Answer:
C. 85%
Explanation:
A cylinder fitted with a piston exists in a high-pressure chamber (3 atm) with an initial volume of 1 L. If a sufficient quantity of a hydrocarbon material is combusted inside the cylinder to produce 1 kJ of energy, and if the volume of the chamber then increases to 1.5 L, what percent of the fuel's energy was lost to friction and heat?
A. 15%
B. 30%
C. 85%
D. 100%
work done by the system will be
W=PdV
p=pressure
dV=change in volume
3tam will be changed to N/m^2
3*1.01*10^5
W=3.03*10^5*(1.5-1)
convert 0.5L to m^3
5*10^-4
W=3.03*10^5*5*10^-4
W=152J
therefore
to find the percentage used
152/1000*100
15%
100%-15%
85% uf the fuel's energy was lost to friction and heat
Range of a projectile motion is given by
R = v cos θ / g (v sin θ + sqrt(v^2 sin^2 θ + 2gy_0)); where R = 188m, θ = 41°, g = 9.8m/s^2, y_0 = 0.9
188 = v cos 41° / 9.8 (v sin 41° + sqrt(v^2 sin^2 41° + 2 x 9.8 x 0.9)) = 0.07701(0.6561v + sqrt(0.4304 v^2 + 17.64)) = 0.05053v + 0.07701sqrt(0.4304v^2 + 17.64)
0.07701sqrt(0.4304v^2 + 17.64) = 188 - 0.05053v
0.005931(0.4304v^2 + 17.64) = 35344 - 19v + 0.002553v^2
0.002553v^2 + 0.1046 = 35344 - 19v + 0.002553v^2
19v = 35344 - 0.1046 = 35343.8954
v = 35343.8954/19 = 1860 m/s
The equation we use is mλ=dsinθ for intensity maximas. We are given at the first maximum (m=1), it occurs at 17.8 degrees. Thus we can solve for d by substituting known values into our equation.
(1) (632.8*10^-9m)=dsin(17.8) => d = 2.07*10^-6m
Next we want to find the angle at the second maximum (m=2) so we need to solve for θ.
(2) (632.8*10^-9m) = (2.07*10^-6m)sinθ
θ=37.69 degrees
Hopes this helps!
P.S. I hope this is right. If not sorry in advance.
