Given a second class lever with a distance of 5.00 feet from the fulcrum to the effort and a distance of 33.0 inches from the re
1 answer:
Answer:
The correct answer is C. 45.5 lbs.
Explanation:
In a second class lever, the load is located between the point in which the force is exerted and the fulcrum.
The formula for any problem involving a lever is:
Where F_e is the effort force, d_e is the total length of the lever, F_l is the load that can be lifted and d_l is the distance between the point of the effort and the fulcrum.
The parameter of the formula that you need is F_l:
The conversion from feet to inches is 1 ft is equal to 12 inches. In this case, 5 ft are equal to 60 inches.
F_l=45.5 lbs
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Explanation:
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Answer: C) divide: distance ÷ velocity
Explanation:
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If we isolate we will have:
Hence, the correct option is C: distance divided by velocity.