LUNES MARTES MIÉRCOLES JUEVES VIERNES SÁBADO DOMINGO

The Micro:bit is considered a what?

Scrat [10]

Sorry need points I'm new

3 0

2 years ago

Which of the following conditions is a good sign of minor

Kryger [21]

Answer:

Explanation:

d

7 0

3 years ago

A cylindrical metal specimen having an original diameter of 12.8 mm and gauge length of 50.80 mm is pulled in tension until frac

Sedaia [141]

Answer:

%Reduction in area = 73.41%

%Reduction in elongation = 42.20%

Explanation:

Given

Original diameter = 12.8 mm

Gauge length = 50.80mm

Diameter at the point of fracture = 6.60 mm (0.260 in.)

Fractured gauge length = 72.14 mm.

%Reduction in Area is given as:

((do/2)² - (d1/2)²)/(do/2)²

Calculating percent reduction in area

do = 12.8mm, d1 = 6.6mm

So,

%RA = ((12.8/2)² - 6.6/2)²)/(12.8/2)²

%RA = 0.734130859375

%RA = 73.41%

Calculating percent reduction in elongation

%Reduction in elongation is given as:

((do) - (d1))/(d1)

do = 72.14mm, d1 = 50.80mm

So,

%RA = ((72.24) - (50.80))/(50.80)

%RA = 0.422047244094488

%RA = 42.20%

3 0

3 years ago

A composite wall is to be used to insulate a freezer chamber at -350C. Two insulating materials are to be used with conductiviti

choli [55]

Answer:

thickness1=1.4m

thickness2=2.2m

convection coefficient=0.33W/m^2K

Explanation:

you must use this equation to calculate the thickness:

L=K(T2-T1)/Q

L=thickness

T=temperature

Q=heat

L1=0.04*(0--350)/10=1.4m

L2=0.1(220-0)/10=2.2m

Then use this equation to calculate the convective coefficient

H=Q/(T2-T1)

H=10/(250-220)=0.33W/m^2K

7 0

3 years ago

An aluminum block weighing 28 kg initially at 140°C is brought into contact with a block of iron weighing 36 kg at 60°C in an in

Anika [276]

Answer:

Equilibrium Temperature is 382.71 K

Total entropy is 0.228 kJ/K

Solution:

As per the question:

Mass of the Aluminium block, M = 28 kg

Initial temperature of aluminium, $T_{a} = 140^{\circ}C$ = 273 + 140 = 413 K

Mass of Iron block, m = 36 kg

Temperature for iron block, $T_{i} = 60^{\circ}C$ = 273 + 60 = 333 K

At 400 k

Specific heat of Aluminium, $C_{p} = 0.949\ kJ/kgK$

At room temperature

Specific heat of iron, $C_{p} = 0.45\ kJ/kgK$

Now,

To calculate the final equilibrium temperature:

Amount of heat loss by Aluminium = Amount of heat gain by Iron

$MC_{p}\Delta T = mC_{p}\Delta T$

$28\times 0.949(140 - T_{e}) = 36\times 0.45(T_{e} - 60)$

Thus

$T_{e} = 109.71^{\circ}C$ = 273 + 109.71 = 382.71 K

where

$T_{e}$ = Equilibrium temperature

Now,

To calculate the changer in entropy:

$\Delta s = \Delta s_{a} + \Delta s_{i}$

Now,

For Aluminium:

$\Delta s_{a} = MC_{p}ln\frac{T_{e}}{T_{i}}$

$\Delta s_{a} = 28\times 0.949\times ln\frac{382.71}{413} = - 2.025\ kJ/K$

For Iron:

$\Delta s_{i} = mC_{i}ln\frac{T_{e}}{T_{i}}$

$\Delta s_{a} = 36\times 0.45\times ln\frac{382.71}{333} = 2.253\ kJ/K$

Thus

$\Delta s =-2.025 + 2.253 = 0.228\ kJ/K$

6 0

2 years ago