Answer:
Layer 1, Rock 2, Rock 1, Fault
Answer:
non linear square relationship
Explanation:
formula for centripetal force is given as
a = mv^2/r
here a ic centripetal acceleration , m is mass of body moving in circle of radius r and v is velocity of body . If m ,and r are constant we have
a = constant × v^2
a α v^2
hence non linear square relationship
Answer:
θ=19.877⁰
Explanation:
Given data
Velocity Va=34.0 km/h
Velocity Va=100 km/h
To find
Angle θ
Solution
We want the bird to fly with velocity Vb=100 km/h with an angle θ relative to the ground so that the bird fly due south relative to the ground.From figure which is attached we got
Sinθ=(Va/Vb)
Sinθ=(34.0/100)
θ=Sin⁻¹(34.0/100)
θ=19.877⁰
Answer:
0.52 Nm
Explanation:
A = 0.12 m^2, N = 200, i = 0.5 A, B = 0.050 T
Angle between the plane of loop and magnetic field = 30 Degree
Angle between the normal of loop and the magnetic field = 90 - 30 = 60 degree
θ = 60°
Torque = N i A B Sinθ
Torque = 200 x 0.5 x 0.12 x 0.050 x Sin 60
Torque = 0.52 Nm
Answer:
The value of the spring constant of this spring is 1000 N/m
Explanation:
Given;
equilibrium length of the spring, L = 10.0 cm
new length of the spring, L₀ = 14 cm
applied force on the spring, F = 40 N
extension of the spring due to applied force, e = L₀ - L = 14 cm - 10 cm = 4 cm
From Hook's law
Force applied to a spring is directly proportional to the extension produced, provided the elastic limit is not exceeded.
F ∝ e
F = ke
where;
k is the spring constant
k = F / e
k = 40 / 0.04
k = 1000 N/m
Therefore, the value of the spring constant of this spring is 1000 N/m
