Answer:
1275J
Explanation:
Given parameters:
Force on box = 85N
Distance moved = 15m
Unknown:
Work done = ?
Solution:
Work done is the amount of force applied on a body to move it through a specific distance.
Work done = Force x distance
Now insert the parameters and solve;
Work done = 85 x 15 = 1275J
Answer:
9)a
10) I think true
11)b
Explanation:
9)a. because it's told that the car is slowing down, the sum of the forces that are towards left, should be more than the ones that are towards right. if the car was gaining speed, "b" would have been correct. and if it was told that the car is moving without a change in the speed, "c" would have been correct.
10) if a moving object has a change of speed or direction, it would have an acceleration. now if a moving object experiences an unbalanced force, it'd either slow down, gain speed or change direction, and in all of the three possibilities it'd have an acceleration.
11) upward and downward forces are equal, and the sum of them would be 0N(because they have opposite directions). so they negate each other.
and the rightward force is 5N more than the leftward force. so the Net Force would be 5N.
-30+30-10+15=5N
if it is unclear or you need more explanation, ask freely.
The answer is species, I hope this helps!
Answer:
The maximum energy stored in the combination is 0.0466Joules
Explanation:
The question is incomplete. Here is the complete question.
Three capacitors C1-11.7 μF, C2 21.0 μF, and C3 = 28.8 μF are connected in series. To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V. Determine the maximum energy stored in the series combination.
Energy stored in a capacitor is expressed as E = 1/2CtV² where
Ct is the total effective capacitance
V is the supply voltage
Since the capacitors are connected in series.
1/Ct = 1/C1+1/C2+1/C3
Given C1 = 11.7 μF, C2 = 21.0 μF, and C3 = 28.8 μF
1/Ct = 1/11.7 + 1/21.0 + 1/28.8
1/Ct = 0.0855+0.0476+0.0347
1/Ct = 0.1678
Ct = 1/0.1678
Ct = 5.96μF
Ct = 5.96×10^-6F
Since V = 125V
E = 1/2(5.96×10^-6)(125)²
E = 0.0466Joules
