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
Before answering this question, first we have to understand the effect of ratio of surface area to volume on the rate of diffusion.
The rate of diffusion for a body having larger surface area as compared to the ratio of surface area to volume will be more than a body having less surface area. Mathematically it can written as-
V∝ R [ where v is the rate of diffusion and r is the ratio of surface area to volume]
As per the question,the ratio of surface area to volume for a sphere is given 
The surface area to volume ratio for right circular cylinder is given 
Hence, it is obvious that the ratio is more for right circular cylinder.As the rate diffusion is directly proportional to the surface area to volume ratio,hence rate of diffusion will be more for right circular cylinder.
Hence the correct option is B. The rate of diffusion would be faster for the right cylinder.
Answer:
(a) 21.36 ohms
(b) 5.62 A
Explanation:
Parameters given:
Potential difference, V = 120 V
Power, P = 674 W
(a) Power is given as:
P = V²/R
Where R is resistance
=> R = V²/P
R = 120²/674
R = 14400/674
R = 21.36 ohms
(b) Power is also given as:
P = I*V
Where I = Current (time rate of flow of Electric charge)
=> I = P/V
I = 674/120
I = 5.62 A
Answer:
lipids are insoluble in water which is why lipids are often found in biological membranes and other waterproof coverings.
(a) The moment of inertia of the wheel is 78.2 kgm².
(b) The mass (in kg) of the wheel is 1,436.2 kg.
(c) The angular speed (in rad/s) of the wheel at the end of this time period is 3.376 rad/s.
<h3>
Moment of inertia of the wheel</h3>
Apply principle of conservation of angular momentum;
Fr = Iα
where;
- F is applied force
- r is radius of the cylinder
- α is angular acceleration
- I is moment of inertia
I = Fr/α
I = (200 x 0.33) / (0.844)
I = 78.2 kgm²
<h3>Mass of the wheel</h3>
I = ¹/₂MR²
where;
- M is mass of the solid cylinder
- R is radius of the solid cylinder
- I is moment of inertia of the solid cylinder
2I = MR²
M = 2I/R²
M = (2 x 78.2) / (0.33²)
M = 1,436.2 kg
<h3>Angular speed of the wheel after 4 seconds</h3>
ω = αt
ω = 0.844 x 4
ω = 3.376 rad/s
Thus, the moment of inertia of the wheel is 78.2 kgm².
The mass (in kg) of the wheel is 1,436.2 kg.
The angular speed (in rad/s) of the wheel at the end of this time period is 3.376 rad/s.
Learn more about moment of inertia here: brainly.com/question/14839816
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