Answer:
0.0059
Explanation:
According to the question the seismic activity density is given by
Here,
Number of Earthquakes over a given time span = 424
The land area affected = 71300 mi²
So,
The seismic activity density is 0.0059
A person weighing 0.70 kn rides in an elevator that has an upward acceleration of 1.5 m/s2. what is the magnitude of the normal
1 answer:
First of all, we can find the mass of the person, since we know his weight W:
And so
We know for Newton's second law that the resultant of the forces acting on the person must be equal to the product between the mass and the acceleration a of the person itself:
There are only two forces acting on the person: his weight W (downward) and the vincular reaction Rv of the floor against the body (upward). So we can rewrite the previous equation as
We know the acceleration of the system,
(upward, so with same sign of Rv), so we can solve to find the value of Rv, the normal force exerted by the elevator's floor on the person:
And so
We know for Newton's second law that the resultant of the forces acting on the person must be equal to the product between the mass and the acceleration a of the person itself:
There are only two forces acting on the person: his weight W (downward) and the vincular reaction Rv of the floor against the body (upward). So we can rewrite the previous equation as
We know the acceleration of the system,
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We can solve the problem by using the law of conservation of energy.
Using the ground as reference point, the mechanical energy of the brick when it is at 5 m from the ground is just potential energy (because the brick is initially at rest, so it doesn't have kinetic energy):
when the brick is at h'=3 m from the ground, its mechanical energy is now sum of kinetic energy and potential energy:
where v is the velocity of the brick. Since E is conserved, it must be equal to the initial energy (98.1 J), so we can solve this equation to find v:
Using the ground as reference point, the mechanical energy of the brick when it is at 5 m from the ground is just potential energy (because the brick is initially at rest, so it doesn't have kinetic energy):
when the brick is at h'=3 m from the ground, its mechanical energy is now sum of kinetic energy and potential energy:
where v is the velocity of the brick. Since E is conserved, it must be equal to the initial energy (98.1 J), so we can solve this equation to find v:
Answer
given,
mass of the person, m = 50 Kg
length of scaffold = 6 m
mass of scaffold, M= 70 Kg
distance of person standing from one end = 1.5 m
Tension in the vertical rope = ?
now equating all the vertical forces acting in the system.
T₁ + T₂ = m g + M g
T₁ + T₂ = 50 x 9.8 + 70 x 9.8
T₁ + T₂ = 1176...........(1)
system is equilibrium so, the moment along the system will also be zero.
taking moment about rope with tension T₂.
now,
T₁ x 6 - mg x (6-1.5) - M g x 3 = 0
'3 m' is used because the weight of the scaffold pass through center of gravity.
6 T₁ = 50 x 9.8 x 4.5 + 70 x 9.8 x 3
6 T₁ = 4263
T₁ = 710.5 N
from equation (1)
T₂ = 1176 - 710.5
T₂ = 465.5 N
hence, T₁ = 710.5 N and T₂ = 465.5 N
