A20 kg wagon is pulled along the level ground by a rope inclined at 30 degree above the horizontal. a friction force of 30 n opposes the motion. how large is the pulling force if the wagon is moving with (a) constant speed and (b) and acceleration of 0.40 m/s2

(a) 34.6 N

To solve the problem, we have to analyze the forces acting along the horizontal direction.

We have:

- Forward: the component of the pull parallel to the ground, which is

where

F is the magnitude of the pull

is the angle

- Backward: the force of friction, which is

So, the equation of motion is

where

m = 20 kg is the mass of the wagon

a is the acceleration

In this part, the wagon is moving at constant speed, so a =0 and the equation becomes

Therefore, we can find the pulling force:

(b) 43.9 N

In this case, the acceleration is

So, the equation of motion in this case is

So this time we have to take into account the term (ma).

Using the  same data as before:

m = 20 kg

We find the new magnitude of F:

found in the nucleus

has mass of one amu

  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