The two conditions are:
1) Application of-force on the body.
2) Displacement of the body in the direction of force.
Hope this helps!
(A) The total initial momentum of the system is
(1.30 kg) (27.0 m/s) + (23.0 kg) (0 m/s) = 35.1 kg•m/s
(B) Momentum is conserved, so that the total momentum of the system after the collision is
35.1 kg•m/s = (1.30 kg + 23.0 kg) <em>v</em>
where <em>v</em> is the speed of the combined blocks. Solving for <em>v</em> gives
<em>v</em> = (35.1 kg•m/s) / (24.3 kg) ≈ 1.44 m/s
(C) The kinetic energy of the system after the collision is
1/2 (1.30 kg + 23.0 kg) (1.44 m/s)² ≈ 25.4 J
and before the collision, it is
1/2 (1.30 kg) (27.0 m/s)² ≈ 474 J
so that the change in kinetic energy is
∆<em>K</em> = 25.4 J - 474 J ≈ -449 J
The working equation to be used for this is written below:
E = kQ/d²
where
E is the electric field
k is a constant equal to 8.99 x 10⁹ N m²/C²
Q is the charge
d is the distance
E = (8.99 x 10⁹ N m²/C²)(17×10⁻⁹ C)/(0.05 m)²
E = 61,132 N/C
The answer is <span>b. False. give the radius, degree of curvature, and length of curve that you would recommend. 3.43 for the h. Give it a try man
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