A ski tow operates on a 15.0° slope of length 300 m. The rope moves at 12.0 km/h and provides power for 50 riders at one time, with an average mass per rider of 70.0 kg. Estimate the power required to operate the tow.

A power of 29562.1 W is required to operate the tow.

Explanation:

Given:

Angle of inclination (x) = 15.0°

Length of slope (L) = 300 m

Mass of each rider = 70 kg

Number of riders = 50

Speed of rope (v) = 12 km/h

Now, mass of the 50 riders (m) = Mass of 1 rider × Number of riders

                                                    = 70 × 50 = 3500 kg

The force acting on the tow is the component of weight of 50 riders in the direction opposite to the motion of the tow.

The component of weight along the slope is given as:

Now, the tow must apply the same force in the direction of motion to move it at constant speed.

Now, speed is in km/h. So, we need to convert it to m/s using the conversion factor.

We know that, 1 km/h = 5/18 m/s

Therefore, 12.0 km/h =

Therefore, the minimum power required to operate the tow is given by the formula:

Power = Force × Velocity

Therefore, a power of 29562.1 W is required to operate the tow.

  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