The answer is electricity
Dart board 3 represents high accuracy but low precision
I don't think that 4m has anything to do with the problem.
anyway. here.
A___________________B_______C
where A is the point that the train was released.
B is where the wheel started to stick
C is where it stopped
From A to B, v=2.5m/s, it takes 2s to go A to B so t=2
AB= v*t = 2.5 * 2 = 5m
The train comes to a stop 7.7 m from the point at which it was released so AC=7.7m
then BC= AC-AB = 7.7-5 = 2.7m
now consider BC
v^2=u^2+2as
where u is initial speed, in this case is 2.5m/s
v is final speed, train stop at C so final speed=0, so v=0
a is acceleration
s is displacement, which is BC=2.7m
substitute all the number into equation, we have
0^2 = 2.5^2 + 2*a*2.7
0 = 6.25 + 5.4a
a = -6.25/5.4 = -1.157
so acceleration is -1.157m/(s^2)
Answer:
22.11 m / s
Explanation:
The falcon catches the prey from behind means both are flying in the same direction ( suppose towards the left )
initial velocity of falcon = 28 cos 35 i - 28 sin 35 j
( falcon was flying in south east direction making 35 degree from the east )
momentum = .9 ( 28 cos 35 i - 28 sin 35 j )
= 20.64 i - 14.45 j
initial velocity of pigeon
= 7 i
initial momentum = .325 x 7i
= 2.275 i
If final velocity of composite mass of falcon and pigeon be V
Applying law of conservation of momentum
( .9 + .325) V = 20.64 i - 14.45 j +2.275 i
V = ( 22.915 i - 14.45 j ) / 1.225
= 18.70 i - 11.8 j
magnitude of V
= √ [ (18.7 )² + ( 11.8 )²]
= 22.11 m / s
