25 Safest Trucks
i need help from a physics genius!! please!!!?
A pick-up truck with two passengers weighs about 14900 N. In good driving conditions around a curve, the maximum friction with the road is equal to the truck’s weight. What is the minimum safe curve radius that the truck could negotiate at 25.4 m/s?
A planet is in a circular orbit around the Sun. Use Newton’s law of gravity and his second law of motion to calculate the period of the planet (in days). Data: mass of Sun = 1.99×1030 kg, mass of planet = 6.00×1024 kg, radius of orbit = 1.496×1011 m.
What is the orbital speed of a satellite of mass 590 kg in a geostationary orbit? Please give your answer in km/s
1)Let M = mass of the truck(plus passengers).
Mg = 14900 N
M = 14900/g = 14900/9.8 = 1520.41 kg
Speed v = 25.4 m/s
Centripetal force F = weight = 14900 N
F = Mv^2/r (where r = radius)
r = Mv^2/F
r = 1520.41 * 25.4^2/14900 = 65.8 m
Ans: 65.8 m
2) Ms = 1.99 * 10^30 kg
Mp = 6.00 * 10^24 kg
r = 1.496 * 10^11 m
Centripetal force F = gravitational force = G*Ms*Mp/r^2
Also F = Mp * v^2/r (where v = orbital speed of the planet)
Therefore Mp * v^2/r = G*Ms*Mp/r^2
Dividing both sides by Mp,
v^2/r = G*Ms/r^2
Multiplying both sides with r,
v^2 = G*Ms/r
v = sqrt(G*Ms/r) = sqrt[6.67 * 10^-11 * 1.99 * 10^30/(1.496 * 10^11)]
v = 29786.79 m/s
Period = time taken to make one revolution = 2 * pi * r/v
= 2 * 3.14 * 1.496 * 10^11/29786.79
= 31540424.46 s
= 31540424.46/3600 hours
= 31540424.46/(24 * 3600) days
= 365.1 days
Ans: 365.1 days
3) Let r = orbital radius,
w = angular speed,
v = orbital speed
For geostationary orbit w = angular speed of rotation of Earth = 2 * pi/(rotational period of Earth)
= 2 * pi/(24 * 3600) = 7.27 * 10^-5 rad/s
Centripetal acceleration = acceleration due to gravity
w^2 * r = GM/r^2 (where M = Earth’s mass)
r^3 = GM/w^2
w^2 * r^3 = GM
w^3 * r^3 = GMw
v^3 = GMw (because v = w*r)
v = (GMw)^(1/3)
= (6.67 * 10^-11 * 6 * 10^24 * 7.27*10^-5)^(1/3)
= 3076 m/s
= 3.076 km/s
Ans: 3.076 km/s
25 Safest Trucks
Fourth Of July Fireworks Displays
Going to a professional fireworks show is truly the safest way to celebrate the 4th of July. Here is a list of some show-stopping locations. If we’re missing your favorite show, please let us know in the comments below!