Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 2.20×106 N , one at an angle 12.0 ∘ west of north,
1 answer:
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We can calculate the length of each spring by using the relationship:
where
F is the force applied to the spring
k is the spring constant
x is the length of the spring (measured with respect to its rest position)
Re-arranging the equation, we have
The force applied to both spring is F=60 N. Spring A has spring constant of k=4 N/m, therefore its length with respect to its rest position is
Spring B has spring constant of k=5 N/m, so its length with respect to its rest position is
Therefore, the correct answer is
<span>D.Spring A is 3 m longer than spring B because 15 – 12 = 3.</span>
where
F is the force applied to the spring
k is the spring constant
x is the length of the spring (measured with respect to its rest position)
Re-arranging the equation, we have
The force applied to both spring is F=60 N. Spring A has spring constant of k=4 N/m, therefore its length with respect to its rest position is
Spring B has spring constant of k=5 N/m, so its length with respect to its rest position is
Therefore, the correct answer is
<span>D.Spring A is 3 m longer than spring B because 15 – 12 = 3.</span>