The maximum velocity in a banked road, ignoring friction, is given by;
v = Sqrt (Rg tan ∅), where R = Radius of the curved road = 2*1000/2 = 1000 m, g = gravitational acceleration = 9.81 m/s^2, ∅ = Angle of bank.
Substituting;
30 m/s = Sqrt (1000*9.81*tan∅)
30^2 = 1000*9.81*tan∅
tan ∅ = (30^2)/(1000*9.81) = 0.0917
∅ = tan^-1(0.0917) = 5.24°
Therefore, the road has been banked at 5.24°.
Answer:
Detailed step wise solution is attached below
Explanation:
(a) wavelength of the initial note 2.34 meters
(b) wavelength of the final note 0.389 meters
(d) pressure amplitude of the final note 0.09 Pa
(e) displacement amplitude of the initial note 4.78*10^(-7) meters
(f) displacement amplitude of the final note 3.95*10^(-8) meters
Three 40w lamp for 6 hours
Answer: 
Explanation:
We are told both planets describe a circular orbit around the star S. So, let's approach this problem begining with the angular velocity
of the planet P1 with a period
:
(1)
Where:
is the velocity of planet P1
is the radius of the orbit of planet P1
Finding
:
(2)
(3)
(4)
On the other hand, we know the gravitational force
between the star S with mass
and the planet P1 with mass
is:
(5)
Where
is the Gravitational Constant and its value is 
In addition, the centripetal force
exerted on the planet is:
(6)
Assuming this system is in equilibrium:
(7)
Substituting (5) and (6) in (7):
(8)
Finding
:
(9)
(10)
Finally:
(11) This is the mass of the star S