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riadik2000 [5.3K]

1 year ago

15

dopasuj wartości pracy z ramki do przedstawionych sytuacji ,a nastepnie wyraż te prace w dżulach uwaga jedna wartośc pracy nie b

edzie pasowała do żadnej sytuacji

1 answer:

Eduardwww [97]1 year ago

3 0

Answer:

A (samolot) - 200 MJ = 200000000 J

B (dźwig) - 800 kJ = 800000 J

C (podnośnik)-1.6 kJ = 1600 J

Explanation:

Całą część pytania można znaleźć na poniższym schemacie.

Z diagramu załączonego poniżej; mamy

A - samolot lotniczy

B - dźwig

C - podnoszenie ciężarów

Wszyscy to wiemy ;

1kJ = 1000 J

1MJ = 1000000 J

Mamy cztery opcje; i.e 200 MJ, 800 kJ, 1.6 kJ and 250 mJ

Z czterech opcji można wykluczyć 250 mJ, ponieważ jest to 0,25 J, co przedstawia bardzo niską energię w porównaniu z trzema warunkami pokazanymi na schemacie.

Więc:

A (samolot) - 200 MJ = 200000000 J

B (dźwig) - 800 kJ = 800000 J

C (podnośnik)-1.6 kJ = 1600 J

Największą pracę wykona samolot. Jest tak, ponieważ ma bardzo dużą masę i bardzo dużą prędkość. W związku z tym istnieje potrzeba wytworzenia ogromnej ilości ciepła i energii.

Z drugiej strony żuraw może podnieść ładunek o wiele większy i przewyższa ciężar ciężaru, więc praca wykonywana przez dźwig musi być zdecydowanie większa niż praca ciężarka.

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$v= \frac{S}{t}$

The total distance covered is
$S=200 km+0+100 km=300 km$
while the total time taken is 2 hours + half an hour (for the rest) + 1 hour and half, so
$t=2h+0.5h+1.5 h=4 h$
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1 year ago

A 50 g tennis ball travels at a velocity of 15 m/s, hits a basketball with a mass of 600 g that is stationary on a frictionless

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Answer:

1.75 m/s

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Momentum is conserved.

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

(50 g) (15 m/s) + (600 g) (0 m/s) = (50 g) (-6 m/s) + (600 g) v

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2 years ago

Calculate the energy in the form of heat (in kJ) required to change 75.0 g of liquid water at 27.0 °C to ice at –20.0 °C. Assume

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Explanation:

The given data is as follows.

mass, m = 75 g

$T_{1} = 0^{o}C$

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First, we will calculate the heat required for water is as follows.

q = $m C \times (T_{1} - T_{2})$

= $75 g \times 4.18 J/g^{o}C \times (0 - 27)^{o}C$

= 8464.5 J/mol

= 8.46 kJ ......... (1)

Also, it is given that $T_{3} = -20^{o}C$ = (20 + 273) K = 293 K and specific heat of ice is 2.108 kJ/kg K.

Now, we will calculate the heat of fusion as follows.

q = $mC \times (T_{3} - T_{1})$

= $0.075 kg \times 2.108 kJ/kg K \times (-293 - 0) K$

= -46.32 kJ ......... (2)

Now, adding both equations (1) and (2) as follows.

8.46 kJ - 46.32 kJ

= -37.86 kJ

Therefore, we can conclude that energy in the form of heat (in kJ) required to change 75.0 g of liquid water at $27.0^{o}C$ to ice at $-20.0^{o}C$ is -37.86 kJ.

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2 years ago

Determine the centripetal force upon a 40-kg child who makes 10 revolution around the cliffhanger in 29.3 seconds.the radius of

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Answer:

The centripetal force acting on the child is 39400.56 N.

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Time taken for 10 revolutions is, $t=29.3\ s$

Therefore, the time period of the child is given as:

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Now, angular velocity is related to time period as:

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Therefore, the centripetal force acting on the child is 39400.56 N.

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1 year ago

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The position function x(t) of a particle moving along an x axis is $x=4.00 - 6.00t^2$

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So dx/dt = 0

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So the particle is starting from rest.

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SO at x = 4m the particle is (momentarily) stop

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2 years ago