The first collision because a greater amount of momentum must be taken and used in order to push the cart back, giving it a greater mass and impulse
Answer:
2.06 m/s
Explanation:
From the law of conservation of linear momentum, the sum of momentum before and after collision are equal. Considering this case where we have frictionless surface, no momentum is lost in the process.
Momentum before collision
Momentum is given by p=mv where m and v represent mass. The initial sum of momentum will be 9v+(27*0.5)=9v+13.5
Momentum after collision
The momentum after collision will be given by (9+27)*0.9=32.4
Relating the two then 9v+13.5=32.4
9v=18.5
V=2.055555555555555555555555555555555555555 m/s
Rounded off, v is approximately 2.06 m/s
It’s just E because ethe positiv and negative current are supposed to flow thorough the bulb in opppsote sides at a equel level.In some them negerive/postive is absent and some of them are connected to the same side
A=Fh
A - work
F - force
h - distance
F=mg
m - mass (god+basket)
so
A=mgh
187 = m*10*4
187=40m
m=187/40
m=4.675 kg
or 4kg and 675g
pretty small dog...
<span>A+B-C
</span><span>A = 6x - 2y
B = -4x - 8y
C = -3x + 9y
(</span>6x - 2y) + (-4x - 8y) - (-3x + 9y)
(6x - 2y) + (-4x - 8y) + (3x - 9y)
2x -10y + (3x - 9y)
5x - 19y
