Consider a particle with initial velocity v that has magnitude 12.0 m/s and is directed 60.0 degrees above the negative x axis.
1 answer:
Answer:
v_x = -6.00 m/s
v_y = 10.4 m/s
Explanation:
To calculate the velocity of the x-component we must use the cosine function:
, where ∝ is the angle 60°, and
is the velocity 12.0 m/s (note: the negative sign due to the question stating it is "above the negative x-axis).
This gives us a value m/s
Similarly for the y-component, we must use the sine function:
which gives us a value ≈ 10.4 m/s (since there are 3 significant figures in the information given in the question, this is rounded to 10.4 m/s).
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Answer:
W(r,out) = 5.81 MW
= 86.1 %
Explanation:
we use here steam table for get value of h1, s1 etc
so use for 6MPa and 600 degree
Enthalphy of steam h1 = 3658.8 kJ/kg
Entropy of steam s 1 is = 7.1693 kJ /kg.K
and
for 50 kPa and 100 degree
Enthalphy of steam h2 = 2682.4 kJ/kg
Entropy of steam s2 is = 7.6953 kJ /kg.K
so we use here energy balance equation that is
..............1
put here value and we get m
m =
solve it we get
m = 5.156 kg/s
so by energy balance equation
m1 = m
2 + W(r,out)
W(r,out) = m(1 -
2)
W(r,out) = h1 - h2 + ΔKE + ΔPE - To(s1-s2)
W(r,out) =
W(r,out) = W(a,out) - m.To.(s1-s2) ........................2
put here value
W(r,out) = 5000 - ( 5.156 × (25 + 273) ×( 7.1693 - 7.6953)
W(r,out) = 5908.19 = 5.81 MW
and
second law deficiency is
=
..............................3
put here value
= \frac{5}{5.81}
= 86.1 %