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AveGali [126]
4 years ago
5

It takes 120 minutes for a man to ride his bicycle up the road to Alpe d'Huez in France. The vertical height of the climb is 1,1

20 m, and the combined mass of the rider and bicycle is 96 kg. What is the bicyclist's average power output, in W?
Physics
1 answer:
Greeley [361]4 years ago
8 0
First solve the potential energy of the biker. using the fomula:
PE = mgh
where m  is the mass of the object
g is the acceleration due to gravity ( 9.81 m/s2)
h is the height

PE = 96 kg ( 1120 m ) ( 9.81 m/s2)
PE = 1054771.2 J
then power = Work / time
P = 1054771.2 J / ( 120 min ) ( 60 s / 1 min)
P = 146.5 W

You might be interested in
while doing a lab , a student found the density of a piece of pure aluminum to be 2.80 g/cm . The accepted value for the density
postnew [5]

Look at the attachment or go online and search the question to get the answers you need


4 0
3 years ago
One end of a thin rod is attached to a pivot, about which it can rotate without friction. Air resistance is absent. The rod has
Mars2501 [29]

Answer:

6.86 m/s

Explanation:

This problem can be solved by doing the total energy balance, i.e:

initial (KE + PE)  = final (KE + PE). { KE = Kinetic Energy and PE = Potential Energy}

Since the rod comes to a halt at the topmost position, the KE final is 0. Therefore, all the KE initial is changed to PE, i.e, ΔKE = ΔPE.

Now, at the initial position (the rod hanging vertically down), the bottom-most end is given a velocity of v0. The initial angular velocity(ω) of the rod is given by ω = v/r , where v is the velocity of a particle on the rod and r is the distance of this particle from the axis.

Now, taking v = v0 and r = length of the rod(L), we get ω = v0/ 0.8 rad/s

The rotational KE of the rod is given by KE = 0.5Iω², where I is the moment of inertia of the rod about the axis of rotation and this is given by I = 1/3mL², where L is the length of the rod. Therefore, KE = 1/2ω²1/3mL² = 1/6ω²mL². Also, ω = v0/L, hence KE = 1/6m(v0)²

This KE is equal to the change in PE of the rod. Since the rod is uniform, the center of mass of the rod is at its center and is therefore at a distane of L/2 from the axis of rotation in the downward direction and at the final position, it is at a distance of L/2 in the upward direction. Hence ΔPE = mgL/2 + mgL/2 = mgL. (g = 9.8 m/s²)

Now, 1/6m(v0)² = mgL ⇒ v0 = \sqrt{6gL}

Hence, v0 = 6.86 m/s

4 0
3 years ago
A toy helicopter takes off and moves 6 m up and then 1 m back down. What
klio [65]

the answer is 5 m up AKA "B"

6 0
4 years ago
The coefficient of kinetic friction for a 22 kg bobsled on a track is 0.10. What force is required to push it down a 5.0 degree
frosja888 [35]

Hi there!

We must begin by converting km/h to m/s using dimensional analysis:

\frac{62km}{1hr} * \frac{1hr}{3600sec}*\frac{1000m}{1km} = 17.22 m/s

Now, we can use the kinematic equation below to find the required acceleration:

vf² = vi² + 2ad

We can assume the object starts from rest, so:

vf² = 2ad

(17.22)²/(2 · 75) = a

a = 1.978 m/s²

Now, we can begin looking at forces.

For an object moving down a ramp experiencing friction and an applied force, we have the forces:

Fκ  = μMgcosθ  = Force due to kinetic friction

Mgsinθ = Force due to gravity

A = Applied Force

We can write out the summation. Let down the incline be positive.

ΣF = A + Mgsinθ - μMgcosθ

Or:

ma = A + Mgsinθ - μMgcosθ

We can plug in the given values:

22(1.978) = A + 22(9.8sin(5)) - 0.10(22 · 9.8cos(5))

A = 46.203 N

6 0
3 years ago
Percent Yield Lab Report
Vlada [557]

Based on the data obtained from the reaction, the following conclusions can be made;

  • according to the law of conservation of mass, the mass of the product was higher than the reactant because of the mass of oxygen added during the combustion.
  • the percent yield is less than 100% because of loss in mass of either reactants or products.
  • the errors could have occurred during the weighing and transfer of reactants and products
  • repeated measurements are required in order to improve accuracy

<h3>What is the percent yield of the reaction?</h3>

Equation of the reaction is given below:

  • 2 Mg + O₂ ----> 2 MgO

Trial 1

Mass of empty crucible with lid = 26.698 (g)

Mass of Mg metal, crucible, and lid  = 27.040 (g)

Mass of MgO, crucible, and lid = 27.198 (g)

Mass of metal = 27.040 - 26.698 = 0.342

Mass of MgO = 27.198 - 26.698 = 0.500

<h3>Moles of Mg used</h3>

moles of Mg = mass/molar mass

  • molar mass of Mg = 24 g

moles of Mg = 0.342/24 = 0.01425

<h3>Moles of MgO expected</h3>

Based on the equation of reaction;

moles of MgO expected = 0.01425

moles of MgO produced =  mass/molar mass

  • molar mass of MgO = 40 g/mol

moles of MgO produced = 0.500/40 = 0.0125

<h3>Percent yield</h3>
  • Percent yield = (moles of MgO produced/moles of MgO expected) * 100%

Percentage yield = 0.0125/0.01425 * 100%

Percent yield of MgO = 87.7%

Trial 2

Mass of empty crucible with lid = 26.691 (g)

Mass of Mg metal, crucible, and lid = 27.099 (g)

Mass of MgO, crucible, and lid = 27.361 (g)

Mass of metal = 27.099 - 26.691 = 0.408

Mass of MgO = 27.361 - 26.691 = 0.670

<h3>Moles of Mg used</h3>

moles of Mg = mass/molar mass

  • molar mass of Mg = 24 g

moles of Mg = 0.408/24 = 0.0170

<h3>Moles of MgO expected</h3>

Based on the equation of reaction;

moles of MgO expected = 0.0170

  • moles of MgO produced =  mass/molar mass

molar mass of MgO = 40 g/mol

moles of MgO produced = 0.670/40 = 0.01675

<h3>Percent yield</h3>
  • Percent yield = (moles of MgO produced/moles of MgO expected) * 100%

Percent yield = 0.01675/0.0170 * 100%

Percent yield of MgO = 98.5%

Average percent yield = (87.7 + 98.5)% / 2

Average percent yield = 89.0%

Based on the data obtained from the reaction, the following conclusion can be made;

  • according to the law of conservation of mass, the mass of the product was higher than the reactant because of the mass of oxygen added during the combustion.
  • the percent yield is less than 100% because of loss in mass of either reactants or products.
  • the errors could have occurred during the weighing and transfer of reactants and products
  • repeated measurements are required in order to improve accuracy

Learn more about law of conservation of mass at: brainly.com/question/1824546

6 0
3 years ago
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