what is the relationship between a tie rod and a wale?

Answer:

98 x 100=9,00

Explanation:

scrity añao devid codicie

Answer:

u need to plan it out

Explanation:

u need to plan it out

Answer:

use the turn 1 degrees option and put a repeat loop on it

Explanation:

u can add sound in ur loop

Answer:

this is your answer. if mistake don't mind.

Answer:

I dont know the answer either

Explanation:

Answer:

flux

Explanation:

Answer:

d

Explanation:

OR criteria, AND criteria

In an OR criteria, it doesn't need all the records to be true. Just one record is enough and all other criterion becomes true.

In an AND criteria, it's unlike the OR criteria and works in opposite. It needs every member of the record to be true to be able to adjudge the whole record as true.

And as such, we have

With OR criteria, only one criterion must evaluate true in order for a record to be selected and with AND criteria, all criteria must be evaluate true in order for a record to be selected.

Solution :

Finding the cohesion of the soil(c) using the relation:

[tex]$c = \frac{q_u}{2}$[/tex]

Here, [tex]$q_u$[/tex] is the unconfined compression strength of the soil;

[tex]$c = \frac{800}{2}$[/tex]

   = 400 psf

∴ The cohesion value is greater than 0

So the use of the angle of internal friction is 0

Referring to the table relation between bearing capacity factors and angle of internal friction.

For the angle of inter friction [tex]$0^\circ$[/tex]

    [tex]$N_c = 5.14$[/tex]

   [tex]$N_q = 1.0$[/tex]

   [tex]$N_r = 0$[/tex]

Therefore,

[tex]$q_{ult} = (400 \times 5.14 )+(110 \times 3 \times 1.0)+(0.5 \times 100 \times 13 \times 0)$[/tex]

     =  2386 psf

∴ Allowable bearing capacity [tex]$q_{a} = \frac{Q_{allow}}{A}$[/tex]

                                                     [tex]$=\frac{30}{B^2}$[/tex]

∴ [tex]$q_a = \frac{q_{ult}}{F.O.S}$[/tex]

  [tex]$\frac{30}{B^2} = \frac{2386}{3}$[/tex]

∴ B = 0.2 ft

Therefore, the dimension of the square footing is 0.2 ft x 0.2 ft

                                                                                 [tex]$=0.04 \ ft^2$[/tex]

Answer:

Explanation:

A scenario is an actual sequence of interactions (i.e., an instance) describing one specific situation; a use case is a general sequence of interactions (i.e., a class) describing all possible scenarios associated with a situation. Scenarios are used as examples and for clarifying details with the client. Use cases are used as complete descriptions to specify a user task or a set of related system features.

Answer:

AC

Explanation:

One situation when alternating current would work better than direct current is if the operator is encountering magnetic arc blow.

Current works best when the operator  encounters magnetic arc blow is AC

Magnetic arc blow is simply defined as the arc deflection due to the warping of the magnetic field that is produced by electric arc current.

This is caused as a result of the following;

- if the material being welded has residual magnetism at an intolerable level

- When the weld root is being made, and the welding current is direct current which indicates constant direction and maintains constant polarity (either positive or negative).

Thus, Current works best when the operator  encounters magnetic arc blow is AC

Read more at; brainly.in/question/38789815?tbs_match=1

Answer:

answer is option (d) all of the above

Explanation:

Dempwolf created by John Augustus, Among the most prominent innovative solutions in Southern California Pennsylvania was established by Dempwolf with  brother Reinhardt or uncle's son Frederick entered the company of J.A. Dozens of structures in 10 states were engineered by Dempwolf.

Answer:

hi there

Explanation:

Answer:

19.81 m/s

Explanation:

The total acceleration of the truck (a) is due to the centripetal acceleration and as a result of the linear acceleration. Therefore the total acceleration (a) is given by:

[tex]a^2=a_n^2+a_t^2\\\\where\ a_n=centripetal\ acceleration=\frac{v^2}{r},a_t=linear \ acceleration\\\\But\ since\ the \ speed\ is \ constant, the \ linear \ acceleration(a_t)\ would\ be\ 0.\\\\a^2=a_n^2+a_t^2\\\\a^2=a_n^2\\\\a=a_n=\frac{v^2}{r} \\\\v^2=ar\\\\v=\sqrt{ar} \\\\a=0.4g=0.4*9.81,r=98\ m+2\ m=100\ m\\\\v=\sqrt{0.4*9.81*100} \\\\v=19.81\ m/s[/tex]

Answer:

d.

Explanation:

Answer:

-2 and 6

Explanation:

Let "x" and "y" be 2 numbers.

The sum of the two numbers is 4. The mathematical expression is:

x + y = 4

y = 4 - x   [1]

The sum of their squares minus three times their product is 76. The mathematical expression is:

x² + y² - 3 x y = 76   [2]

If we substitute [1] in [2], we get:

x² + (4 - x)² - 3 x (4 - x) = 76

x² + 16 - 8 x + x² - 12 x + 3 x² = 76

5 x² - 20 x - 60 = 0

We apply the solving formula for second order equations and we get x₁ = 6 and x₂ = -2.

If we replace these x values in [1], we get:

y₁ = 4 - x₁ = 4 - 6 = -2

y₂ = 4 - x₂ = 4 - (-2) = 6

As a consequence, one of the numbers is 6 and the other is -2.

Answer:

Both

Explanation:

Answer:

Explanation:

From the given information:

The equation for applied stress can be expressed as:

[tex]\sigma_{app} = \dfrac{\tau_{CRSS}}{cos \phi \ cos \lambda}[/tex]

where;

[tex]\phi[/tex] = angle between the applied stress [100] and [111]

To determine the [tex]\phi[/tex] and [tex]\lambda[/tex] for the system

Using the equation:

[tex]\phi= cos^{-1}\Big [\dfrac{l_1l_2+m_1m_2+n_1n_2}{\sqrt{(l_1^2+m_1^2+n_1^2)(l_2^2+m_2^2+n_2^2)}}\Big][/tex]

for [100]

[tex]l_1 = 1, m_1 = 0, n_1 = 0[/tex]

for [111]

[tex]l_1 = 1 , m_1 = 1, n_1 = 1[/tex]

Thus;

[tex]\phi= cos^{-1}\Big [\dfrac{1*1+0*1+0*1}{\sqrt{(1^2+0^2+0^2)(1^2+1^2+1^2)}}\Big][/tex]

[tex]\phi= cos^{-1}\Big [\dfrac{1}{\sqrt{(3)}}\Big][/tex]

[tex]\phi= 54.74^0[/tex]

To determine  [tex]\lambda[/tex]  for [tex][1 \overline 1 0][/tex]

where;

for [100]

[tex]l_1 = 1, m_1 = 0, n_1 = 0[/tex]

for [tex][1 \overline 1 0][/tex]

[tex]l_1 = 1 , m_1 = -1, n_1 = 0[/tex]

Thus;

[tex]\lambda= cos^{-1}\Big [\dfrac{1*1+0*1+0*0}{\sqrt{(1^2+0^2+0^2)(1^2+(-1)^2+0^2)}}\Big][/tex]

[tex]\phi= cos^{-1}\Big [\dfrac{1}{\sqrt{(2)}}\Big][/tex]

[tex]\phi= 45^0[/tex]

Thus, the magnitude of the applied stress can be computed as:

[tex]\sigma_{app} = \dfrac{\tau_{CRSS}}{cos \phi \ cos \lambda }[/tex]

[tex]\sigma_{app} = \dfrac{2.00}{cos (54.74) \ cos (45) }[/tex]

[tex]\mathbf{\sigma_{app} =4.89 \ MPa}[/tex]

Answer:

shrinkae limit = 20.35%

shrinkage ratio = 1.45

Explanation:

1. to get the shrinkage limit we would first calculate the moisture content w.

w = (37-28)/28

= 9/28

= 0.3214

then the formula for shrinkage limit is

[tex][w-\frac{(V-Vd}{wd} ]*100[/tex]

w = 0.3214

V = 19.3

Vd = 16

Wd = 28

when we put these values into the formula:

[tex][0.3214-\frac{(19.3-16)}{28} ]*100\\[/tex]

= 20.35%

2. the shrinkage limit = Wd/V

= 28/19.3

= 1.45

Answer:

[tex]u = 260.22m/s[/tex]

[tex]S_{max} = 1141.07ft[/tex]

Explanation:

Given

[tex]S_0 = 89.6ft[/tex] --- Initial altitude

[tex]S_{16.5} = 0ft[/tex] -- Altitude after 16.5 seconds

[tex]a = -g = -32.2ft/s^2[/tex] --- Acceleration (It is negative because it is an upward movement i.e. against gravity)

Solving (a): Final Speed of the rocket

To do this, we make use of:

[tex]S = ut + \frac{1}{2}at^2[/tex]

The final altitude after 16.5 seconds is represented as:

[tex]S_{16.5} = S_0 + ut + \frac{1}{2}at^2[/tex]

Substitute the following values:

[tex]S_0 = 89.6ft[/tex]       [tex]S_{16.5} = 0ft[/tex]     [tex]a = -g = -32.2ft/s^2[/tex]    and [tex]t = 16.5[/tex]

So, we have:

[tex]0 = 89.6 + u * 16.5 - \frac{1}{2} * 32.2 * 16.5^2[/tex]

[tex]0 = 89.6 + u * 16.5 - \frac{1}{2} * 8766.45[/tex]

[tex]0 = 89.6 + 16.5u- 4383.225[/tex]

Collect Like Terms

[tex]16.5u = -89.6 +4383.225[/tex]

[tex]16.5u = 4293.625[/tex]

Make u the subject

[tex]u = \frac{4293.625}{16.5}[/tex]

[tex]u = 260.21969697[/tex]

[tex]u = 260.22m/s[/tex]

Solving (b): The maximum height attained

First, we calculate the time taken to attain the maximum height.

Using:

[tex]v=u + at[/tex]

At the maximum height:

[tex]v =0[/tex] --- The final velocity

[tex]u = 260.22m/s[/tex]

[tex]a = -g = -32.2ft/s^2[/tex]

So, we have:

[tex]0 = 260.22 - 32.2t[/tex]

Collect Like Terms

[tex]32.2t = 260.22[/tex]

Make t the subject

[tex]t = \frac{260.22}{ 32.2}[/tex]

[tex]t = 8.08s[/tex]

The maximum height is then calculated as:

[tex]S_{max} = S_0 + ut + \frac{1}{2}at^2[/tex]

This gives:

[tex]S_{max} = 89.6 + 260.22 * 8.08 - \frac{1}{2} * 32.2 * 8.08^2[/tex]

[tex]S_{max} = 89.6 + 260.22 * 8.08 - \frac{1}{2} * 2102.22[/tex]

[tex]S_{max} = 89.6 + 260.22 * 8.08 - 1051.11[/tex]

[tex]S_{max} = 1141.0676[/tex]

[tex]S_{max} = 1141.07ft[/tex]

Hence, the maximum height is 1141.07ft

Solution :

Assuming that the wire has an uniform temperature, the equivalent convective heat transfer coefficient is given as :

[tex]$h_T= \epsilon \sigma (T_s+T_{surr})(T_s^2 +T^2_{surr})$[/tex]

[tex]$h_T= 0.20 \times 5.67 \times 10^{-8} (1473+323)(1473^2 +323^2)$[/tex]

[tex]$h_T=46.3 \ W/m^2 .K$[/tex]

The total heat transfer coefficient will be :

[tex]$h_T=(250+46.3) \ W/m^2 .K$[/tex]

    [tex]$=296.3 \ W/m^2 .K$[/tex]

Now calculating the maximum volumetric heat generation :

[tex]$\dot {q}_{max}=\frac{2h_t}{r_0}(T_s-T_{\infty})$[/tex]

[tex]$\dot {q}_{max}=\frac{2\times 296.3}{0.0005}(1200-50)$[/tex]

        [tex]$= 1.362 \times 10^{9} \ W/m^3$[/tex]

The heat generation inside the wire is given as :

[tex]$\dot{q} = \frac{I^2R}{V}$[/tex]

Here, R is the resistance of the wire

         V is the volume of the wire

∴ [tex]$\dot{q} = \frac{I^2\left( \rho \times \frac{L}{A} \right)}{A \times L}$[/tex]

     [tex]$=\frac{I^2 \rho}{\left(\frac{\pi}{4}D^2 \right)}$[/tex]

where, ρ is the resistivity.

[tex]$I_{max}= \left(\frac{\dot{q}_{max}}{\rho} \right)^{1/2} \times \frac{\pi}{4}D^2$[/tex]

[tex]$I_{max}= \left(\frac{1.36 \times 10^9}{10^{-6}} \right)^{1/2} \times \frac{3.14}{4}(1 \times 10^{-3})^2$[/tex]

        = 28.96 A

Now considering the relation for the current flow through the finite potential difference.

[tex]$E=I_{max} \times R$[/tex]

[tex]$E=I_{max} \times \rho \times \frac{L}{A}$[/tex]

[tex]$L=\frac{AE}{I_{max} \ \rho}$[/tex]

[tex]$L=\frac{\frac{\pi}{4} \times (1 \times 10^{-3})^2 \times 110}{28.96 \times 10^{-6}}$[/tex]

   = 2.983 m

Now calculating the power rating of the heater:

[tex]$P= E \times I_{max}$[/tex]

[tex]$P= 110 \times 28.96}$[/tex]

   = 3185.6 W

   = 3.1856 kW

Answer:

0.6727

-0.02017

Explanation:

diameter = 15.7

l = 178

E =elastic modulus = 67.1 Gpa

poisson ratio = 0.34

p = force = 49100N

first we calculate the area of the cross section

[tex]A=\frac{\pi }{4} d^{2}[/tex]

[tex]A=\frac{\pi }{4} (15.7)^{2} \\A = \frac{774.683}{4} \\[/tex]

A = 193.6mm²

1. Change in directon of the applied stress

[tex]= \frac{pl}{AE}[/tex]

= 49100*178/193.6*67.1*10³

= [tex]=\frac{8739800}{12990560}[/tex]

δl = 0.6727  mm

2. change in diameter of the specimen

equation for poisson distribution =

m = -(δd/d) / (δl/l)

0.34 = (δd/15.7) / (0.6727/178)

0.34 = (-δd * 178) / 15.7 * 0.6727

0.34 = -178δd / 10.56139

we cross multiply

10.56139*0.34 =-178δd

3.5908726 = -178δd

δd = 3.5908726/-178

δd = -0.02017 mm

the change in dimeter has a negative sign. it decreases

Answer

The skateboarder applies pressure to the trucks and gives/releases pressure on the levers. Second, the wheels and the axles are also examples of simple machines. They help the skater ride, spin, grind, and do a bunch of other radical movements on a skateboard.:

Explanation:

Answer:

The correct option is - A categorical syllogism argues that A and B are both members of C whereas a conditional syllogism argues that if A is true then B is also true.

Explanation:

As,

Categorical syllogisms follow an "If A is part of C, then B is part of C" logic.

Conditional syllogisms follow an "If A is true, then B is true" pattern of logic.

So,

The correct option is - A categorical syllogism argues that A and B are both members of C whereas a conditional syllogism argues that if A is true then B is also true.

Answer:

day if you workout without Zero billing that means you're not sweating. Sweating you're not losing anything that means you have zero outcomes

Explanation:

Solution :

Given :

[tex]$V_1 = 7 \ m/s$[/tex]

Operation time, [tex]$T_1$[/tex] = 3000 hours per year

[tex]$V_2 = 10 \ m/s$[/tex]

Operation time, [tex]$T_2$[/tex] = 2000 hours per year

The density, ρ = [tex]$1.25 \ kg/m^3$[/tex]

The wind blows steadily. So, the K.E. = [tex]$(0.5 \dot{m} V^2)$[/tex]

                                                             [tex]$= \dot{m} \times 0.5 V^2$[/tex]

The power generation is the time rate of the kinetic energy which can be calculated as follows:

Power = [tex]$\Delta \ \dot{K.E.} = \dot{m} \frac{V^2}{2}$[/tex]

Regarding that [tex]$\dot m \propto V$[/tex]. Then,

Power [tex]$ \propto V^3$[/tex] → Power = constant x [tex]$V^3$[/tex]

Since, [tex]$\rho_a$[/tex] is constant for both the sites and the area is the same as same winf turbine is used.

For the first site,

Power, [tex]$P_1= \text{const.} \times V_1^3$[/tex]

            [tex]$P_1 = \text{const.} \times 343 \ W$[/tex]

For the second site,

Power, [tex]$P_2 = \text{const.} \times V_2^3 \ W$[/tex]

           [tex]$P_2 = \text{const.} \times 1000 \ W$[/tex]

Answer:

The answer is "583.042533 MPa".

Explanation:

Solve the following for the real state strain 1:

[tex]\varepsilon_{T}=\In \frac{I_{il}}{I_{01}}[/tex]

Solve the following for the real stress and pressure for the stable.[tex]\sigma_{r1}=K(\varepsilon_{r1})^{n}[/tex]

[tex]K=\frac{\sigma_{r1}}{[\In \frac{I_{il}}{I_{01}}]^n}[/tex]

Solve the following for the true state stress and stress2.

[tex]\sigma_{r2}=K(\varepsilon_{r2})^n[/tex]

     [tex]=\frac{\sigma_{r1}}{[\In \frac{I_{il}}{I_{01}}]^n} \times [\In \frac{I_{i2}}{I_{02}}]^n\\\\=\frac{399 \ MPa}{[In \frac{54.4}{47.7}]^{0.2}} \times [In \frac{57.8}{47.7}]^{0.2}\\\\ =\frac{399 \ MPa}{[ In (1.14046122)]^{0.2}} \times [In (1.21174004)]^{0.2}\\\\ =\frac{399 \ MPa}{[ In (1.02663509)]} \times [In 1.03915873]\\\\=\frac{399 \ MPa}{0.0114161042} \times 0.0166818905\\\\= 399 \ MPa \times 1.46125948\\\\=583.042533\ \ MPa[/tex]