That depends on a few things that you haven't told us about the setup.
So I'm going to assume one of them, and then give you the answer
in terms of another one:
-- Assume a Class-I lever . . . the fulcrum is between the load and the effort.
-- Then the effort needed to lift the load is
(the weight of the load) x (13 / the distance between the fulcrum and the effort)
Its the second option if you have Suttons Bay Virtual Online School
F=ma
We know from this that a=10,000m/s2
V=at
X=vt
You end up with v^2=ax
Plug in 10,000 and 0.04 and solve for v =20m/s
Answer:
34.51
Explanation:
k=1/2mv² is the kenetic energy equation to fill is in
k=[1/2(0.235)×50]²
Answer:i One way to solve the quadratic equation x2 = 9 is to subtract 9 from both sides to get one side equal to 0: x2 – 9 = 0. The expression on the left can be factored:
Explanation:
