Work done = 0.5*m*[(v2)^2 - (v1)^2]
where m is mass,
v2 and v1 are the velocities.
Given that m = 1.50 x 10^3 kg, v2 = -15 m/s (decelerates), v1 = 25 kg,
Work done = 0.5 * 1.50 x 10^3 * ((-15)^2 - 25^2) = 3 x 10^5 joules
Just ignore the negative value for the final result because work is a scalar quantity.
About 4.6 billion years ago
Immediately after the switch is closed, the capacitor acts like a short circuit for a very small time. We can calculate the currents through each resistor by first calculating the total resistance:
Req = 1/(1/3+1/6) + 8 = 1/(3/6) + 8 = 2 + 8 = 10 ohms.
IR1 = E / Req = 42 / 10 = 4.2 A.
VR2 = VR3 = E * R23 / Req = 42 * 2 / 10 = 8.4 V
IR2 = VR2 / R2 = 8.4 / 6 = 1.4 A.
IR3 = VR3 / R3 = 8.4 / 3 = 2.8 A.
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