in this since your volume remains at a constant you'll need to use Gay-Lussacs law, p1/t1=p2/t2.
your temp should be converted in kelvin
variables:
p1=3.0×10^6 n/m^2
t1= 270k
just add 273 to your celcius
p2= ? your solving for this
t2= 315k
then you set up the equation
(3.0×10^6)/270= (x)(315)
you then cross multiply
(3.0×10^6)315=270x
distribute the 315 to the pressure.
9.45×10^8=270x then you divide 270 o both sides to get
answer
3.5×10^6 n/m^2
Answer:
a = 4.9(1 - sinθ - 0.4cosθ)
Explanation:
Really not possible without a complete setup.
I will ASSUME that this an Atwood machine with two masses (m) connected by an ideal rope passing over an ideal pulley. One mass hangs freely and the other is on a slope of angle θ to the horizontal with coefficient of friction μ. Gravity is g
F = ma
mg - mgsinθ - μmgcosθ = (m + m)a
mg(1 - sinθ - μcosθ) = 2ma
½g(1 - sinθ - μcosθ) = a
maximum acceleration is about 2.94 m/s² when θ = 0
acceleration will be zero when θ is greater than about 46.4°
The answer is C (the same number of valence electrons)
Because the coefficient of friction depends on the surface
