Answer: 7.38 km
Explanation: The attachment shows the illustration diagram for the question.
The range of the bomb's motion as obtained from the equations of motion,
H = u(y) t + 0.5g(t^2)
U(y) = initial vertical component of velocity = 0 m/s
That means t = √(2H/g)
The horizontal distance covered, R,
R = u(x) t = u(x) √(2H/g)
Where u(x) = the initial horizontal component of the bomb's velocity = 287 m/s, H = vertical height at which the bomb was thrown = 3.24 km = 3240 m, g = acceleration due to gravity = 9.8 m/s2
R = 287 √(2×3240/9.8) = 7380 m = 7.38 km
Explanation:
'What is the magnitude of the force needed to stop the horses and bring the box into equilibrium?' ≈42N; according to the vectors rules.
'Where would you locate the rope to apply the force?' - in point D.
PS. zoom out the attached picture.
It takes work to push charge through a change of potential.
There's no change of potential along an equipotential path,
so that path doesn't require any work.
Answer:
maximum speed of the elevator is 6.54 m/s
Explanation:
given,
force exerted on the elevator is 1.59 times passenger weight.
when the elevator accelerates upward with rate of a.
total force
F = m × (a+g)
1.59× mg = ma + mg
0.59 mg = ma
a = 0.59 g
we also have that
=
= 6.54 m/s
maximum speed of the elevator is 6.54 m/s