A pile driver lifts a 450 kg weight and then lets it fall onto the end of a steel pipe that needs to be driven into the ground.
A fall of 1.5 m drives the pipe in 45 cm.
What is the average force exerted on the pipe?
2 answers:
The gravitational potential energy becomes work driving the pipe
<span>m g h = f d </span>
<span>450 * g * 1.5 = f * .45</span>
Answer:
F = 14700N
Explanation:
Hello! Let's solve this!
We have the following data:
vi = 0m / s
dy = 1.5m
m = 450kg
dx = 0.45m
First we need to know the speed of the impeller.
We can use the following formula

We get the vf


vf = 4.42m / s
Then the kinetic energy is equal to the work done.

KE=1/2*
*450kg
We cleared KE
KE = W = 6615J
From the working formula we will clear the force exerted.
W = F * dx
F = W / dx = 6615J / 0.45m
F = 14700N
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F=ma
F=QE = 1.602e-19C*700N/C = 1.1214e-16N
1.1214e-16N = ma = 1.6726e-27kg * a
a = 6.702e10 m/s² along the direction of the field line
True............................................
Answer:
- we need unbalance force to lift objects
- we need unbalance force to drag objects