Total thermal energy is the answer to your question.
Answer:
Moving a unit "positive" test charge from A to B will result in a reduction in potential
V = K Q / R potential at a point
V2 - V1 = K Q (1 / .4 - 1 / .15) = = k Q (.15 - .4) / .06 = -4.17 K Q
V2 - V1 = -4.17 * 9 & 10E9 * 6.25 E-8
V2 - V1 = -4.17 * 562.5 J/C
V = - 2346 Volts
Answer:
I believe the correct answer is B..<u>" 1 "</u>
Explanation:
hope it helps! ;)
Answer:
A) 10 m/s
Explanation:
We know that according to conservation of momentum,
m1v1 + m2v2 = m1u1 + m2u2 ..............(equation 1)
where m1 and m2 are masses of two bodies, v1 and v2 are initial velocity before collision and u1 and u2 are final velocities after collision respectively.
From the given data
If truck and car are two bodies
truck : m1 = 2000 Kg v1 = 5 m/s u1 = 0
car : m2 = 1000 kg v2 = 0 u2 = ?
final velocity of truck and initial velocity of car are static because the objects were at rest in the respective time.
substituting the values in equation 1, we get
(2000 x 5) + 0 = 0 + (1000 x u2)
u2 =
x 5
= 10 m/s
Hence after collision, car moves at a velocity of 10 m/s