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NISA [10]
1 year ago
14

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
Physics
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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Consider a simple ideal Rankine cycle with fixed turbine inlet conditions. What is the effect of lowering the condenser pressure
mr Goodwill [35]

Answer:

The effect of lowering the condenser pressure on different parameters is explained below.

Explanation:

The simple ideal Rankine cycle is shown in figure.

Effect of lowering the condenser pressure on

(a). Pump work input :- By lowering the condenser pressure the pump work increased.

(b) Turbine work output :- By lowering the condenser pressure the turbine work increased.

(c). Heat supplied :- Heat supplied increases.

(d). Heat rejected :- The heat rejected may increased  or decreased.

(e). Efficiency :- Cycle  efficiency is increased.

(f). Moisture content at turbine exit :- Moisture content increases.

8 0
1 year ago
A 15.0-gram lead ball at 25.0°C was heated with 40.5 joules of heat. Given the specific heat of lead is 0.128 J/g∙°C, what is th
mr Goodwill [35]

Answer:

T=4985.5^{\circ}K

Explanation:

The equation that relates heat Q with the temperature change T-T_0 of a substance of mass <em>m </em>and specific heat <em>c </em>is Q=mc(T-T_0).

We want to calculate the final temperature <em>T, </em>so we have:

T=\frac{Q}{mc}+T_0

Which for our values means (in this case we do not need to convert the mass to Kg since <em>c</em> is given in g also and they cancel out, but we add 273^{\circ} to our temperature in ^{\circ}C to have it in ^{\circ}K as it must be):

T=\frac{Q}{mc}+T_0=\frac{40.5J}{(15g)(0.128J/g^{\circ}C)}+(298^{\circ}K)=4985.5^{\circ}K

3 0
1 year ago
A positively-charged particle is released near the positive plate of a parallel plate capacitor. a. Describe its path after it i
il63 [147K]

Answer:

a. The electric field lines are linear and perpendicular to the plates inside a parallel-plate capacitor, and always from positive plate to the negative plate. If a positive charge is released near the positive plate, then<em> it will follow a linear path towards the negative plate under the influence of electrostatic force, F = Eq</em>, where q is the charge of the particle. The electric field inside a parallel plate capacitor is constant and equal to

This can be calculated by Gauss' Law.

A positive charge always follow the electric field lines when released. Another approach is that the positive plate repels the positive charge and negative plate attracts the positive charge. Therefore, the positive charge follows a path towards the negative charge.

b. The particle moves from the higher potential to the lower potential. <em>The direction of motion is the same as the direction of the force that moves the particle, so the work done on the particle by that force is positive.</em>

8 0
2 years ago
16) A wheel of moment of inertia of 5.00 kg-m2 starts from rest and accelerates under a constant torque of 3.00 N-m for 8.00 s.
KiRa [710]

Answer:

57.6Joules

Explanation:

Rotational kinetic energy of a body can be determined using the expression

Rotational kinetic energy = 1/2Iω²where;

I is the moment of inertia around axis of rotation. = 5kgm/s²

ω is the angular velocity = ?

Note that torque (T) = I¶ where;

¶ is the angular acceleration.

I is the moment of inertia

¶ = T/I

¶ = 3.0/5.0

¶ = 0.6rad/s²

Angular acceleration (¶) = ∆ω/∆t

∆ω = ¶∆t

ω = 0.6×8

ω = 4.8rad/s

Therefore, rotational kinetic energy = 1/2×5×4.8²

= 5×4.8×2.4

= 57.6Joules

6 0
2 years ago
Read 2 more answers
The weights of a large number of miniature poodles are approximately normally distributed with a mean of 99 kilograms and a stan
Karo-lina-s [1.5K]

Answer:

See explanation below

Explanation:

As I say in the comments, the question is incomplete, however, I will try to answer this by using data that I found on another site.

This is the part of the question that is not here:

If measurements are recorded to the nearest

tenth of a kilogram, find the fraction of these poodles

with weights

(a) over 9.5 kilograms;

(b) of at most 8.6 kilograms;

So, assuming a mean of 8 kg, and 0.9 of standard deviation, let X represents the weight of the poodles

The expression to calculate the fraction of poodle needed is:

Z = X - u / d

u: weight of the large number of poodle

d: standard deviation

Replacing data of a) wer have:

Z = 9.5 - 8 / 0.9

Z = 1.67

With this value, we need to take the value of Z, and see the area under the curve of standard deviation (see table attached)

Therefore:

P (X > 9.5) = P(Z > 1.67) = 0.5 - P (Z < 1.67) = 0.5 - 0.4525 = 0.0475

b) In this part, is the same as part a) so:

Z = 8.6 - 8 / 0.9 = 0.67

The value for area in the curve is 0.2486 so:

P = 0.5 + 0.2486 = 0.7486

Hope this helps

8 0
2 years ago
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