How does the use of a scanning electron microscope differ from that of a transmission electron microscope?. the options are....
2 answers:
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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
Answer:
(a) 1.093 rad/s^2
(b) 4.375 rad/s
(c) 8.744 rad/s
(d) 67.845 rad
Explanation:
initial angular velocity, ωo = 0
time, t = 8s
angular displacement, θ = 35 rad
(a) Let α be the angular acceleration.
Use second equation of motion for rotational motion
By substituting the values
35 = 0 + 0.5 x α x 8 x 8
α = 1.093 rad/s^2
(b) The average angular velocity is defined as the ratio of total angular displacement to the total time taken .
Average angular velocity = 35 / 8 = 4.375 rad/s
(c) Let ω be the instantaneous angular velocity at t = 8 s
Use first equation of motion for rotational motion
ω = ωo + αt
ω = 0 + 1.093 x 8 = 8.744 rad/s
(d) Let in next 5 seconds the angular displacement is θ.
By substituting the values
θ = 8.744 x 5 + 0.5 x 1.093 x 5 x 5
θ = 67.845 rad