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Papessa [141]
3 years ago
7

A brass alloy is known to have a yield strength of 275 MPa (40,000 psi), a tensile strength of 380 MPa (55,000 psi), and an elas

tic modulus of 103 GPa (15.0×106 psi). A cylindrical specimen of this alloy 5.4 mm (0.21 in.) in diameter and 225 mm (8.87 in.) long is stressed in tension and found to elongate 6.8 mm (0.27 in.). On the basis of the information given, is it possible to compute the magnitude of the load that is necessary to produce this change in length? If not, explain why.
Engineering
1 answer:
zlopas [31]3 years ago
3 0

Answer:

The magnitude of the load can be computed  because it is mandatory in order to produce the change in length ( elongation )

Explanation:

Yield strength = 275 Mpa

Tensile strength = 380 Mpa

elastic modulus = 103 GPa

The magnitude of the load can be computed  because it is mandatory in order to produce the change in length ( elongation ) .

Given that the yield strength, elastic modulus and strain that is experienced by the test spectrum are given

strain = yield strength / elastic modulus

          = 0.0027

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It is said that Archimedes discovered his principle during a bath while thinking about how he could determine if KingHiero‘s cro
Rudiy27

Answer:

the crown is false densty= 12556kg/m^3[/tex]

Explanation:

Hello! The first step to solve this problem is to find the mass of the crown, this is found using the weight of the crown in the air by means of the equation for the weight.

W=mg

W=weight(N)=31.4N

M=Mass

g=gravity=9.81m/S^2

solving for M

m=W/g

m=\frac{31.4N}{9.81m/S^2}=3.2kg

The second step is find the volume of crown  remembering that when an object is weighed in the water the result is the subtraction between the weight of the object and the buoyant force of the water which is the product of the volume of the crown by gravity by density of water

F=mg-\alpha  V g

Where

F=weight in water=28.9N

m=mass of crown=3.2kg

g=gravity=9.81m/S^2

α=density of water=1000kg/m^3

V= crown´s volume

solving for V

V=\frac{mg-F }{g \alpha } =\frac{(3.2)(9.81)-28.9}{9.81(1000)} =0.000254m^3

finally, we remember that the density is equal to the index between mass and volume

\alpha =\frac{m}{v} =\frac{3.2}{0.000254} =12556kg/m^3

To determine the density of the crown without using the weight in the water and with a bucket we can use the following steps.

1.weigh the crown in the air and find the mass

2. put water in a cylindrical bucket and measure its height with a ruler.

3. Put the crown in the bucket and measure the new water level with a ruler.

4. Subtract the heights, and find the volume of a cylinder knowing the difference in heights and the diameter of the bucket, in order to determine the volume of the crown.

5. find density by dividing mass by volume

7 0
3 years ago
Question 9.1 from the textbook. Consider the following workload: Process Burst Time Priority Arrival Time P1 50 4 0 P2 20 1 20 P
Marizza181 [45]

Answer:

Explanation:

The schedule using shortest remaining time, non-preemptive priority and round Robin with quantum number 30 is shown in the attached file, please kindly go through it to access the answer.

5 0
4 years ago
At what times should you use your headlights?
SVETLANKA909090 [29]

Answer:

Headlights are required to be used 1/2 hour after sunset to 1/2 hour before sunrise, when windshield wipers are being used, when visibility is less than 1000 feet, or when there is insufficient light or adverse weather.

Explanation:

hope this helps

8 0
3 years ago
Read 2 more answers
1. A 260 ft (79.25 m) length of size 4 AWG uncoated copper wire operating at a tem-
Murljashka [212]

A 260 ft (79.25m) length of size 4 AWG uncoated copper wire operating at a temperature of 75°c has a resistance of 0.0792 ohm.

Explanation:

From the given data the area of size 4 AWG of the code is 21.2 mm², then K is the Resistivity of the material at 75°c is taken as ( 0.0214 ohm mm²/m ).

To find the resistance of 260 ft (79.25 m) of size 4 AWG,

R= K * L/ A

K = 0.0214 ohm mm²/m

L = 79.25 m

A = 21.2 mm²

R = 0.0214 * \frac{79.25}{21.2}

  = 0.0214 * 3.738

  = 0.0792 ohm.

Thus the resistance of uncoated copper wire is 0.0792 ohm

5 0
4 years ago
The roof of a refrigerated truck compartment consists of a layer of foamed urethane insulation (t2 = 21 mm, ki = 0.026 W/m K) be
lakkis [162]

Answer:

Tso = 28.15°C

Explanation:

given data

t2 = 21 mm

ki = 0.026 W/m K

t1 = 9 mm

kp = 180 W/m K

length of the roof is L = 13 m

net solar radiation into the roof = 107 W/m²

temperature of the inner surface Ts,i = -4°C

air temperature is T[infinity] = 29°C

convective heat transfer coefficient h = 47 W/m² K

solution

As when energy on the outer surface at roof of a refrigerated truck that is balance as

Q = \frac{T \infty - T si }{\frac{1}{hA}+\frac{t1}{AKp}+\frac{t2}{AKi}+\frac{t1}{aKp}}       .....................1

Q = \frac{T \infty - Tso}{\frac{1}{hA}}                         .....................2

now we compare both equation 1 and 2 and put here value

\frac{29-(-4)}{\frac{1}{47}+\frac{2\times0.009}{180}+\frac{0.021}{0.026}} = \frac{29-Tso}{\frac{1}{47}}            

solve it and we get

Tso = 28.153113

so Tso = 28.15°C

3 0
3 years ago
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