Answer:
A. Mrŵ² = ųMg
Ŵ = (ųg/r)^½
B.
Ŵ =[ (g /r)* tan á]^½
Explanation:
T.v.= centrepetal force = mrŵ²
Where m = mass of block,
r = radius
Ŵ = angular momentum
On a horizontal axial banking frictional force supplies the Pentecostal force is numerically equal.
So there for
Mrŵ² = ųMg
Ŵ = (ųg/r)^½
g = Gravitational pull
ų = coefficient of friction.
B. The net external force equals the horizontal centerepital force if the angle à is ideal for the speed and radius then friction becomes negligible
So therefore
N *(sin á) = mrŵ² .....equ 1
Since the car does not slide the net vertical forces must be equal and opposite so therefore
N*(cos á) = mg.....equ 2
Where N is the reaction force of the car on the surface.
Equ 2 becomes N = mg/cos á
Substituting N into equation 1
mg*(sin á /cos á) =mrŵ²
Tan á = rŵ²/g
Ŵ =[ (g /r)* tan á]^½
Option (B)is correct.
Alternating electric and magnetic fields has electromagnetic radiation that travel in the form of a wave.
The varying or alternating electric field produces the varying magnetic field.This varying magnetic field in turn produces varying electric field . Thus these varying electric and magnetic fields travel in the form of electromagnetic waves. These electromagnetic waves travel with the velocity of light.These are transverse in nature.
Answer:
v=0.04m/s
Explanation:
To solve this problem we have to take into account the expression
where v and r are the magnitudes of the velocity and position vectors.
By calculating the magnitude of r and replacing w=0.02rad/s in the formula we have that
the maximum relative velocity is 0.04m/s
hope this helps!!
Answer:
The velocity of block = 0.188 
Explanation:
Mass m = 5.6 kg
k = 1040 
= 0.26
0.035 m , 
= 0
0.02 m
From work energy theorem
--------- (1)
Kinetic energy
------- (1)
Potential energy
------- (2)
Work done
W = F.s ------ (3)
From Newton's second law
= mg
= 5.6 × 9.81 = 54.9 N
Friction force = 0.4 × 54.9 = 21.9 N
Now the work done by the friction
= - 21.9 × 0.015
= - 0.329 J
Now kinetic energy
At point 1
0.637 J
At point 2
Potential energy
J
From equation (1) we get
0 + 0.637 - 0.329 = 2.8 
+ 0.208
2.8 = 0.1
0.188
This is the velocity of block.
Answer:
Explanation:
If v be the velocity just after the rebound
Kinetic energy will be converted into potential energy
1/2 m v² = mgh
v² = 2gh
v = √ 2gh
= √ 2 x 9.8 x .96
= 4.33 m / s
