A) average acceleration = final velocity - initial velocity / time
= 7700 - 0 / 11
= 700ms^-2
B) force = mass x acceleration
= (3.05 x 105) x 700
= 320.25 x 700
= 224,175N
Consider velocity to the right as positive.
First mass:
m₁ = 4.0 kg
v₁ = 2.0 m/s to the right
Second mass:
m₂ = 8.0 kg
v₂ = -3.0 m/s to the left
Total momentum of the system is
P = m₁v₁ + m₂v₂
= 4*2 + 8*(-3)
= -16 (kg-m)/s
Let v (m/s) be the velocity of the center of mass of the 2-block system.
Because momentum of the system is preserved, therefore
(m₁+m₂)v= -16
(4+8 kg)*(v m/s) = -16 (kg-m)/s
v = -1.333 m/s
Answer:
The center of mass is moving at 1.33 m/s to the left.
In this problem we have the electric field intensity E:
E = 6.5 × 
newtons/coulomb
We have the magnitude of the load:
q = 6.4 × 
coulombs
We also have the distance d that the load moved in a direction parallel to the field 1.2 × 
meters.
We know that the electric potential energy (PE) is:
PE = qEd
So:
PE = (6.4 × 
)(6.5 × )(1.2 × )
PE = 5.0 x 
joules
None of the options shown is correct.
Newton' 1st Law of Motion
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