A textbook is sitting at rest on a desk. Compared to the magnitude of the force of the textbook on the desk, how

One end of a metal rod is in contact with a thermal reservoir at 745. K, and the other end is in contact with a thermal reservoi

Masteriza [31]

Answer:

a) $S_1=-9.65}\ J/K$

b) $S_2=71.18\ J/K$

c) 0 J/K

d)S= 61.53 J/K

Explanation:

Given that

T₁ = 745 K

T₂ = 101 K

Q= 7190 J

a)

The entropy change of reservoir 745 K

$S_1=-\dfrac{7190}{745}\ J/K$

Negative sign because heat is leaving.

$S_1=-9.65}\ J/K$

b)

The entropy change of reservoir 101 K

$S_2=\dfrac{7190}{101}\ J/K$

$S_2=71.18\ J/K$

c)

The entropy change of the rod will be zero.

d)

The entropy change of the system

S= S₁ + S₂

S = 71.18 - 9.65 J/K

S= 61.53 J/K

3 0

4 years ago

To model time-variant data, one must create a new entity in an m:n relationship with the original entity.

Ksenya-84 [330]

To model time-variant data, one must create a new entity in an m:n relationship with the original entity, is a False statement.

Like the majority of software engineering initiatives, the ER process begins with gathering user requirements. What information must be retained, what questions must be answered, and what business rules must be implemented (For instance, if the manager column in the DEPARTMENT table is the only column, we have simply committed to having one manager for each department.)

The end result of the E-R modeling procedure is an E-R diagram that can be roughly mechanically transformed into a set of tables. Tables will represent both entities and relationships; entity tables frequently have a single primary key, but the primary key for relationship tables nearly invariably involves numerous characteristics.

To know more about entity AND relationship visit : brainly.com/question/28232864

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6 0

1 year ago

In a container of negligible mass, 0.400 kg of ice at an initial temperature of -29.0 ∘C is mixed with a mass m of water that ha

Pie

Answer:

1 kg

Explanation:

The container has negligible mass and no heat is loss to the surrounding.

Mass of ice = 0.4kg, initial temperature of ice = -29oC, final temperature of the mixture = 26oC, mass of water (m2) = ?kg, initial temperature of water = 80oC, c ( specific heat capacity of water ) = 4200J/kg.K, Lf = heat of fusion of water = 3.36 × 10^5 J/kg

Using the formula:

Quantity of heat gain by ice = Quantity of heat loss by water

Quantity of heat gain by ice = mass of ice × heat of fusion of ice + mass of water × specific heat capacity of water = (0.4 × 3.36 × 10^ 5) + (0.4 × 4200 × (26- (-29) = 13.44 × 10^4 + 9.24 × 10^ 4 = 22.68 × 10^4 J

Quantity of heat loss by water = m2cΔT

Quantity of heat loss by water = m2 ×4200× (80 - 26) = m(226800)

since heat gain = heat loss

22.68 × 10^4 = 226800 m2

divide both side by 226800

226800 / 226800 = m2

m2 = 1 kg

5 0

3 years ago

Two mating steel spur gears are 20 mm wide, and the tooth profiles have radii of curvature at the line of contact of 10 and 15 m

Rom4ik [11]

Answer:

a) The maximum contact pressure is 274.58 MPa and the width of contact is 0.058 mm

b) The maximum shear stress is 82.37 MPa at a distance of 0.023 mm

Explanation:

Given data:

L = 20 mm

F = 250 N

r₁ = 10 mm

r₂ = 15 mm

v = 0.3

E = 2.07x10⁵ MPa

$A=\frac{1-V_{1}^{2} }{E_{1} }-\frac{1-V_{2}^{2} }{E_{2} } =\frac{1-0.3^{2} }{2.07x10^{5} } *2=8.79x10^{-6}$

a) The maximum contact pressure is:

$P=0.564*\sqrt{\frac{F*(\frac{1}{r_{1} }+\frac{1}{r_{2} }) }{LA} } =0.564*\sqrt{\frac{250*(\frac{1}{10} +\frac{1}{15} )}{20*8.79x10^{-6} } } =274.58MPa$

The width of contact is:

$b=1.13*\sqrt{\frac{FA}{L(\frac{1}{r_{1} }+\frac{1}{r_{2} }) } } =1.13*\sqrt{\frac{250*8.79x10^{-6} }{20*(\frac{1}{10} +\frac{1}{15} )} } =0.029mm\\2*b=0.058mm$

b) According the graph elastic stresses below the surface, for v = 0.3, the maximum shear stress is

T = 0.3*P = 0.3 * 274.58 = 82.37 MPa

At a distance of

0.8*b = 0.8*0.029 = 0.023 mm

6 0

3 years ago

How large must the coefficient of static friction be between the tires and the road if a car is to round a level curve of radius

maria [59]

Answer:

897

Explanation:

Speed of the car, v = 126 km/h, converting to m/s, we have v = 35 m/s and

Radius of the curve, R = 150 mm = 0.15 m

The centripetal acceleration a(c) is given by the formula = v² / R so that

a(c) = 35² / 0.15

a(c) = 1225 / 0.15

a(c) = 8167 m/s²

The force that causes the acceleration is frictional force = µ m g, where

µ = coefficient of friction

m = the mass of the car and

g = acceleration due to gravity, 9.81

From Newton's law:

µ m g = m a(c) , so that

µ = a(c) / g

µ = 8167 / 9.81

µ = 897

Therefore, the coefficient of static friction must be as big as 897

5 0

3 years ago