Answer:
the center of mass is 7.07 cm apart from the bend
Explanation:
the centre of mass of a wire of length L is L/2 ( assuming uniform density). Then initially the x coordinate of the centre of mass is
x₁ = L/2 = 20 cm /2 = 10 cm
when the wire is bent in a right angle the coordinates of the new centre of mass will be
x₂ = L₂/2
y₂= L₂/2
where L₂ is the length of the horizontal piece and vertical piece . Then L₂=L/2
x₂ = L₂/2 = L/4 = 20 cm/4 = 5 cm
y₂= L₂/2 = L/4 = 20 cm/4 = 5 cm
x₂=y₂=X
locating the bend in the origin (0,0) the distance to the centre of mass is
d = √(x₂²+y₂²) = √(2X²) = √2*X=√2*5cm = 7.07 cm
d = 7.07 cm
Answer:
By 16.7% or 0.167 IPM
Explanation:
Substracting the final IPM (6.088) to the initial IPM (5.921) gives us the net difference, which is how much did it increase in IPM. Multiplying this number by 100 gives us the percentual increase in the feed rate.
A pesticide is something that is used to kill / deter pests that eat / destroy crop.
The formula is Ke = 1/2 m v^2
The two of them together have a Ke of mv^2. So you either increase m or v. That's what makes the problem difficult. He can do D or B. We have to choose.
A is no solution. The Ke goes down because Paul loses Ivan's mass.
C is out of the question 3 meters/sec is a big reduction from 5 m/s. So now what do we do about B and D?
The question is what does the third person add. The tandoms I've peddled only allow for 1 or 2 people to add to the motion. So the third person only adds mass. He does not have a v that he is contributing to. To say that he is going 5m/s is true, but he's not contributing anything to that motion.
I pick B, but it is one of those questions that the correctness of it is in the head of the proposer. Be prepared to get it wrong. Argue the point politely if you agree with me, but back off as soon as you have presented your case.
B <<<<====== answer.
Answer:
Final speed of the crate is 15 m/s
Explanation:
As we know that constant force F = 80 N is applied on the object for t = 12 s
Now we can use definition of force to find the speed after t = 12 s
so here we know that object is at rest initially so we have
Now for next 6 s the force decreases to ZERO linearly
so we can write the force equation as
now again by same equation we have
put t = 6 s
