Explanation:
Sum of forces in the x direction:
∑Fx = ma
Rx − 250 N = 0
Rx = 250 N
Sum of forces in the y direction:
∑Fy = ma
Ry − 120 N − 300 N = 0
Ry = 420 N
Sum of forces in the z direction:
∑Fz = ma
Rz − 50 N = 0
Rz = 50 N
Sum of moments about the x axis:
∑τx = Iα
Mx + (-50 N)(0.2 m) + (-120 N)(0.1 m) = 0
Mx = 22 Nm
Sum of moments about the y axis:
∑τy = Iα
My = 0 Nm
Sum of moments about the z axis:
∑τz = Iα
Mz + (250 N)(0.2 m) + (-120 N)(0.16 m) = 0
Mz = -30.8 Nm
The answer is choice C
Explanation:
As during construction ,the site is cleared for all debris before laying out the foundation. Even the sewer lines are dug out .
So it will be useful for the construction crews to connect the pipes to the sewer lines before the foundation is poured.
But usually the steps take in construction activity is:- first the site is cleared for the foundation to be poured and once the foundation wall is set , then all utilities , including plumbing and electrical activities are done.,
After this process is over, the city inspector comes to check whether the foundation has been laid down as per the code of construction.
Only after that the rest of the construction activity follows through.
Answer:
The break force that must be applied to hold the plane stationary is 12597.4 N
Explanation:
p₁ = p₂, T₁ = T₂
The heat supplied = × Heating value of jet fuel
The heat supplied = 0.5 kg/s × 42,700 kJ/kg = 21,350 kJ/s
The heat supplied = ·
= 20 kg/s
The heat supplied = 20* = 21,350 kJ/s
= 1.15 kJ/kg
T₃ = 21,350/(1.15*20) + 485.03 = 1413.3 K
p₂ = p₁ × p₂/p₁ = 95×9 = 855 kPa
p₃ = p₂ = 855 kPa
T₃ - T₄ = T₂ - T₁ = 485.03 - 280.15 = 204.88 K
T₄ = 1413.3 - 204.88 = 1208.42 K
T₅ = 1208.42*(2/2.333) = 1035.94 K
= √(1.333*287.3*1035.94) = 629.87 m/s
The total thrust = × = 20*629.87 = 12597.4 N
Therefore;
The break force that must be applied to hold the plane stationary = 12597.4 N.
Answer:
The distance between the station A and B will be:
Explanation:
Let's find the distance that the train traveled during 60 seconds.
We know that starts from rest (v(0)=0) and the acceleration is 0.6 m/s², so the distance will be:
Now, we need to find the distance after 25 min at a constant speed. To get it, we need to find the speed at the end of the first distance.
Then the second distance will be:
The final distance is calculated whit the decelerate value:
The final velocity is zero because it rests at station B. The initial velocity will be v(1).
Therefore, the distance between the station A and B will be:
I hope it helps you!
Answer:
The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.
Explanation: