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3 years ago

Explain 3 ways that people in sports use engineering to increase their performance?

1 answer:

7 0

Designing systems for manufacturing, motion analysis or impact testing;
building and testing prototypes;
analyzing the human body to prevent injury;
developing or designing new light weight materials that will be more comfortable and withstand greater impacts or forces;

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Vector A extends from the origin to a point having polar coordinates (7, 70ᵒ ) and vector B extends from the origin to a point h

yaroslaw [1]

Answer:

13.95

Explanation:

Given :

Vector A polar coordinates = ( 7, 70° )

Vector B polar coordinates = ( 4, 130° )

To find A . B we will

A ( r , ∅ ) = ( 7, 70 )

A = rcos∅ + rsin∅

therefore ; A = 2.394i + 6.57j

B ( r , ∅ ) = ( 4, 130° )

B = rcos∅ + rsin∅

therefore ; B = -2.57i + 3.06j

Hence ; A .B

( 2.394 i + 6.57j ) . ( -2.57 + 3.06j ) = 13.95

8 0

3 years ago

A horizontal curve on a two-lane road is designed with a 2,300-ft radius, 12-ft lanes, and a 65-mph design speed. Determine the

Ierofanga [76]

Answer:

distance = 22.57 ft

superelevation rate = 2%

Explanation:

given data

radius = 2,300-ft

lanes width = 12-ft

no of lane = 2

design speed = 65-mph

solution

we get here sufficient sight distance SSD that is express as

SSD = 1.47 ut + $\frac{u^2}{30(\frac{a}{g}\pm G)}$ ..............1

here u is speed and t is reaction time i.e 2.5 second and a is here deceleration rate i.e 11.2 ft/s² and g is gravitational force i.e 32.2 ft/s² and G is gradient i.e 0 here

so put here value and we get

SSD = 1.47 × 65 ×2.5 + $\frac{65^2}{30(\frac{11.2}{32.2}\pm 0)}$

solve it we get

SSD = 644 ft

so here minimum distance clear from the inside edge of the inside lane is

Ms = Rv ( 1 - $cos (\frac{28.65 SSD}{Rv})$ ) .....................2

here Rv is = R - one lane width

Rv = 2300 - 6 = 2294 ft

put value in equation 2 we get

Ms = 2294 ( 1 - $cos (\frac{28.65 \times 664}{2294})$ )

solve it we get

Ms = 22.57 ft

and

superelevation rate for the curve will be here as

R = $\frac{u^2}{15(e+f)}$ ..................3

here f is coefficient of friction that is 0.10

put here value and we get e

2300 = $\frac{65^2}{15(e+0.10)}$

solve it we get

e = 2%

3 0

3 years ago

Which of the following ranges depicts the 2% tolerance range to the full 9 digits provided?

Lyrx [107]

Answer:

the only one that meets the requirements is option C .

Explanation:

The tolerance of a quantity is the maximum limit of variation allowed for that quantity.

To find it we must have the value of the magnitude, its closest value is the average value, this value can be given or if it is not known it is calculated with the formula

x_average = ∑ $x_{i}$ / n

The tolerance or error is the current value over the mean value per 100

Δx₁ = x₁ / x_average

tolerance = | 100 -Δx₁ 100 |

bars indicate absolute value

let's look for these values for each case

x_average = (2.1700000+ 2.258571429) / 2

x_average = 2.2142857145

fluctuation for x₁

Δx₁ = 2.17000 / 2.2142857145

Tolerance = 100 - 97.999999991

Tolerance = 2.000000001%

fluctuation x₂

Δx₂ = 2.258571429 / 2.2142857145

Δx2 = 1.02

tolerance = 100 - 102.000000009

tolerance 2.000000001%

x_average = (2.2 + 2.29) / 2

x_average = 2,245

fluctuation x₁

Δx₁ = 2.2 / 2.245

Δx₁ = 0.9799554

tolerance = 100 - 97,999

Tolerance = 2.00446%

fluctuation x₂

Δx₂ = 2.29 / 2.245

Δx₂ = 1.0200445

Tolerance = 2.00445%

x_average = (2.211445 +2.3) / 2

x_average = 2.2557225

Δx₁ = 2.211445 / 2.2557225 = 0.9803710

tolerance = 100 - 98.0371

tolerance = 1.96%

Δx₂ = 2.3 / 2.2557225 = 1.024624

tolerance = 100 -101.962896

tolerance = 1.96%

x_average = (2.20144927 + 2.29130435) / 2

x_average = 2.24637681

Δx₁ = 2.20144927 / 2.24637681 = 0.98000043

tolerance = 100 - 98.000043

tolerance = 2.000002%

Δx₂ = 2.29130435 / 2.24637681 = 1.0200000017

tolerance = 2.0000002%

x_average = (2 +2,3) / 2

x_average = 2.15

Δx₁ = 2 / 2.15 = 0.93023

tolerance = 100 -93.023

tolerance = 6.98%

Δx₂ = 2.3 / 2.15 = 1.0698

tolerance = 6.97%

Let's analyze these results, the result E is clearly not in the requested tolerance range, the other values may be within the desired tolerance range depending on the required precision, for the high precision of this exercise the only one that meets the requirements is option C .

4 0

3 years ago

What are the benefits of NSTP CWTS?

geniusboy [140]

Explanation:

The purpose of this program is to recognize the Youth's vital role in nation- building, promote consciousness among youth and develop their physical, moral, spiritual, intellectual and social well-being. It shall inculcate in the youth patriotism, nationalism, and advance their involvement in public and civic affairs.

8 0

3 years ago

Viefleur [7K]

Answer:

Both come from the sun

Both are reusable sources

and both don't cause pollution

Explanation:

3 0

2 years ago