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3 years ago

A race car starts from rest and accelerates uniformly to 69 mph in 4.5 s.

1 answer:

5 0

For this question we should apply
a = v^2 - u^2 by t
a = 69 - 0 by 4.5
a = 69 by 4.5
a = 15.33
a = 6.85 m/s^2

If the answer in option is near to answer then , you can mark it as correct.
.:. The acceleration is 6.9 m/s^2

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3 years ago

Example 3 :

kherson [118]

The friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 × 10^8 respectively. Also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 × 10^8 respectively.

<h3>How to determine the friction factor</h3>

Using the formula

μ = viscosity = 0. 06 Pas

d = diameter = 120mm = 0. 12m

V = velocity = 1m/s and 3m/s

ρ = density = 0.9

a. Velocity = 1m/s

friction factor = 0. 52 × $\frac{0. 06}{0. 12* 1* 0. 9}$

friction factor = 0. 52 × $\frac{0. 06}{0. 108}$

friction factor = 0. 52 × 0. 55

friction factor $= 0. 289$

b. When V = 3mls

Friction factor = 0. 52 × $\frac{0. 06}{0. 12 * 3* 0. 9}$

Friction factor = 0. 52 × $\frac{0. 06}{0. 324}$

Friction factor = 0. 52 × 0. 185

Friction factor $= 0.096$

Loss When V = 1m/s

Head loss/ length = friction factor × 1/ 2g × velocity^2/ diameter

Head loss = 0. 289 × $\frac{1}{2*6. 6743 * 10^-11}$ × $\frac{1^2}{0. 120}$ × $\frac{1}{100}$

Head loss = 1. 80 × 10^8

Head loss When V = 3m/s

Head loss = $0. 096$ × $\frac{1}{1. 334 *10^-10}$ × $\frac{3^2}{0. 120}$ × $\frac{1}{100}$

Head loss = 5. 3× 10^8

Thus, the friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 ×10^8 respectively also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 ×10^8 respectively.

Learn more about friction here:

brainly.com/question/24338873

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