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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Answer:
Explanation:
Given data
Q₁=200cm³/s
We know that:
can be written as:
ΔP=F/A=n×v/L
And
Q=ΔP/R
As
n₂=6.0n₁
So
Q=ΔP/R
Answer:
K.E = 100 J
Final P.E = 100 J
Explanation:
The kinetic energy of any object can be given by the following formula:
where,
K.E = Kinetic Energy
m = mass of ball = 2 kg
v = speed of ball
Initially, v = 10 m/s. Therefore, the initial K.E is given as:
<u>K.E = 100 J</u>
Now, at the highest point the K.E of the ball becomes zero. because the ball stops for a moment at the highest point and its velocity becomes zero. So, from Law of Conservation of energy:
Initial K.E + Initial P.E = Final K.E + Final P.E
Initial P.E is also zero due to zero height initially.
K.E + 0 = 0 + Final P.E
<u>Final P.E = 100 J</u>