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:
PE = mgh where m is the mass, g is gravity, and h is the height from which the object can potentially fall. We are given enough info here to solve for h. Filling in:
42 = 1.8(9.8)h and
and rounding to 2 sig fig's as we need:
h = 2.4 m, choice D
10.7 rad/s is the final angular velocity of the stick.
Given:
Mass of the stick = 4.42 kg
Length of the stick = 1.23m
Force of impulse (I) = 12.8 N s
The linear velocity of the stick,
Therefore, the final linear velocity of the stick is 2.89 m/s
∴
Therefore, 10.7 rad/s is the final angular velocity of the stick.
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We have centripetal acceleration =
So, v =
Now by equation of motion we have S= ut +0.5a
S =displacement, u = initial velocity, a= acceleration and t = time
Here S = 2r and a = g , u = 0
Distance traveled in horizontal direction =