Answer:
L = 1.545 m
Explanation:
Let the total length of the rod is L
now the torque must applied on the other end of the rod so that it will balance the torque due to weight of rock on other side of fulcrum
so we will have
so we have
F = 663 N
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
Answer
Maximum speed at 75 m radius will be 22.625 m /sec
Explanation:
We have given radius of the curve r = 150 m
Maximum speed
Coefficient of friction
Now new radius r = 75 m
So maximum speed at new radius
Explanation:
It is given that,
Uncertainty in the speed of an electron,
According to Heisenberg uncertainty principle,
is the uncertainty in the position of an electron
Since,
So, the uncertainty in its position is . Hence, this is the required solution.