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:
M g H = 1/2 M v^2 potential energy = kinetic energy
v^2 = 2 g H = 2 * 9.80 * 6 = 117.6 m/s^2
v = 10.8 m/s
(C)
When Sam presses the brake lever, a pair of rubber shoes clamps onto the metal inner rim of the front and back wheels. As the brake shoes rub against the wheels, friction is caused and the kinetic energy possessed by the vehicle is converted into heat which slows down the vehicle.
Answer:
on the first shell (ring) there will be 2 electrons
and on the 2nd shell there will be only one electron
while in the nucleus (the middle of the diagram) there will be 4 neutrons and 3 protons
Explanation:
you can see the picture attached
