Answer:
a) 10.7° ≈ 11°
b) 0.19
Explanation:
If the road is banked at an angle, without seeking the help of friction, (i.e. frictionless road), the forces acting on the car are shown in the attached free body diagram to the question
In the y - direction
mg = N cos θ (eqn 1)
mg = weight of the car.
N = normal reaction of the plane on the car
And in the direction parallel to the inclined plane,
(mv²/r) = N sin θ (eqn 2)
(mv²/r) = force keeping the car in circular motion
Divide (eqn 2) by (eqn 1)
(v²/gr) = Tan θ
v = velocity of car = 60 km/h = 16.667 m/s
g = acceleration due to gravity
r = 150 m
(16.667²/(9.8×150)) = Tan θ
θ = Tan⁻¹ (0.18896)
θ = 10.7° ≈ 11°
b) In the absence of banking, the frictional force on the road has to balance the force keeping the car in circular motion
That is,
Fr = (mv²/r)
Fr = μN = μ mg
μ mg = mv²/r
μ = (v²/gr) = (16.667²/(9.8×150)) = 0.19
Hope this Helps!!!
Answer:
Series equivalent resistance:
Parallel equivalent resistance:
Explanation:
The equivalent resistance in series circuit is calculated as
The equivalent resistance in parallel circuit is calculated as
Which can be simplified to
The parallel circuit has the higher equivalent resistance as compared to series circuit.
Answer:
Horizontal component,
Explanation:
It is given that,
Displacement of the submarine, d = 185 m
It is at an angle of 15 degrees below the horizontal. We need to find the horizontal component of the submarine's displacement. It is given by :
Substituting the values in above equation as :
So, the horizontal component of the submarine’s displacement is 178.69 meters. Hence, this is the required solution.