As the temperature of the lead and helium is the same. Thus the average kinetic energy is also the same for lead and helium.
Reason:
It is given that a 5.0-kg bar of lead is placed inside a 12-L chamber filled with helium gas. The temperature of the lead and helium is the same. It is required to compare the average kinetic energy of the lead atoms and helium atoms.
The average kinetic energy is calculated as, 
.
Here K is the average kinetic energy, R is the gas constant, N is the Avogadro's number, and T is the temperature.
As the temperature is the same for both lead and helium. As a result, the average kinetic energy is also the same for lead and helium.
Learn more about average kinetic energy here,
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2HCl + MgO ---> MgCl2 + H2O
Answer:
Total pressure in atmosphere IS 4.36 atm
Explanation:
Given Data:
Depth of lake is 400 mtr
surface tension is 147 kilopascals
surface temperature 94 kelvin
Gravity = 1.35 m/s^2
specific gravity of methane(liquid) = 0.415
specific gravity of ethane(liquid) = 0.546
density of methane can be obtained as
Density of ethane can be obtained as
Total pressure in atmosphere can be obtained as
= 441.84 kPa
= 4.36 atm
Answer:
w = vR/3
Explanation:
The centre of mass of the loop to bullet system is given by D / 4 from centre of loop, which is equivalent to R / 2 from its centre.
From the principle of conservation of linear momentum
, we have
m*v = 2*m* Vcm
Where v = velocity of bullet, Vcm = velocity of wood
Hence, we have
Vcm = v2
Also, from the conservation of angular momentum about the centre of mass.
M*V*(R/2) = Ic*w - equation (I)
where Ic = moment of inertia and w = angular velocity
Ic for a ring is given by 
Ic of a bullet is given by 
Hence, the moment of inertia of the system is given by the summation of the two moments of inertia Ic(ring) + Ic(bullet) which gives
Ic(system) = 
Substituting back into equation (I), we have
Hence, we obtain w =vR/3
w=v3R
