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ki77a [65]
3 years ago
8

The stair stepper is a novel exercise machine that attempts to reproduce the work done against gravity by walking up stairs. Wit

h each step, Brad (of mass 60 kg) simulates stepping up a vertical distance of 0.25 m with this machine. If Brad exercises for 20 min per day with a stair stepper at a frequency of 60 steps per minute, what total work he does he do each day? The acceleration of gravity is 9.8 m/s 2 .
Physics
1 answer:
nika2105 [10]3 years ago
6 0

Answer:

The total work done by Brad each day is 176400 J

Explanation:

Hi there!  The work done by a force (F) pointed in the same direction as the displacement (d) is calculated as follows:

W = F · d

The force applied is equal to the weight of Brad, that is calculated as follows:

Weight = m · g

Where:

m =  mass of Brad

g = acceleration due to gravity (9.8 m/s²)

Then:

Weight = 60 kg · 9.8 m/s² = 588 N

 

Let´s find the vertical distance traveled by Brad each day:

He exercises 20 min per day. Each minute Brad does 60 steps. In total, Brad steps up (20 min · 60 steps/min) 1200 steps. If each step has a height of 0.25 m, the total distance traveled by Brad will be

(1200 steps · 0.25 m/step) 300 m.

Then, the total work done by Brad is

W = F · d

W =  588 N · 300 m

W = 176400 J

The total work done by Brad each day is 176400 J

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A closed, rigid tank fitted with a paddle wheel contains 2 kg of air, initially at 300 K. During an interval of 5 minutes, the p
anzhelika [568]

Answer:

The final temperature of the air is T_2= 605 K

Explanation:

We can start by doing an energy balance for the closed system

\Delta KE+\Delta PE+ \Delta U = Q - W

where

\Delta KE = the change in kinetic energy.

\Delta PE = the change in potential energy.

\Delta U = the total internal energy change in a system.

Q = the heat transferred to the system.

W = the work done by the system.

We know that there are no changes in kinetic or potential energy, so \Delta KE = 0 and \Delta PE=0

and our energy balance equation is \Delta U = Q - W

We also know that the paddle-wheel transfers energy to the air at a rate of 1 kW and the system receives energy by heat transfer at a rate of 0.5 kW, for 5 minutes.

We use this information to calculate the total internal energy change \Delta U=W+Q using the energy balance equation.

We convert the interval of time to seconds t = 5 \:min = 300\:s

\Delta \dot{U}=\dot{W}+ \dot{Q}\\=\Delta U=(W+ Q)\cdot t

\Delta U=(1 \:kW+0.5\:kW)\cdot 300\:s\\\Delta U=450 \:kJ

We can use the change in specific internal energy \Delta U = m(u_2-u_1) to find the final temperature of the air.

We are given that T_1=300 \:K and the air can be describe by ideal gas model, so we can use the ideal gas tables for air to determine the initial specific internal energy u_1

u_1=214.07\:\frac{kJ}{kg}

Next, we will calculate the final specific internal energy u_2

\Delta U = m(u_2-u_1)\\\frac{\Delta U}{m} =u_2-u_1

\frac{\Delta U}{m} =u_2-u_1\\u_2=u_1+\frac{\Delta U}{m}

u_2=214.07 \:\frac{kJ}{kg} +\frac{450 \:kJ}{2 \:kg}\\u_2= 439.07 \:\frac{kJ}{kg}

With the value u_2=439.07 \:\frac{kJ}{kg} and the ideal gas tables for air we make a regression between the values u = 434.78 \:\frac{kJ}{kg},T=600 \:K and u = 442.42 \:\frac{kJ}{kg}, T=610 \:K and we find that the final temperature T_2 is 605 K.

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Answer:

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B. +150KJ.

C. 250KJ.

Explanation:

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Enthalphy change (ΔH) =..?

Enthalphy change = Heat of product – Heat of reactant.

ΔH = Hp – Hr

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