Answer:
R = 148.346 N
M₀ = - 237.2792 N-m
Explanation:
Point O is selected as a convenient reference point for the force-couple system which is to represent the given system
We can apply
∑Fx = Rx = - 60N*Cos 45° + 40N + 80*Cos 30° = 66.8556 N
∑Fy = Ry = 60N*Sin 45° + 50N + 80*Sin 30° = 132.4264 N
Then
R = √(Rx²+Ry²) ⇒ R = √((66.8556 N)²+(132.4264 N)²)
⇒ R = 148.346 N
Now, we obtain the moment about the origin as follows
M₀ = (0 m*40 N)-(7 m*60 N*Sin 45°)+(4 m*60 N*Cos 45°)-(5 m*50 N)+ 140 N-m + (0 m*80 N*Cos 30°) + (0 m*80 N*Sin 30°) = - 237.2792 N-m (clockwise)
We can see the pic shown in order to understand the question.
Answer:
final volume V2 = 0.71136 m³
work done in process W = -291.24 kJ
heat transfer Q = 164 kJ
Explanation:
given data
mass = 1.5 kg
pressure p1 = 200 kPa
temperature t1 = 150°C
final pressure p2 = 600 kPa
final temperature t2 = 350°C
solution
we will use here superheated water table that is
for pressure 200 kPa and 150°C temperature
v1 = 0.95964 m³/kg
u1 = 2576.87 kJ/kg
and
for pressure 600 kPa and 350°C temperature
v2 = 0.47424 m³/kg
u2 = 2881.12 kJ/kg
so v1 is express as
V1 = v1 × m ............................1
V1 = 0.95964 × 1.5
V1 = 1.43946 m³
and
V2 = v2 × m ............................2
V2 = 0.47424 × 1.5
final volume V2 = 0.71136 m³
and
W = P(avg) × dV .............................3
P(avg) =
=
= 400 × 10³
put here value
W = 400 × 10³ × (0.71136 - 1.43946 )
work done in process W = -291.24 kJ
and
heat transfer is
Q = m × (u2 - u1) + W .............................4
Q = 1.5 × (2881.12 - 2576.87) + 292.24
heat transfer Q = 164 kJ
Answer:
have you heard of gnoogle?
Explanation:have you heard of goongle?
Answer:
I think D is correct
Explanation:
C is decreasing function, probably worst
A is arctan -> in radian, the rate of increasing is very slow-> second worst
B(14) = ln(9*14) = 4.8
D(14) = sqrt(8+14^2)=14.2