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Serga [27]
4 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]4 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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A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle
ra1l [238]

Answer:

a) t = 4.16 s

b) x = 141.51 m

Explanation:

Given

v = 21.5 m/s

x0 = 52.0 m

a = 6.0 m/s²

a) Motorcycle

x = v0*t + (a*t²/2)

x = 21.5t + (6*t²/2)

x = 21.5t + 3t²   <em>(I)</em>

Car

x = x0 + v0*t

x = 52 + 21.5t  <em>(II)</em>

<em />

then we can apply <em>I = II</em>

21.5t + 3t² = 52 + 21.5t

⇒ 3t² = 52

⇒ t = 4.16 s

b) We can use <em>I</em> or <em>II</em>, then

x = 52 + 21.5*(4.16)

⇒ x = 141.51 m

8 0
3 years ago
Two resistors, A and B, are connected in series to a 6.0 V battery. A voltmeter connected across resistor A measures a potential
mestny [16]

Answer:

Resistance of resistor A = 6.0 Ω and resistance of resistor B = 3.0 Ω

Explanation:

When the two resistors are in series, let V₁ = voltage in resistor A and R₁ = resistance of resistor A and V₂ = voltage in resistor B and R₂ = resistance of resistor B.

Given that V₁ + V₂ = 6.0 V and V₁ = 4.0 V,

V₂ = 6.0 V - V₁ = 6.0 V - 4.0 V = 2.0 V

Also, let the current in series be I.

So, V₁ = IR₁ and V₂ = IR₂

I = V₁/R₁ and I = V₂/R₂

equating both expressions, we have

V₁/R₁ = V₂/R₂

4.0 V/R₁ = 2.0 V/R₂

dividing through by 2.0 V, we have

2/R₁ = 1/R₂

taking the reciprocal, we have

R₂ = R₁/2

R₁ = 2R₂

From the parallel connection, let V₁ = voltage in resistor A and R₁ = resistance of resistor A and V₂ = voltage in resistor B and R₂ = resistance of resistor B. Since it is parallel, V₁ = V₂ = V = 6.0 V

Also, V₂ = I₂R₂ where I₂ = current in resistor B = 2.0 A and R₂ = resistance of resistor B

So, R₂ = V₂/I₂

= 6.0 V/2.0 A

= 3.0 Ω

R₁ = 2R₂

= 2(3.0 Ω)

= 6.0 Ω

So, resistance of resistor A = 6.0 Ω and resistance of resistor B = 3.0 Ω

6 0
3 years ago
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