A motor is connected to the smaller 0.2 m radius pulley and cable, lifting 150 kg mass. Neglecting the mass/weight of the cable and pulley(s), find the torque necessary to accelerate the mass from 0 ms to 10 ms in 10 s (assume linear). Q4. 198.1 [Nm](a) 236.2 [Nm](b) 324.3 [Nm](c) 491.1 [Nm]

324.3Nm

Explanation:

The torque is given by the equation

in this case the vectors r and F are perpendicular between them, thus:

The forces acting on the mass are:

   (1)

where T is the tension of the cable, M is the mass and a is the acceleration.

Furthermore, we have that the acceleration is:

By replacing in (1) we can obtain:

The force T produces the torque on the pulley, hence:

the answer is 324.4Nm

hope this helps!

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz

a person with a mass of 106kg has a weight of 1060n