A person is pushing a fully loaded 24.50 kg wheelbarrow at constant velocity along a plank that makes an angle =42.0∘ with the horizontal. The load is distributed such that the center of mass of the wheelbarrow is exactly halfway along its length . What is the magnitude of the total force the person must apply so that the wheelbarrow is parallel to that plank? You may neglect the radius of the wheel in your analysis. The gravitational acceleration is =9.81 m/s2.

We have that for the Question, it can be said that

From the question we are told

A person is pushing a fully loaded 24.50 kg wheelbarrow at constant velocity along a plank that makes an angle =42.0∘ with the horizontal. The load is distributed such that the center of mass of the wheelbarrow is exactly halfway along its length . What is the magnitude of the total force the person must apply so that the wheelbarrow is parallel to that plank? You may neglect the radius of the wheel in your analysis. The gravitational acceleration is =9.81 m/s2.

Generally the equation for the Force  is mathematically given as

For x axis

F_net=183.75

Therefore

the magnitude of the total force the person must apply so that the wheelbarrow is parallel to that plank is

F_net=183.75

For more information on this visit

link

184 N

Explanation:

Draw a free body diagram of the wheelbarrow.  There are three forces on the wheelbarrow:

Weight force mg pulling straight down, at the center.

Normal force N pushing perpendicular to the plank, at the wheel.

Applied force F at the handle.

Take x direction to be parallel to the plank, and the y direction to be perpendicular to the plank.

Sum of torques about the wheel:

∑τ = Iα

Fᵧ L − (mg cos α) (L/2) = 0

Fᵧ L = (mg cos α) (L/2)

Fᵧ = mg cos α / 2

Fᵧ = (24.50 kg) (9.81 m/s²) (cos 42.0°) / 2

Fᵧ = 89.3 N

Sum of forces in the x direction:

∑F = ma

Fₓ − mg sin α = 0

Fₓ = mg sin α

Fₓ = (24.50 kg) (9.81 m/s²) (sin 42.0°)

Fₓ = 160.8 N

So the magnitude of the force is:

F = √(Fₓ² + Fᵧ²)

F = √((160.8 N)² + (89.3 N)²)

F = 184 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