From tables, the density of mercury is
13545 kg/m^3 at 20°C,
13472 kg/m^3 at 50°C.
Because mass = density * volume, the mass of mercury at 20°C is
m = (13545 kg/m^3)*(0.002 m^3) = 27.09 kg
Let V = volume of mercury at 50°C.
Because the mass of mercury does not change, therefore at 50°C,
(13472 kg/m^3)*(V m^3) = 27.09
V = 27.09/13472 = 0.0020108 m^3
Answer: B. 0.002010812 m³
Answer: -7J
Explanation:
To determine the change in potential energy, use the equation ΔPE=mgΔh
Δ
PE=mg
Δ
h
, where m is the mass, g= 9.8 m/s2
g= 9.8 m/s
2
, and Δh is the change in height. Hence, we have that the potential energy PE=4.0 kg×9.8 m/s2×-2.0 m=-78 J
PE=4.0 kg
×
9.8 m/s
2
×
-2.0 m=-78 J
. To verify that this is correct, note that since the cat changes the potential energy to kinetic energy by jumping, the potential energy decreases. Hence, the potential energy should be negative.
T is the time for a whole round.
centripetal acceleration = V^2/R,
20 = 40^2 / R, find R = 40^2/20 = 40*40/20 = 80 m, right?
Now, one round is L = 2*pi*R = 2*pi*80 = 160*pi
And T = L/v (distance/speed) = 160*pi/40 = 4*pi seconds, or ~ 12.57 s
Answer:
i dont know
Explanation:im sorry to do this to you but you dont have to watch ads if you answer questions
Answer:
Zero or +2
Explanation:
The noble gases already have a avplete outermost shell. They are the least reactive elements of earth?
Their normal oxidation number is zero but some have been shown to be reactive.
