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3 years ago

5

Frank wrote several statements to summarize the relationship between the first and second laws of thermodynamics.

2 answers:

Gekata [30.6K]3 years ago

6 0

The second law of thermodynamics states that the total entropy of a system does not decrease with time. This is because not all heat can be converted into useful work. Therefore, the best option is the 2nd choice.
The other choices relate to conservation of energy and the first law of thermodynamics.

Kaylis [27]3 years ago

3 0

First law is Conservation of Energy

Second is that entropy of an isolated system will always increase with time.

Entropy is the change of disorder through time. The best statement which relates to the 2nd law is C. Thermal energy flows from areas of higher to lower temperature

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A 5-m steel beam is lowered by means of two cables unwinding at the same speed from overhead cranes. As the beam approaches the

ArbitrLikvidat [17]

Answer:

a) The angular acceleration of the beam is 0.5 rad/s²CW (direction clockwise due the tangential acceleration is positive)

b) The acceleration of point A is 3.25 m/s²

The acceleration of point E is 0.75 m/s²

Explanation:

a) The relative acceleration of B with respect to D is equal:

$a_{B} =a_{D} +(a_{B/D} )_{n} +(a_{B/D} )_{t}$

Where

aB = absolute acceleration of point B = 2.5 j (m/s²)

aD = absolute acceleration of point D = 1.5 j (m/s²)

(aB/D)n = relative acceleration of point B respect to D (normal direction BD) = 0, no angular velocity of the beam

(aB/D)t = relative acceleration of point B respect to D (tangential direction BD)

$a_{B} =a_{D} +(a_{B/D} )_{t}$

$2.5j=1.5j +(a_{B/D} )_{t}\\(a_{B/D} )_{t}=j=1m/s^{2}$

We have that

(aB/D)t = BDα

Where α = acceleration of the beam

BDα = 1 m/s²

Where

BD = 2

$2\alpha =1\\\alpha =0.5rad/s^{2}CW$

b) The acceleration of point A is:

$a_{A} =a_{D} +(a_{A/D} )_{t}$

(aA/D)t = ADαj

$a_{A} =a_{D} +AD\alpha j\\a_{A}=1.5j+(3.5*0.5)j\\a_{A}=3.25jm/s^{2}$

The acceleration of point E is:

(aE/D)t = -EDαj

$a_{E} =a_{D} -ED\alpha j\\a_{E}=1.5j-(1.5*0.5)j\\a_{E}=0.75jm/s^{2}$

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3 years ago

A spherical gas-storage tank with an inside diameter of 8.0 m is being constructed to store gas under an internal pressure of 1.

Mariulka [41]

Answer:

The minimum wall thickness Tmin required for the spherical tank is 65.90mm

Explanation:

Solution

Recall that,

Tmin = The minimum wall thickness =PD/2бp

where D = diameter of 8.0 m

Internal pressure = 1.62 MPa

Then

The yield strength = 295MPa/3.0 = 98.33

thus,

PD/2бp = 1.62 * 8000/ 2 *98.33

= 12960/196.66 = 65.90

Therefore the wall thickness Tmin required for the spherical tank is 65.90mm

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