Answer:
Capacitive Reactance is 4 times of resistance
Solution:
As per the question:
R = 
where
R = resistance
f = fixed frequency
Now,
For a parallel plate capacitor, capacitance, C:
where
x = separation between the parallel plates
Thus
C ∝ 
Now, if the distance reduces to one-third:
Capacitance becomes 3 times of the initial capacitace, i.e., x' = 3x, then C' = 3C and hence Current, I becomes 3I.
Also,
Also,
Z ∝ I
Therefore,
Solving the above eqn:
Answer:
F₁ / F₂ = 10
therefore the first out is 10 times greater than the second barrier
Explanation:
For this exercise let's use the relationship between momentum and momentum.
I = F t = Δp
in this case the final velocity is zero
F t = 0 -m v₀
F = m v₀ / t
in order to answer the question we must assume that the two vehicles have the same mass and speed
concrete barrier
F₁ = -p₀ / 0.1
F₁ = - 10 p₀
barrier collapses
F₂ = -p₀ / 1
let's look for the relationship of the forces
F₁ / F₂ = 10
therefore the first out is 10 times greater than the second barrier
solution:
radius of steel ball(r)=5cm=0.05m
density of ball =8000kgm
terminal velocity(v)=25m/s^2
density of air( d) =1.29 kgm
now
volume of ball(V)=4/3pir^3=1.33×3.14×0.05^3=0.00052 m^3
density of ball= mass of ball/Volume of ball
or, 8000=m/0.00052
or, m=4.16 kg
weight of the ball (W)= mg=4.16×10=41.6 N
viscous force(F)=6 × pi × eta × r × v
=6×3.14×eta×0.05×25
=23.55×eta
To attain the terminal velocity,
Fiscous force=Weight
or, 23.55× eta = 41.6
or, eta = 1.76
whete eta is the coefficient of viscosity.
