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3 years ago

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A 20 N force is applied to an object causing it to move 10 m. How much work was done on the object? How much energy was needed t

o move the object?

1 answer:

olga2289 [7]3 years ago

3 0

Answer:

Here, force=20N and displacement=10m

Work=Force×Displacement=20N×10m=200Nm

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Problem 6.056 Air enters a compressor operating at steady state at 15 lbf/in.2, 80°F and exits at 275°F. Stray heat transfer and

vova2212 [387]

To solve this process it is necessary to consider the concepts related to the relations between pressure and temperature in an adiabatic process.

By definition the relationship between pressure and temperature is given by

$(\frac{P_2}{P_1})=(\frac{T_2}{T_1})^{(\frac{\gamma}{\gamma-1})}$

Here

P = Pressure

T = Temperature

$\gamma =$ The ratio of specific heats. For air normally is 1.4.

Our values are given as,

$P_1 = 15lb/in^2\\T_1= 80\°F = 299.817K\\T_2 =400\°F = 408.15K$

Therefore replacing we have,

$(\frac{P_2}{P_1})=(\frac{T_2}{T_1})^{(\frac{\gamma}{\gamma-1})}$

$(\frac{P_2}{15})=(\frac{408.15}{299.817})^{(\frac{1.4}{1.4-1})}$

Solving for $P_2,$

$P_2 = 15*(\frac{408.15}{299.817})^{(\frac{1.4}{1.4-1})}$

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Therefore the maximum theoretical pressure at the exit is $44.15Lbf/in^2$

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3 years ago

An aluminum calorimeter with a mass of 100 g contains 250 g of water. The calorimeter and water are in thermal equilibrium at 10

Nadya [2.5K]

Answer:

1822.14 J/kg C

Explanation:

Mass of aluminum = 0.100 kg

Mass of Water = 0.250 kg

Mass of copper = 0.300 kg

Mass of unknown object = 0.70 kg

Initial temperature of Aluminum = 10 C

Initial temperature of water = 10 C

Initial temperature of copper = 30 C

Initial temperature of = 100 C

Final temperature of all substances = 20 C

Change in temperature of each of them Aluminum, water , copper and unknown material respectively = 10 C , 10 C , -10 C, -80 C respectively

Specific heat of each of them Aluminum, water , copper and unknown material respectively = 900,4186,387 and c respectively.

Heat gained by aluminum and water = (0.1)(900)(10) + (0.250)(4186)(10) = 11365 j

heat lost by copper and unknown material =(0.3)(387)(-10) = -1161

heat lost by unknown material = (0.070)(c)(-80) = -5.6 c

Now solve for c using the calorimetry principle.

Heat lost + Heat gained = 0

⇒ -5.6 c = -11365 - (-1161) = -10204 j

Specific heat of unknown material = - 10204 / -5.6

= 1822.14 j/kg C

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