Steps 1 and 2)
The variables are W = work, P = power, and t = time. In this case, W = 9514 joules and P = 347 watts.
The goal is to solve for the unknown time t.
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Step 3)
Since we want to solve for the time, and we have known W and P values, we use the equation t = W/P
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Step 4)
t = W/P
t = 9514/347
t = 27.4178674351586
t = 27.4 seconds
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Step 5)
The lawn mower ran for about 27.4 seconds. I rounded to three sig figs because this was the lower amount of sig figs when comparing 9514 and 347.
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Note: we don't use the mass at all
Answer:
T
Explanation:
= magnitude of current in each wire = 2.0 A
= length of the side of the square = 4 cm = 0.04 m
= length of the diagonal of the square =
a =
(0.04) = 0.057 m
= magnitude of magnetic field by wires at A and C


T
= magnitude of magnetic field by wire at B


T
Net magnitude of the magnetic field at D is given as



T
Answer:
1/2
Explanation:
The energy stored in a capacitor is given by

where
C is the capacitance
V is the potential difference
Calling
the capacitance of capacitor 1 and
its potential difference, the energy stored in capacitor 1 is

For capacitor 2, we have:
- The capacitance is half that of capacitor 1: 
- The voltage is twice the voltage of capacitor 1: 
so the energy stored in capacitor 2 is

So the ratio between the two energies is

Answer:
0.29D
Explanation:
Given that
F = G M m / r2
F = GM(6m) / (D-r)2
G Mm/r2 = GM(6m) / (D-r)2
1/r2 = 6 / (D-r)2
r = D / (Ö6 + 1)
r = 0.29 D
See diagram in attached file