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
Answer:
W = 24.28 kN
Explanation:
given,
Mass of satellite = 5850 Kg
height , h = 4.1 x 10⁵ m
Radius of planet = 4.15 x 10⁶ m
Time period = 2 h
= 2 x 3600 = 7200 s
Time period of satellite

R is the radius of planet
h is the height of satellite

now calculation of acceleration due to gravity


g = 4.15 m/s²
True weight of satellite
W = m g
W = 5850 x 4.15
W = 24277.5 N
W = 24.28 kN
True weight of the satellite is W = 24.28 kN
The air drag is a force that depends on the speed of an object relative to the wind. Under certain conditions, it can be modeled as:

Where b is a constant.
As a falling object reaches a speed so that its weight is cancelled out by the air drag, the object will reach a maximum velocity.
In a speed vs time gaph, the speed would approach the maximum speed like an asymptote.
On the other hand, since the object falls from rest, the initial speed on the graph must be zero.
Taking these considerations into account, the correct graph for the movement of an object that falls from rest if air drag is not ignored, is option B.
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