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
The field is
<em><u>E</u></em> = 1 / (4 pi ε₀) Q / <em><u>R</u></em>² directed radially outward from
the center of the shell.
R is the radius of the spherical shell.
Notice that the field is exactly the same as the field due to a point-charge
with magnitude 'Q' that's located at the center of the sphere.
Sometimes, you can play sport for a longer time than you should. You should always stop and take a little break and always drink water. Also, if something starts to hurt, stop, let someone responsible know and take a break.
The answer is A. conduction
Answer:
d = 329.81m
Explanation:
V_f = V_0+a*t
V_f = Velocity final
V_0 = Velocity initial
a = acceleration
t = time
V_f = (0m/s)+(9.81m/s²)*(8.2s)
V_f = 80.442m/s
d = ((V_f-V_0)/2)*t
d = distance
d = ((80.442m/s-0m/s)/2)*(8.2s)
d = 329.81m
