Answer:
f = 878,080 N
Explanation:
mass of pile driver (m) = 2100 kg
distance of pile driver to steel beam (s) = 5 m
depth of steel driven (d) = 12 cm = 0.12 m
acceleration due to gravity (g0 = 9.8 m/s^{2}
calculate the average force exerted on the pile driver by the beam.
- from work done = force x distance
- work done = change in potential energy of the pile driver
- equating the two equations above we have
force x distance = m x g x (s - d)
f x 0.12 = 2100 x 9.8 x (5- (-0.12))
d = - 0.12 because the steel beam went down at we are taking its
initial position to be an origin point which is 0
f = ( 2100 x 9.8 x (5- (-0.12)) ) ÷ 0.12
f = 878,080 N
Answer:
∆PE = 749.7 J
At 0.9 m high, PE = 793.8 J
At 1.75 m high, PE = 1543.5 J
Answer:
94.1 m
Explanation:
From Coulombs law,
F = Gm1m2/r²................... Equation 1
where F = force, m1 = first mass, m2 = second mass, G = universal constant, r = distance of separation.
Make r the subject of the equation,
r = √(Gm1m2/F)................. Equation 2
Given: F = 7×10² N, m1 = 15×10⁷ kg, m2 = 62×10⁷ kg,
Constant: G = 6.67×10⁻¹¹ Nm²/kg²
Substitute into equation 2
r = √( 6.67×10⁻¹¹×15×10⁷×62×10⁷/7×10²)
r √(886.16×10)
r √(88.616×10²)
r = 9.41×10
r = 94.1 m.
Hence the distance of separation = 94.1 m
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