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Mashcka [7]
2 years ago
13

1) What output force (Fout) is produced if the lever arm length (rout) is 100 mm?

Engineering
2 answers:
valentina_108 [34]2 years ago
8 0
Im not sure but I need points I’m sorry
Ierofanga [76]2 years ago
6 0
The length of the arm is the main part of natur
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The Eads Bridge, which crosses the Mississippi River near St Louis, Missouri, was one of the first all steel bridges built in th
MrMuchimi

Answer: At 520 feet between the piers, the center arch of Eads Bridge was the longest rigid span ever built at the time of its construction (only a few suspension bridges had longer spans).

Explanation:

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2 years ago
Are all transmissions fluids interchangeable
KiRa [710]

Answer:

<u>No</u>.

Explanation:

They are not all the same. Moreover, using a fluid that is not approved by the vehicle manufacturer will void the transmission warranty.

6 0
3 years ago
If a hoist lifts a 4500lb load 30ft in 15s, the power delivered to the load is a) 18.00hp b) 9000hp c) 16.36hp d) None of the ab
12345 [234]

Answer:

Explanation:

load = 4500lb                   lift height= 30 ft

time =15 s

velocity=\frac{30}{15} ft/s

velocity=2 ft/s

power = force\times velocity

power={4500}\times2

power= 9000 lb ft/s

1 hp= 550 lb ft/s

power= \frac{9000}{550} =16.36 hp

5 0
3 years ago
The Micro:bit is considered a what?
Scrat [10]

Sorry need points I'm new

3 0
3 years ago
For a cylindrical annulus whose inner and outer surfaces are maintained at 30 ºC and 40 ºC, respectively, a heat flux sensor mea
miskamm [114]

Answer:

k=0.12\ln(r_2/r_1)\frac {W}{ m^{\circ} C}

where r_1 and r_2 be the inner radius, outer radius of the annalus.

Explanation:

Let r_1, r_2 and L be the inner radius, outer radius and length of the given annulus.

Temperatures at the inner surface, T_1=30^{\circ}C\\ and at the outer surface, T_2=40^{\circ}C.

Let q be the rate of heat transfer at the steady-state.

Given that, the heat flux at r=3cm=0.03m is

40 W/m^2.

\Rightarrow \frac{q}{(2\pi\times0.03\times L)}=40

\Rightarrow q=2.4\pi L \;W

This heat transfer is same for any radial position in the annalus.

Here, heat transfer is taking placfenly in radial direction, so this is case of one dimentional conduction, hence Fourier's law of conduction is applicable.

Now, according to Fourier's law:

q=-kA\frac{dT}{dr}\;\cdots(i)

where,

K=Thermal conductivity of the material.

T= temperature at any radial distance r.

A=Area through which heat transfer is taking place.

Here, A=2\pi rL\;\cdots(ii)

Variation of temperature w.r.t the radius of the annalus is

\frac {T-T_1}{T_2-T_1}=\frac{\ln(r/r_1)}{\ln(r_2/r_1)}

\Rightarrow \frac{dT}{dr}=\frac{T_2-T_1}{\ln(r_2/r_1)}\times \frac{1}{r}\;\cdots(iii)

Putting the values from the equations (ii) and (iii) in the equation (i), we have

q=\frac{2\pi kL(T_1-T_2)}{\LN(R_2/2_1)}

\Rightarrow k= \frac{q\ln(r_2/r_1)}{2\pi L(T_2-T_1)}

\Rightarrow k=\frac{(2.4\pi L)\ln(r_2/r_1)}{2\pi L(10)} [as q=2.4\pi L, and T_2-T_1=10 ^{\circ}C]

\Rightarrow k=0.12\ln(r_2/r_1)\frac {W}{ m^{\circ} C}

This is the required expression of k. By putting the value of inner and outer radii, the thermal conductivity of the material can be determined.

7 0
3 years ago
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