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Serga [27]
3 years ago
5

Zad. 1. Jaką pracę wykona dźwig, który pracował przez 1,5 h napędzany silnikiem o mocy 200 kW?

Physics
1 answer:
Alekssandra [29.7K]3 years ago
7 0

Answer:

Explanation:

1) Moc to wskaźnik czasu pracy w stosunku do czasu.

Moc = czas pracy / czas

Czas pracy = moc * czas

Podana moc = 200 kW = 200 000 Watów

Czas = 1,5 godziny = 1,5 * 3600 = 5400 sekund

workdone = 200 000/5400

Obróbka = 37,04 dżula

2) Moc = Gotowe / czas

Od zakończenia = Siła * Odległość

Moc = siła * Odległość / czas

Podana siła = 300 N.

Odległość = 2,5 m

czas = 0,5 min = 0,5 * 60 = 30 sekund

Moc = 300 * 2,5 / 30

Moc = 25 watów

3) Moc = Gotowe / czas

Czas potrzebny = praca / moc

Biorąc pod uwagę pracę wykonaną przez ciągnik = 10 MJ = 10 * 10 ^ 6 Dżuli

Moc = 100 kW = 100 000 watów

Czas wzięty = 10 ^ 7/10 ^ 5

Zajęty czas = 100 sekund

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Answer

given,

height of window = 4 m

time taken to travel = 1 s

acceleration due to gravity = 9.8 m/s²

s = ut + \dfrac{1}{2}at^2

u = \dfrac{s - \dfrac{1}{2}at^2}{t}

u = \dfrac{4 - \dfrac{1}{2}\times 9.8\times 1^2}{1}

u = -0.905 m/s

initial velocity of ledge v = 0

now,

v² = u² + 2 a s

(-0.905)² = 0 + 2 × 9.8 ×s

s = 0.042 m

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A car accelerates from rest at 3.6 m/s 2 . How much time does it need to attain a speed of 5 m/s?
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car starts from rest

v_i = 0

final speed attained by the car is

v_f = 5 m/s

acceleration of the car will be

a = 3.6 m/s^2

now the time to reach this final speed will be

t = \frac{v_f - v_i}{a}

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A steel pipe of 12-in. outer diameter is fabricated from 1 4 -in.-thick plate by welding along a helix that forms an angle of 25
Varvara68 [4.7K]

Answer:

Normal stress = 66/62.84 = 1.05kips/in²

shearing stress  = T/2 = 0.952/2 = 0.476 kips/in²

Explanation:

A steel pipe of 12-in. outer diameter  d₂ =12in  d₁= 12 -4in = 8in

4 -in.-thick  

angle of 25°

Axial force P = 66 kip axial force

determine the normal and shearing stresses

Normal stress б = force/area = P/A

           = 66/ (П* (d₂²-d₁²)/4

           =66/ (3.142* (12²-8²)/4

          = 66/62.84 = 1.05kips/in²

Tangential stress T = force* cos ∅/area = P/A

           = 66* cos 25/ (П* (d₂²-d₁²)/4

           =59.82/ (3.142* (12²-8²)/4

          = 59.82/62.84 = 0.952kips/in²

shearing stress  = tangential stress /2

                           = T/2 = 0.952/2 = 0.476 kips/in²

8 0
3 years ago
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