Aboy pulls his toy on a smooth horizontal surface with a rope inclined at 60 degrees to the horizontal. if the effective force pulling the toy along the horizontal surface is 5n, calculate the tension in the rope

As per the question the rope is inclined at an angle of 60 degree with the horizontal.

Let the tension produced in the rope is T.

The effective force which pulls the toy along horizontal direction is 5 N.

Resolving T into two components we get-

 in vertical upward direction.

in horizontal direction.

The vertical component is balanced by the weight of the body i.e

Hence the toy moves due to the horizontal component.

                                      [   ]

Hence the tension produced in the rope is 10 N.

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂

490 n

f=125n a=2.5m/s2