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Zinaida [17]
3 years ago
7

Free runners jump long distances and land on the ground or a wall. How do they apply Newton’s second law to lessen the force of

impact?
Physics
1 answer:
Veseljchak [2.6K]3 years ago
5 0

As we know that as per Newton's II law we have

F = \frac{dP}{dt}

here we will have

dP = change in momentum

dt = time interval in which momentum is changed

now in order to have least injury during jumping we need to have least force on the jumper

so in order to have least force we can say that the momentum must have to change in maximum time so that amount of force must be least

So we need to increase the time in which momentum of the system is changed

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All organisms have adaptations. The list below describes how different adaptations might affect different species.
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A bag of sand has a density of 45 g/cm3 and a mass of 15 kg. How much space does the sand take up?
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Volume = mass/density

 

Volume = 15000 g/45 g/cm3 ≈ 333.3 cm<span>3</span>

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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

#SPJ1

4 0
2 years ago
Crystallization is a separation technique used to produce rock candy from a sugar solution. true or false
Zinaida [17]
This is true due to the reaction that happens from water evaporating and leaving the sugar crystals behind to form.
3 0
3 years ago
What is the electric field strength between two parallel plates 5cm apart with a potential difference of 25V across them?
Butoxors [25]

Use the following formula for the electric field strength between two parallel plates:

E = V/d

where,

V: potential difference = 25V

d: distance between plates = 5 cm = 0.05 m

Replace the previous values of the parameters into the formula for E:

E=\frac{25V}{0.05m}=500\frac{V}{m}

Hence, the electric field strength is 500V/m

5 0
1 year ago
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