Answer:
v = 8.09 m/s
Explanation:
For this exercise we use that the work done by the friction force plus the potential energy equals the change in the body's energy.
Let's calculate the energy
starting point. Higher
Em₀ = U = m gh
final point. To go down the slope
Em_f = K = ½ m v²
The work of the friction force is
W = fr L cos 180
to find the friction force let's use Newton's second law
Axis y
N - W_y = 0
N = W_y
X axis
Wₓ - fr = ma
let's use trigonometry
sin θ = y / L
sin θ = 11/110 = 0.1
θ = sin⁻¹ 0.1
θ = 5.74º
sin 5.74 = Wₓ / W
cos 5.74 = W_y / W
Wₓ = W sin 5.74
W_y = W cos 5.74
the formula for the friction force is
fr = μ N
fr = μ W cos θ
Work is friction force is
W_fr = - μ W L cos θ
Let's use the relationship of work with energy
W + ΔU = ΔK
-μ mg L cos 5.74 + (mgh - 0) = 0 - ½ m v²
v² = - 2 μ g L cos 5.74 +2 (gh)
v² = 2gh - 2 μ gL cos 5.74
let's calculate
v² = 2 9.8 11 - 2 0.07 9.8 110 cos 5.74
v² = 215.6 -150.16
v = √65.44
v = 8.09 m/s
Answer:
the molecular formula for the gas is NO₂
Explanation:
since it contains
Nitrogen = n → 30.45%
Oxygen = o → 69.55%
and 30.45%+69.55% = 100% , then the gas only contains nitrogen and oxygen
Also we know that the proportion of oxygen over nitrogen is
proportion of oxygen over nitrogen = moles of oxygen / moles of nitrogen
since
moles = mass / molecular weight
then for a sample of 100 gr of the unknown gas
mass of oxygen = 69.55%*100 gr = 69.55 gr
mass of Nitrogen = 30.45%*100 gr = 30.45 gr
proportion of oxygen over nitrogen = (mass of oxygen/ molecular weight)/(mass of nitrogen / molecular weight of nitrogen ) = (69.55 gr/ 16 gr/mol) /( 30.45 gr /14 gr/mol) = 1.998 mol of O/ mol of N≈ 2 mol of O/ mol of N
therefore there are 2 atoms of oxygen per atom of nitrogen
thus the molecular formula for the gas is:
NO₂
The ideal gas law.
PV=nRT
P=presure
V=volume
n=number of moles
R=Gas costant
T=temperature.
Answer: a. Number of moles.
