

is the water mass,

is the specific heat of the water, and

is the variation of temperature of the water.
Similarly, the heat absorbed by the iron is:

where

is the iron mass,

is the iron specific heat, and

is the variation of temperature of the iron.
Writing

and replacing the numbers, we can solve to find mw, the mass of the water:
Answer:
The current is changing at the rate of 0.20 A/s
Explanation:
Given;
inductance of the inductor, L = 5.0-H
current in the inductor, I = 3.0 A
Energy stored in the inductor at the given instant, E = 3.0 J/s
The energy stored in inductor is given as;
E = ¹/₂LI²
E = ¹/₂(5)(3)²
E = 22.5 J/s
This energy is increased by 3.0 J/s
E = 22.5 J/s + 3.0 J/s = 25.5 J/s
Determine the new current at this given energy;
25.5 = ¹/₂LI²
25.5 = ¹/₂(5)(I²)
25.5 = 2.5I²
I² = 25.5 / 2.5
I² = 10.2
I = √10.2
I = 3.194 A/s
The rate at which the current is changing is the difference between the final current and the initial current in the inductor.
= 3.194 A/s - 3.0 A/s
= 0.194 A/s
≅0.20 A/s
Therefore, the current is changing at the rate of 0.20 A/s.