Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.20×10^6 N, one at an angle 14.0∘ west of north,
and the other at an angle 14.0∘ east of north, as they pull the tanker a distance 0.750 km toward the north. What is the total work they do on the supertanker?
1 answer:
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Answer:
66.2 sec
Explanation:
C₁ = 1.0 F
C₂ = 1.0 F
ΔV = Potential difference across the capacitor = 6.0 V
C = parallel combination of capacitors
Parallel combination of capacitors is given as
C = C₁ + C₂
C = 1.0 + 1.0
C = 2.0 F
R = resistance = 33 Ω
Time constant is given as
T = RC
T = 33 x 2
T = 66 sec
V₀ = initial potential difference across the combination = 6.0 Volts
V = final potential difference = 2.2 volts
Using the equation
t = 66.2 sec
Answer:
final displacement lf = 0.39 m
Explanation:
from change in momentum equation:
given: m = 0.4kg, y/x = 19/85, li = 1.9 m,
\delta p = 1.27 kg*m/s.
putting all value to get the final displacement value
final displacement lf = 0.39 m