The first opiton is the answer A)<span>Rahul’s weight
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Isothermal Work = PVln(v₂/v₁)
PV = nRT = 2 mole * 8.314 J/ (k.mol) * 330 k = 5487.24 J
Isothermal Work = PVln(v₂/v₁) v₂ = ? v₁ = 19L,
1.7 kJ = (5487.24)In(v₂/19)
1700 = (5487.24)In(v₂/19)
In(v₂/19) = (1700/5487.24) = 0.3098
In(v₂/19) = 0.3098
(v₂/19) =

v₂ = 19*

v₂ = 25.8999
v₂ ≈ 26 L Option b.
Answer:
1) Addition of a catalyst
2) To change the reaction rate of slope B to look like slope A, simply add a catalyst to speed up the rate of reaction, giving you a higher amount of products in a shorter amount of time (line A)
Explanation:
1 and 2)Two things can alter the rate of a reaction, either the addition of a catylist which will not alter the composition of the products or reactants, but will accelerate the reaction time, or an increase in temperature will also increase the rate at which a reaction will occur.
You could choose temperature also and have the same result, it's your choice both are correct, but catalyst is the easiest.
Mass is indirectly proportional to acceleration, so, lighter the object greater would be it's acceleration...
A) 0.10 kg is lightest among them, so it's your answer
Answer:
Induced emf, 
Explanation:
Given that,
Length of the helicopter, l = 4 m
Angular speed of the helicopter, 
The vertical component of the Earth’s magnetic field is, 
We need to find the induced emf between the tip of a blade and the hub. The induced emf in terms of angular velocity of an rotating object is given by :



So, the induced emf between the tip of a blade and the hub is
. Hence, this is the required solution.