Explanation:
It is given that,
Speed of the sports car, v = 85 mph = 37.99 m/s
The radius of curvature, r = 525 m
Let 
is the normal weight and 
is the apparent weight of the person. Its apparent weight is given by :
So, 
or
Hence, this is the required solution.
Answer:
I can't see the picture
Explanation:
Im sorry can you right whats on the picture down in the comments plz.
Answer:
laws of motion relate an object’s motion to the forces acting on it. In the first law, an object will not change its motion unless a force acts on it. In the second law, the force on an object is equal to its mass times its acceleration. In the third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.
Answer:
<h2>135,000 J</h2>
Explanation:
The work done by an object can be found by using the formula
workdone = force × distance
From the question we have
workdone = 900 × 150
We have the final answer as
<h3>135,000 J</h3>
Hope this helps you
Answer:
Tension= 21,900N
Components of Normal force
Fnx= 17900N
Fny= 22700N
FN= 28900N
Explanation:
Tension in the cable is calculated by:
Etorque= -FBcostheta(1/2L)+FT(3/4L)-FWcostheta(L)= I&=0 static equilibrium
FTorque(3/4L)= FBcostheta(1/2L)+ FWcostheta(L)
Ftorque=(Fcostheta(1/2L)+FWcosL)/(3/4L)
Ftorque= 2/3FBcostheta+ 4/3FWcostheta
Ftorque=2/3(1350)(9.81)cos55° + 2/3(2250)(9.81)cos 55°
Ftorque= 21900N
b) components of Normal force
Efx=FNx-FTcos(90-theta)=0 static equilibrium
Fnx=21900cos(90-55)=17900N
Fy=FNy+ FTsin(90-theta)-FB-FW=0
FNy= -FTsin(90-55)+FB+FW
FNy= -21900sin(35)+(1350+2250)×9.81=22700N
The Normal force
FN=sqrt(17900^2+22700^2)
FN= 28.900N
