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arlik [135]
3 years ago
9

Determine the maximum mass of the crate so

Engineering
1 answer:
Tpy6a [65]3 years ago
7 0

Answer:

293 kg

Explanation:

Let's say the tension in each cable is Tb, Tc, and Td.

First, find the length of cable AD:

r = √(2² + 2² + 1²)

r = 3

Using similar triangles:

Tdx = 2/3 Td

Tdy = 2/3 Td

Tdz = 1/3 Td

Sum of the forces in the x direction:

∑F = ma

Tb − 2/3 Td = 0

Td = 3/2 Tb

Sum of the forces in the y direction:

∑F = ma

2/3 Td − Tc = 0

Td = 3/2 Tc

Sum of the forces in the z direction:

∑F = ma

1/3 Td − mg = 0

Td = 3mg

From the first two equations, we know Td is greater than Tb or Tc.  So we need to set Td to 8.6 kN, or 8600 N.

8600 N = 3mg

m = 8600 N / (3 × 9.8 m/s²)

m ≈ 292.5 kg

Rounded to three significant figures, the maximum mass of the crate is 293 kg.

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Greeley [361]

Answer:

The essence including its problem is listed throughout the clarification section following.

Explanation:

Projects build deliverable that seem to be the products of the venture or indeed the implementation of the project. This ensures that perhaps the agile methodology may be as broad as either the goal of the study itself as well as the coverage that would be part of a much larger venture.

For every other production to have been marked as "deliverable" within the same project, this should satisfy a few eligibility requirements:

  • It should be within the development of the work.
  • The interested parties-external or internal-must consent to the above. This is perhaps the product of hard effort.

So that the above seems to be the right answer.

7 0
3 years ago
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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
DRIVERS ED
forsale [732]

Answer:

b

Explanation:

only if there signal is turned on

8 0
3 years ago
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Plot the function for . Notice that the function has two vertical asymptotes. Plot the function by dividing the domain of x into
elena-s [515]
This is a very very difficult one for me, let me get back to you with the proper answer.
8 0
3 years ago
Estimate the endurance strength, Se, of a 37.5-mm- diameter rod of AISI 1040 steel having a machined finish and heat-treated to
7nadin3 [17]

Answer:

endurance length is 236.64 MPa

Explanation:

data given:

d = 37.5 mm

Sut = 760MPa

endurance limit is

Se = 0.5 Sut

   = 0.5*760 = 380 MPa

surface factor is

Ka = a*Sut^b

where

Sut is ultimate strength

for AISI 1040 STEEL

a = 4.51, b = -0.265

Ka = 4.51*380^{-0.265}

Ka = 0.93

size factor is given as

Kb =1.29 d^{-0.17}

Kb = 0.669

Se = Sut *Ka*Kb

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Se = 236.64 MPa

therefore endurance length is 236.64 MPa

4 0
3 years ago
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