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1 year ago

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Consider as a system the gas in a vertical cylinder; the cylinder is fitted with a piston on which a number of small weights are

placed. The initial pressure is 200 kPa, and the initial volume of the gas is 0.04 m3 . (a) Let a Bunsen burner be placed under the cylinder, and let the volume of the gas increase to 0.1 m3 while the pressure remains constant. Calculate the work done by the system during this process. (b) Consider the same system and initial conditions, but at the same time that the Bunsen burner is under the cylinder and the piston is rising, remove weights from the piston at a rate such that, during the process, the temperature of the gas remains co

1 answer:

Amiraneli [1.4K]1 year ago

5 0

Answer:

Explanation:

a ) At constant pressure , work done = P x Δ V

= 200 x 10³ x ( .1 - .04 )

= 12 x 10³ J .

b )

At constant temperature work done

= n RT ln v₂ / v₁

= PV ln v₂ / v₁

= 200 x 10³ x .04 ln .1 / .04

8 x 10³ x .916

= 7.33 x 10³ J .

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__________ curves help lessen the effect of the force of the forward motion on your vehicle as it enters the curve.

Rainbow [258]

Answer:

Banked

Explanation:

Banked curves are formed when the inner edge is below the outer edge.

It is done in order to ensure the reliability of the frictional force as it varies when the road is wet wet or oily. Thus in order to avoid these problems the curved roads are banked.

Banking of the curve provides the necessary centripetal force, i.e., the horizontal component of the normal reaction force to keep the vehicle i motion and thus helps in reducing the effect of the forward motion force on the vehicle.

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1 year ago

The block in the diagram below is AT REST. However, the tension in the cable is not the only thing holding the block back. Stati

Vedmedyk [2.9K]

Answer:

The tension in the rope is 229.37 N.

Explanation:

Given:

Mass of the block is, $m=33.2\ kg$

Coefficient of static friction is, $\mu = 0.214$

Angle of inclination is, $\theta = 31.5$ °

Draw a free body diagram of the block.

From the free body diagram, consider the forces in the vertical direction perpendicular to inclined plane.

Forces acting are $mg\cos \theta$ and normal $N$ . Now, there is no motion in the direction perpendicular to the inclined plane. So,

$N=mg\cos \theta\\N=(33.2)(9.8)\cos (31.5)\\N=277.415\ N$

Consider the direction along the inclined plane.

The forces acting along the plane are $mg\sin \theta$ and frictional force, $f$ , down the plane and tension, $T$ , up the plane.

Now, as the block is at rest, so net force along the plane is also zero.

$T=mg\sin \theta+f\\T=mg\sin \theta +\mu N\\T= (33.2)(9.8)(\sin (31.5)+(0.214\times 277.415)\\T= 170+59.37\\T=229.37\ N$

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1 year ago

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sineoko [7]

Answer:

Total number of electrons

$N = 7.25 \times 10^{24}$

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$N = 5.27 \times 10^{15}$

Fraction of electrons transferred is given as

$f = 7.27 \times 10^{-10}$

Explanation:

As we know that moles is defined as

$n =\frac{mass}{molar mass}$

$n = \frac{0.0250}{0.026982}$

$n = 0.93$

so number of atoms of Al in each sphere is given as

$N = 0.93(6.02 \times 10^{23})$

$N = 5.58 \times 10^{23}$

Now number of electrons in each atom is given as

atomic number = number of electrons in each atom = 13

total number of electrons in each sphere is

$N = 13 \times (5.58 \times 10^{23})$

$N = 7.25 \times 10^{24}$

Also we know that force of attraction between them is given as

$F= \frac{kq_1q_2}{r^2}$

$1.00 \times 10^4 = \frac{(9\times 10^9)q^2}{0.80^2}$

$q = 8.4 \times 10^{-4} C$

now we have

$q = Ne$

$8.4 \times 10^{-4} = N(1.6 \times 10^{-19}$

$N = \frac{(8.4 \times 10^{-4})}{1.6 \times 10^{-19}}$

$N = 5.27 \times 10^{15}$

Fraction of electrons transferred is given as

$f = \frac{5.27 \times 10^{15}}{7.25 \times 10^{24}}$

$f = 7.27 \times 10^{-10}$

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1 year ago

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Y_Kistochka [10]

Answer:

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Explanation:

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m = Mass of sulfur

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The speed with which the liquid sulfur left the volcano is 529.15 m/s

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2 years ago

A marble is dropped straight down from a distance h above the floor.

Mrac [35]

Fm=Fe and am>ae
Hopefully this helps

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2 years ago