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2 years ago

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A gas is contained in a vertical piston-cylinder assembly by a piston with a face area of 40 in2 and weight of 100 lbf. The atmo

sphere exerts a pressure of 14.7 lbf/in2 on top of the piston. A paddle wheel transfers 3 Btu of energy to the gas during a process in which the elevation of the piston increases by 1 ft. The piston and cylinder are poor thermal conductors, and friction between them can be neglected. Determine the change in internal energy of the gas, in Btu.

1 answer:

kompoz [17]2 years ago

6 0

Answer:

ΔU=0.8834 Btu

Explanation:

Given data

Area of piston A=40 in²

The weight W=100 lbf

Atmospheric Pressure P=14.7 lbf/in²

Work added E=3 Btu

The change in elevation Δh=1 ft =12 inch

To find

Change in internal energy of the gas ΔU

Solution

For Piston

ΔPE=| W+(P×A)×Δh |

ΔPE=| 100+(14.7×40)×12 |

ΔPE=8256 lbf.in

ΔPE=8256×0.000107

ΔPE=0.8834 Btu

From law of conservation of energy then ,the charging in the potential energy of the piston is made by exerting force by gas

Wgas= -ΔPE

Wgas= -0.8834 Btu

For the gas as a system and by applying first law of thermodynamics

Q-W=ΔU

0-(-0.8834 Btu)=ΔU

So

ΔU=0.8834 Btu

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A 5.0-kg rock and a 3.0 × 10−4-kg pebble are held near the surface of the earth.(a)Determine the magnitude of the gravitational

a_sh-v [17]

Answer:

a). Determine the magnitude of the gravitational force exerted on each by the earth.

Rock: $F = 49.06N$

Pebble: $F = 29.44N$

(b)Calculate the magnitude of the acceleration of each object when released.

Rock: $a =9.8m/s^{2}$

Pebble: $a =9.8m/s^{2}$

Explanation:

The universal law of gravitation is defined as:

$F = G\frac{m1m2}{r^{2}}$ (1)

Where G is the gravitational constant, m1 and m2 are the masses of the two objects and r is the distance between them.

<em>Case for the rock </em> $m = 5.0 Kg$ <em>:</em>

m1 will be equal to the mass of the Earth $m1 = 5.972×10^{24} Kg$ and since the rock and the pebble are held near the surface of the Earth, then, r will be equal to the radius of the Earth $r = 6371000m$ .

$F = (6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2})\frac{(5.972x10^{24} Kg)(5.0 Kg)}{(6371000 m)^{2}}$

$F = 49.06N$

Newton's second law can be used to know the acceleration.

$F = ma$

$a =\frac{F}{m}$ (2)

$a =\frac{(49.06 Kg.m/s^{2})}{(5.0 Kg)}$

$a =9.8m/s^{2}$

<em>Case for the pebble </em> $m = 3.0 Kg$ <em>:</em>

$F = (6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2})\frac{(5.972x10^{24} Kg)(3.0 Kg)}{(6371000 m)^{2}}$

$F = 29.44N$

$a =\frac{F}{m}$

$a =\frac{(29.44 Kg.m/s^{2})}{(3.0 Kg)}$

$a =9.8m/s^{2}$

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2 years ago

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Physics Help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

djverab [1.8K]

Good work on solving part a).
b) may look complicated, but it's not too bad.

It says that the body is 25% efficient in converting fat to mechanical energy.
In other words, only 25% of the energy we get from our stored fat shows up
in the physical, mechanical moving around that we do. (The rest becomes
heat, which dissipates into the environment as we keep our bodies warm,
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You already know how much mechanical energy the climber needed to lift
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That's 25% of what he needs to convert in order to accomplish the climb.
He needs to pull 4 times as much energy out of fat.

-- Fat energy required = 4 x (2.4 x 10⁶) = 9.6 x 10⁶ joules.

-- Amount stored in 1kg of fat = 3.8 x 10⁷ joules

-- Portion of a kilogram he needs to use = (9.6 x 10⁶) / (3.8 x 10⁷)

Note:
That much of a kilogram weighs about 8.9 ounces ... which shows why it's so
hard to lose weight with physical exercise alone. It also helps you appreciate
that fat is much more efficient at storing energy than batteries are ... that one
kilogram of fat stores the amount of energy used by a 100-watt light bulb, to
burn for 105 hours (more than 4-1/2 days ! ! !)

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2 years ago

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Three wires are made of copper having circular cross sections. Wire 1 has a length l and radius r. Wire 2 has a length l and rad

Alex73 [517]

Explanation:

Below is an attachment containing the solution.

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2 years ago

A machine part is vibrating along the x-axis in simple harmonic motion with a period of 0.27 s and a range (from the maximum in

Gnoma [55]

Answer:

$x = -1.437 cm$

Explanation:

The general equation for position of Simple harmonic motion is given as:

$x = A sin(\omega t)$ ........(1)

where,

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A = Amplitude of the wave

ω = Angular velocity

t = time

In this case, the amplitude is just half the range,

thus,

$A =\frac{3cm}{2}=1.5cm$ (Given range = 3cm)

A = 1.5 cm

Now, The angular velocity is given as:

$\omega=\frac{2\pi}{T}$

Where, T = time period of the wave =0.27s (given)

$\omega=\frac{2\pi}{0.27s}$

or

$\omega=23.27s^{-1}$

so, at time t = 55 s, the equation (1) becomes as:

$x = 1.5 sin(23.27\times 55)$

on solving the above equation we get,

$x = -1.437 cm$

here the negative sign depicts the position in the opposite direction of +x

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2 years ago

Alex goes cruising on his dirt bike. He rides 700m north, 300m east, 400 m north, 600m west, 1200m south, 300m east and finally

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Find Displacement and Distance

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south=1200m

1200-1200=0

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2 years ago