Answer:
Bank angle = 35.34o
Explanation:
Since the road is frictionless,
Tan (bank angle) = V^2/r*g
Where V = speed of the racing car in m/s, r = radius of the arc in metres and g = acceleration due to gravity in m/s^2
Tan ( bank angle) = 40^2/(230*9.81)
Tan (bank angle) = 0.7091
Bank angle = tan inverse (0.7091)
Bank angle = 35.34o
Felectric = q*E
<span> Ftranslational = m*a
</span><span> Felectric = Ftranslational
</span> <span>q*E = m*a
</span><span> Solve for a
</span><span> a = q/m*E </span>
<span> Our sign convention is "up is positive"
</span><span> q = 1.6*10^-19 C
</span><span> m = 1.67*10^-27 kg
</span><span> E = -150 N/C (- because it is down and up is positive)
</span> a =<span>
-6,4*10^5</span><span> m/s^2 (downward)
</span> answer
a = -6,4*10^5 m/s^2 (downward)
<span>Assuming pulley is frictionless. Let the tension be ‘T’. See equation below.</span>
<span> </span>
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