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maw [93]
3 years ago
5

Please help on questions D AND E and please show working out as well so I can understand better thank you only QUESTIONS D AND E

Physics
1 answer:
tester [92]3 years ago
5 0

Answer:

d. 87,500 J

e. 49,600 J

Explanation:

The total energy is the heat absorbed by the copper plus the heat absorbed by the water.

d)

E = m₁C₁ΔT + m₂C₂ΔT

E = (1 kg) (390 J/kg/°C) (10 °C) + (2 kg) (4180 J/kg/°C) (10 °C)

E = 87,500 J

e)

E = m₁C₁ΔT + m₂C₂ΔT

E = (2 kg) (390 J/kg/°C) (10 °C) + (1 kg) (4180 J/kg/°C) (10 °C)

E = 49,600 J

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Two objects are being lifted by a machine. One object has a mass of 2 kg, and is lifted at a speed of 2
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Answer:

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First object: (1/2) (2 kg) (2 m/s)² = 4 joules .

Second object: (1/2) (4 kg) (3 m/s)² = 18 joules .

The second object had more kinetic energy than the first one had.

Explanation:

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3 years ago
How do I solve this problem?<br> Can someone please help?
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Answer: Out of all the options presented above the ones that represent the difficulties that the builders of the transcontinental railroad found ways to overcome are answer choices A) providing supplies to build the tracks and support the workers and B) natural barriers such as mountains, rivers, and forests. It was also the same reason why the transcontinental railroad was being constructed. It would help the transportation of  if this is hitory i dont know what subject this is but if its history heres the answers

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3 years ago
The hydraulic oil in a car lift has a density of 8.53 x 102 kg/m3. The weight of the input piston is negligible. The radii of th
navik [9.2K]

Answer:

(a) the input force is 36.56 N

(b) the input force is 37.49 N

Explanation:

Given;

density of hydraulic oil, ρ =  8.53 x 10² kg/m³

radius of plunger, r₁ = 0.135 m

radius of piston, r₂ = 5.43 x 10⁻³ m

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P =\frac{F}{A}

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F is force

A is circular area = πr²

\frac{F_1}{A_1} =\frac{F_2}{A_2} \\\\F_2 = \frac{F_1*A_2}{A_1} =\frac{F_1* \pi r_2^2}{\pi r_1^2} = \frac{F_1*  r_2^2}{ r_1^2} \\\\F_2 = \frac{22600*(5.43*10^{-3})^2 }{(0.135)^2}\\\\F_2 = 36.56 \ N

Part (b) The input force needed to support 22600-N weight, when the  bottom surface of the output plunger is 1.20 m above that of the input plunger

P_2 = P_1 + \rho gh

But, F = PA  and  A = πr²

F_2 = F_1(\frac{A_2}{A_1} ) + \rho gh*A_2\\\\F_2 = F_1(\frac{r_2^2}{r_1^2} )+\rho gh(\pi r_2^2)\\\\F_2 = 22600(\frac{5.43*10^{-3}}{0.135})^2 \ + 853*9.8*1.2*\pi (5.43*10^{-3})^2\\\\F_2=36.56 + 0.93\\\\F_2 = 37.49 \ N

4 0
3 years ago
Limestone cave can develop when Limestone rock is weathered. The weathering of the rock leaves an empty space that forms the cav
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Answer:

Wind blowing across the surface eroded the limestone.

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