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polet [3.4K]
2 years ago
15

A sprinter reaches his maximum speed in 2.5sec from rest with constant acceleration. He then maintains that speed and finishes t

he 100m in the overall time of 9.60 seconds. Determine his maximum speed . 5. The cone falling with a speed strikes and penetrates the block of packing material. The acceleration of the cone after impact is = −
Engineering
1 answer:
liubo4ka [24]2 years ago
3 0

The maximum Speed of the Sprinter from the velocity time graph of his motion is; 11.98 m/s

<h3>How to find the maximum speed?</h3>

We are given;

Initial Speed; u = 2.5 s

Total distance; d = 100 m

Total time; T = 9.6 s

The total distance is;

d = ¹/₂(9.6 + (9.6 - 2.5) * v

where v is maximum speed.

Thus;

¹/₂(9.6 + (9.6 - 2.5) * v = 100

16.7v = 200

v = 200/16.7

v = 11.98 m/s

Read more about Maximum Speed at; brainly.com/question/4931057

#SPJ1

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Answer:

a) w_{out} = 281.55\,\frac{kJ}{kg}, b) s_{gen} = 0.477\,\frac{kJ}{kg\cdot K}

Explanation:

a) The process within the turbine is modelled after the First Law of Thermodynamics:

-q_{out} - w_{out} + h_{in}-h_{out} = 0

w_{out} = h_{in} - h_{out}-q_{out}

w_{out} = c_{p}\cdot (T_{in}-T_{out})-q_{out}

w_{out} = \left(1.005\,\frac{kJ}{kg\cdot K}\right)\cdot (980\,K-670\,K)-30\,\frac{kJ}{kg}

w_{out} = 281.55\,\frac{kJ}{kg}

b) The entropy production is determined after the Second Law of Thermodynamics:

-\frac{q_{out}}{T_{surr}} + s_{in}-s_{out} + s_{gen} = 0

s_{gen} = \frac{q_{out}}{T_{surr}}+s_{out}-s_{in}

s_{gen} = \frac{q_{out}}{T_{surr}}+c_{p}\cdot \ln\left(\frac{T_{out}}{T_{in}} \right)

s_{gen} = \frac{30\,\frac{kJ}{kg} }{315\,K} + \left(1.005\,\frac{kJ}{kg\cdot K} \right)\cdot \ln\left(\frac{980\,K}{670\,K} \right)

s_{gen} = 0.477\,\frac{kJ}{kg\cdot K}

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A laboratory in the Y building keep a vacuum pressure of 0.1 kPa abs. What is the net force acting on the door considering the a
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Answer:

net force acting on the floor is 100 kN

Explanation:

Given data:

P_{vaccum} = 0.1 kPa

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we know that

Net force can be calculated as follow

f_{net} = P_{vaccum} \times area

f_{net} = 0.1\times 10^3 \times 2\times 0.5

f_{net} = 0.1\times 10^3 \times 1

f_{net} = 100 kN

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Answer:

attached below

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Find the second derivative of: y = sqrt(x + 4)/4; x ≥ -4.
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Answer: \dfrac{-1}{16(x+4)^{\frac{3}{2}}}

Explanation:

Given

y=\dfrac{\sqrt{x+4}}{4}\\\text{differentitate w.r.t x}\\\dfrac{\mathrm{d} y}{\mathrm{d} x}=\dfrac{1}{4}\times \dfrac{1}{2\sqrt{x+4}}=\dfrac{1}{8\sqrt{x+4}}

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