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:
1) the final temperature is T2 = 876.76°C
2) the final volume is V2 = 24.14 cm³
Explanation:
We can model the gas behaviour as an ideal gas, then
P*V=n*R*T
since the gas is rapidly compressed and the thermal conductivity of a gas is low a we can assume that there is an insignificant heat transfer in that time, therefore for adiabatic conditions:
P*V^k = constant = C, k= adiabatic coefficient for air = 1.4
then the work will be
W = ∫ P dV = ∫ C*V^(-k) dV = C*[((V2^(-k+1)-V1^(-k+1)]/( -k +1) = (P2*V2 - P1*V1)/(1-k)= nR(T2-T1)/(1-k) = (P1*V1/T1)*(T2-T1)/(1-k)
W = (P1*V1/T1)*(T2-T1)/(1-k)
T2 = (1-k)W* T1/(P1*V1) +T1
replacing values (W=-450 J since it is the work done by the gas to the piston)
T2 = (1-1.4)*(-450J) *308K/(101325 Pa*650*10^-6 m³) + 308 K= 1149.76 K = 876.76°C
the final volume is
TV^(k-1)= constant
therefore
T2/T1= (V2/V1)^(1-k)
V2 = V1* (T2/T1)^(1/(1-k)) = 650 cm³ * (1149.76K/308K)^(1/(1-1.4)) = 24.14 cm³
Answer:
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<u>Explanation:</u>
Task 1 time period = 200ms, Task 2 time period = 300ms
Task ticked = 
→ 5 times
Task 2 ticked =
→ 3 times
At 600 ms → 200ms 200ms 200ms
300ms → 
Largest time period = H.C.M of (200ms, 300ms)
= 600ms
