When a 100 metric ton flying saucer reaches an altitude of 75 km over Earth\'s s
ID: 1631381 • Letter: W
Question
When a 100 metric ton flying saucer reaches an altitude of 75 km over Earth's surface. At that moment the saucer has a velocity of 9000 km/hr. If the friction force is 500,000 N, how much force was needed to propel the saucer upward? At a ski resort a chair lift takes skiers up a 30 degree hillside to an elevation of 180 m as shown in figure 1. How much horsepower must the engine have to pull 50 skiers at a time, at a speed of 4.0 m/s, if the average mass of each skier and their equipment is 82 kg each and the frictional energy loss is 3.6 MJ (Mega-Joules)? [120.360 W = 163.6 hp]. Starting from rest a 1500 kg car moves down a level street under the influence of two forces: a 1000 N forward force exerted on the drive wheels and a 800 N resistive force. a. What is the speed of the car after it has traveled 40 m? [3.266 m/s] b. At the 40 m point the engine switches off and the car coasts up an incline of 7.0 degree. If the resistive force (not the energy) on the incline is 20% less than on the level street, how far along the incline will the car travel? [3.29 m]Explanation / Answer
From the given question,
mass(m)=100000 kg
height(h)=75km=75000 m
velocity(v)=9000 km/hr= 2500 m/s
Frictional force(F)=500,000 N
Driving force * displacement= change in kinetic energy + work done against friction
Fd*h= (1/2)mv2+ Ff*h + mgh
Fd * 75000= (1/2)(100000)(25002) + (500000)(75000) + 100000*9.8*75000
Fd=5.647 x 106 N
Driving force is 5.647 x 106 N
4.mass(m)=1500 kg
net force(F)=Driving force- resistive force
=1000-800
=200N
acceleration= F/m=200/1500
=2/15
initial velocity(u)=0
v2=u2+2as
v2=02+2(2/15)(40)
v=3.266 m/s
On the incline plane, resistive force Fr= 0.8*800=640 N
a= -(gsin7+640/1500)
v2=u2+ 2as
0=(3.266)2 - 2(9.8*sin7+(640/1500))(s)
s=3.29 m
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