physics.. please help. 
?
i have physics summer hw and i really don’t understand this..
1. how large must the coefficient of friction be between the tires and the road if a car is to round a level curve of radius 85m at a speed of 95km/h?
2. a 1000-kg sports car moving at 20 m/s crosses the rounded top of a hill (radius = 100m). Determine (a) the normal force on the car, (b) the normal force on the 70 kg driver, (c) the car speed at which the normal force equals zero.
3. a pilot performs an evasive maneuver by diving vertically at 310 m/s. if he can withstand an acceleration of 9.0 g’s without blacking out, at what altitude must he begin to pull out of the dive to avoid crashing into the sea?
4. calculate the force of gravity on a spacecraft 12,800 km (2 earth radii) above the Earth’s surface if its mass is 1400 kg?
5. a hypothetical planet has a radius 2.5 times that of Earth, but has the same mass. What is the acceleration due to gravity near its surface?
thank you. please reply soon.
All the below answers are taken from the links …please go through the links for better symbols…used in each problem…i copied some of them are not copied properly so you just look into that
1.
First, convert the annoying km/hr to m/s:
95km/h/3.6 = 26.38888889 m/s
The force of friction is making the car go around the corner, so we will set the formula for friction:
Ffr = µFN = µmg
Equal to the centripetal force:
F = m4π2r/T2
Finally:
µmg = mv2/r
So, let’s solve for µ:
µmg = mv2/r
µg = v2/r
µ = v2/rg
µ = (26.38888889 m/s)2/(85 m)(9.80 m/s/s) = 0.836 = .84 or larger
2.
This is a well disguised motion in a vertical circle problem. The centripetal acceleration of the car is given by v2/r :
a = v2/r = (20. m/s)2/(100. m) = 4 m/s/s. At the crest of the hill, the centripetal acceleration would be downward because the center of the circle would be below the car, and centripetal acceleration is toward the center.
The car has a weight of (1000 kg)(9.80 N/kg) = 9800 N, and so the expression of Newton’s second law at the top of the hill is shaping up. We have the weight down (-), the acceleration down (-) and the normal force of the road FN (up???) so F = ma looks like:
The driver has a weight of (70 kg)(9.80 N/kg) = 686 N, and so we have the weight down (-), the acceleration down (-) and the normal force of the car FN (up???) so F = ma looks like:
The normal force will be zero (and the car will take some air!!!) when its centripetal acceleration is equal to 9.8 m/s/s
a = v2/r
ar = v2
Ö(ar) = v
So v = Ö(ar) = Ö[(9.80 m/s/s)(100. m)] = 31.3050 = 31.3 m/s
3.
The centripetal acceleration is 9.0 “g”s which is (9.0)(9.8 m/s/s) = 88.2 m/s/s. Basically, the question is what radius circle has a centripetal acceleration of 88.2 m/s/s when you are flying at 310 m/s:
a = v2/r
r = v2/a = (310 m/s)2/(88.2 m/s/s) = 1089.5692 m = 1100 m = 1.1 km
4.
SOLUTION:
The gravitational force on the spacecraft is
F= GMm/r2
= (6.6710–11 N • m2/kg2)(5.981024 kg)(1400 kg)/[3(6.38106 m)]2 = 1.52103 N.
5.
SOLUTION:
The acceleration due to gravity on the surface of a planet is
g = F/m = GM/R2 = (6.67 x 10-11)(5.98 x 1024 kg)/[2.5(6.38 x 106 m)]2 = 1.6 m/s2
I
The acceleration due to gravity on the surface of a planet is
g = F/m = GM/R2 = (6.67 x 10-11)(2.5 x 5.98 x 1024 kg)/[(6.38 x 106 m)]2 = 24.5 m/s2
please check the following links for better understanding
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