Ask question

Ask question

All categories

3 years ago

8

A world class runner can run long distances at a pace of 15 km/hour. That runner expends 800 kilocalories of energy per hour. a)

Find the power (in Watts [W]) the runner is exerting while running. b) Find the total energy (in Joules [J]) exerted by the runner in a 15 km run. c) How many Milky Way (Original Single 52.2g) chocolate bars does the runner need to buy to supply the amount of energy to complete a half-marathon (13.1 miles)? (Cite your source for the number of calories in a Milky Way bar)

1 answer:

maks197457 [2]3 years ago

7 0

Answer: a) 1.05kW b) 3.78MJ c) 5.3 bars

Explanation :

A)

Conversions give 900 kcal as 900000 x 4.2 J/cal {4.2 J/cal is the standard factor}

= 3780kJ

And 1 hour = 3600s

Therefore, Power in watts = 3780/3600 = 1.05kW = 1050W

B)

At 15km/hour a 15km run takes 1 hour.

1 hour is 3600s and the runner burns 1050 joule per second.

Energy used in 1 hour = 3600 x 1050 J/s

= 3780000 J or 3.78MJ

C)

1 mile = 1.61km so 13.1 mile is 13.1 x 1.61 = 21.1km

15km needs 3.78 MJ of energy therefore 21.1km needs 3.78 x 21.1/15 = 5.32MJ =5320 kJ

Finally,

1 Milky Way = 240000 calories = 4.2 x 240000 J = 1008000J or 1008kJ

This means that the runner needs 5320/1008 = 5.3 bars

You might be interested in

A 150-lbm astronaut took his bathroom scale (aspring scale) and a beam scale (compares masses) to themoon where the local gravit

Nonamiya [84]

Answer:

a) W = 25.5 lbf

b) W = 150 lbf

Explanation:

Given data:

Mass of astronaut = 150 lbm

local gravity = 5.48 ft/s^2

a) weight on spring scale

it can be calculated by measuring force against local gravitational force which is equal to weight of body

W = mg

$W = (150 \times 5.48)\times \frac{1 lbm}{32.32 lbm. ft/s^2} = 25.5 lbf$

b) As we know that beam scale calculated mass only therefore no change in mass due to variation in gravity

thus W= 150 lbf

7 0

3 years ago

A silicon carbide plate fractured in bending when a blunt load was applied to the plate center. The distance between the fractur

Amanda [17]

Question in order:

A silicon carbide plate fractures in bending when a blunt load was applied to the plate center. The distance between the fracture origin and the mirror/mist boundary on the fracture surface was 0.796 mm. To determine the stress used to break the plate, three samples of the same material were tested and produced the following. What is the estimate of the stress present at the time of fracture for the original plate?

Mirror Radius (mm) Bending Failure Stress (MPa)

0.603 225

0.203 368

0.162 442

Answer:

191 MPa

Explanation:

Failure stress of bending is Inversely proportional to the mirror radius

$Bending Stress = \frac{1}{(Mirror Radius)^{n}}$

At mirror radius 1 = 0.603 mm Bending stress 1 = 225 Mpa..............(1)

At mirror radius 2 = 0.203 mm Bending stress 2 = 368 Mpa...............(2)

At mirror radius 3 = 0.162 mm Bending stress 3 = 442 Mpa...............(3)

comparing case 1 and 2 using the above equation

$\frac{Stress 1}{Stress 2} = ({\frac{Radius 2}{Radius 1}})^{n_1}$

$\frac{225}{368} = ({\frac{0.203}{0.603}})^{n_1}$

$0.6114 = (0.3366)^{n_1}$

Taking the natural logarithm of both side

ln(0.6114) = n ln(0.3366)

n₁ = ln(0.6114)/ln(0.3366)

n₁ = 0.452

comparing case 2 and 3 using the above equation

$\frac{Stress 2}{Stress 3} = ({\frac{Radius 3}{Radius 2}})^{n_2}$

$\frac{368}{442} = ({\frac{0.162}{0.203}})^{n_2}$

$0.8326 = (0.7980)^{n_2}$

Taking the natural logarithm of both side

ln(0.8326) = n ln(0.7980)

n₂ = ln(0.8326)/ln(0.7980)

n₂ = 0.821

comparing case 1 and 3 using the above equation

$\frac{Stress 1}{Stress 3} = ({\frac{Radius 3}{Radius 1}})^{n_3}$

$\frac{225}{442} = ({\frac{0.162}{0.603}})^{n_3}$

$0.5090 = (0.2687)^{n_3}$

Taking the natural logarithm of both side

ln(0.5090) = n ln(0.2687)

n₃ = ln(0.5090)/ln(0.2687)

n₃ = 0.514

average for n

$n = \frac{n_1 + n_2 + n_3}{3}$

$n = \frac{0.452 +0.821 + 0.514}{3}$

n = 0.596

Hence to get bending stress x at mirror radius 0.796

$\frac{Stress x}{Stress 3} = ({\frac{Radius 3}{Radius x}})^{0.596}$

$\frac{Stress x}{225} = ({\frac{0.603}{0.796}})^{0.596}$

$\frac{Stress x}{225} = 0.8475$

stress x = 191 MPa

3 0

3 years ago

A gas mixture containing 3 moles CO2, 5 moles H2 and 1 mole water is undergoing the following reactions CO2+3H2 →cH3OH + H2O Dev

Ilya [14]

Aaaaaaaaaaaaaaaaaaaaaa

7 0

2 years ago

An engineer measures a sample of 1200 shafts out of a certain shipment. He finds the shafts have an average diameter of 2.45 inc

Vadim26 [7]

Answer: 78.89%

Explanation:

Given : Sample size : n= 1200

Sample mean : $\overline{x}=2.45$

Standard deviation : $\sigma=0.07$

We assume that it follows Gaussian distribution (Normal distribution).

Let x be a random variable that represents the shaft diameter.

Using formula, $z=\dfrac{x-\mu}{\sigma}$ , the z-value corresponds to 2.39 will be :-

$z=\dfrac{2.39-2.45}{0.07}\approx-0.86$

z-value corresponds to 2.60 will be :-

$z=\dfrac{2.60-2.45}{0.07}\approx2.14$

Using the standard normal table for z, we have

P-value = $P(-0.86$

$=P(z$

Hence, the percentage of the diameter of the total shipment of shafts will fall between 2.39 inch and 2.60 inch = 78.89%

7 0

3 years ago

Unit for trigonometric functions is always "radian". 1. 10 points: Do NOT submit your MATLAB code for this problem (a) Given f(x

RoseWind [281]

Answer:

Below is the required code.

Explanation:

%% Newton Raphson Method

clear all;

clc;

x0=input('Initial guess:\n');

x=x0;

f=exp(-x)-sin(x)-0.2;

g=-exp(-x)-cos(x);

ep=10;

i=0;

cc=input('Condition of convergence:\n');

while ep>=cc

i=i+1;

temp=x;

x=x-(f/g);

f=exp(-x)-sin(x)-0.2;

g=-exp(-x)-cos(x);

ep=abs(x-temp);

fprintf('x = %6f and error = %6f at iteration = %2f \n',x,ep,i);

end

fprintf('The solution x = %6f \n',x);

%% End of MATLAB Program

Command Window:

(a) First Root:

Initial guess:

1.5

Condition of convergence:

0.01

x = -1.815662 and error = 3.315662 at iteration = 1.000000

x = -0.644115 and error = 1.171547 at iteration = 2.000000

x = 0.208270 and error = 0.852385 at iteration = 3.000000

x = 0.434602 and error = 0.226332 at iteration = 4.000000

x = 0.451631 and error = 0.017029 at iteration = 5.000000

x = 0.451732 and error = 0.000101 at iteration = 6.000000

The solution x = 0.451732

>>

Second Root:

Initial guess:

3.5

Condition of convergence:

0.01

x = 3.300299 and error = 0.199701 at iteration = 1.000000

x = 3.305650 and error = 0.005351 at iteration = 2.000000

The solution x = 3.305650

>>

(b) Guess x=0.5:

Initial guess:

0.5

Condition of convergence:

0.01

x = 0.450883 and error = 0.049117 at iteration = 1.000000

x = 0.451732 and error = 0.000849 at iteration = 2.000000

The solution x = 0.451732

>>

Guess x=1.75:

Initial guess:

1.75

Condition of convergence:

0.01

x = 227.641471 and error = 225.891471 at iteration = 1.000000

x = 218.000998 and error = 9.640473 at iteration = 2.000000

x = 215.771507 and error = 2.229491 at iteration = 3.000000

x = 217.692636 and error = 1.921130 at iteration = 4.000000

x = 216.703197 and error = 0.989439 at iteration = 5.000000

x = 216.970438 and error = 0.267241 at iteration = 6.000000

x = 216.971251 and error = 0.000813 at iteration = 7.000000

The solution x = 216.971251

>>

Guess x=3.0:

Initial guess:

3

Condition of convergence:

0.01

x = 3.309861 and error = 0.309861 at iteration = 1.000000

x = 3.305651 and error = 0.004210 at iteration = 2.000000

The solution x = 3.305651

>>

Guess x=4.7:

Initial guess:

4.7

Condition of convergence:

0.01

x = -1.916100 and error = 1.051861 at iteration = 240.000000

x = -0.748896 and error = 1.167204 at iteration = 241.000000

x = 0.162730 and error = 0.911626 at iteration = 242.000000

x = 0.428332 and error = 0.265602 at iteration = 243.000000

x = 0.451545 and error = 0.023212 at iteration = 244.000000

x = 0.451732 and error = 0.000187 at iteration = 245.000000

The solution x = 0.451732

>>

Explanation:

The two solutions are x =0.451732 and 3.305651 within the range 0 < x< 5.

The initial guess x = 1.75 fails to determine the solution as it's not in the range. So the solution turns to unstable with initial guess x = 1.75.

7 0

3 years ago