Answer:
Lifting force, F = 21240 N
Explanation:
It is given that,
Mass of the helicopter, m = 1800 kg
It rises with an upward acceleration of 2 m/s². We need to find the lifting force supplied by its rotating blades. It is given by :
F = mg + ma
Where
mg is its weight
and "ma" is an additional acceleration when it is moving upwards.
So, 
F = 21240 N
So, the lifting force supplied by its rotating blades is 21240 N. Hence, this is the required solution.
20/45=0.4*100= 44.4 so the answer is..................................................
Answer: 44.4%
Answer:
If a vertical line extending down from an object's CG extends outside its area of support, the object will topple
Explanation:
We can understand better this situation using a diagram with the forces acting on it.
In the attached image we can see that when the gravity center is bouncing outside from the area of the pedestal, the object will be out of balance and will fall.
Answer:
Explanation:
Let the velocity after first collision be v₁ and v₂ of car A and B . car A will bounce back .
velocity of approach = 1.5 - 0 = 1.5
velocity of separation = v₁ + v₂
coefficient of restitution = velocity of separation / velocity of approach
.8 = v₁ + v₂ / 1.5
v₁ + v₂ = 1.2
applying law of conservation of momentum
m x 1.5 + 0 = mv₂ - mv₁
1.5 = v₂ - v₁
adding two equation
2 v ₂= 2.7
v₂ = 1.35 m /s
v₁ = - .15 m / s
During second collision , B will collide with stationary A . Same process will apply in this case also. Let velocity of B and A after collision be v₃ and v₄.
For second collision ,
coefficient of restitution = velocity of separation / velocity of approach
.5 = v₃ + v₄ / 1.35
v₃ + v₄ = .675
applying law of conservation of momentum
m x 1.35 + 0 = mv₄ - mv₃
1.35 = v₄ - v₃
adding two equation
2 v ₄= 2.025
v₄ = 1.0125 m /s
v₃ = - 0 .3375 m / s
