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3 years ago

9

A particle moves along a straight line with a velocity V=(200s) mm/s, where s is in millimeters. Determine acceleration of the p

article at S = 2000mm.

1 answer:

iragen [17]3 years ago

8 0

Answer:

200 mm/s²

Explanation:

See it in the pic

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Zanzabum

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3 0

3 years ago

For a steel alloy it has been determined that a carburizing heat treatment of 15 h duration will raise the carbon concentration

Free_Kalibri [48]

Answer:

135 hour

Explanation:

It is given that a carburizing heat treatment of 15 hour will raise the carbon concentration by 0.35 wt% at a point of 2 mm from the surface.

We have to find the time necessary to achieve the same concentration at a 6 mm position.

we know that $\frac{x_1^2}{Dt}=constant$ where x is distance and t is time .As the temperature is constant so D will be also constant

So $\frac{x_1^2}{t}=constant$

then $\frac{x_1^2}{t_1}=\frac{x_2^2}{t_2}$ we have given $x_1=2 mm\ ,t_1=15 hour\ ,x_2=6\ mm$ and we have to find $t_2$ putting all these value in equation

$\frac{2^2}{15}=\frac{6^2}{t_2}$

so $t_2=135\ hour$

5 0

3 years ago

A hydraulic cylinder is to be used to move a workpiece in a manufacturing operation through a distance of 50 mm in 10 s. A force

svetlana [45]

Answer:

The answer to this question is 1273885.3 ∅

Explanation:

<em>The first step is to determine the required hydraulic flow rate liquid if working pressure and if a cylinder with a piston diameter of 100 mm is available.</em>

<em>Given that,</em>

<em>The distance = 50mm</em>

<em>The time t =10 seconds</em>

<em>The force F = 10kN</em>

<em>The piston diameter is = 100mm</em>

<em>The pressure = F/A</em>

<em> 10 * 10^3/Δ/Δ </em>

<em> P = 1273885.3503 pa</em>

<em>Then</em>

<em>Power = work/time = Force * distance /time</em>

<em> = 10 * 1000 * 0.050/10</em>

<em>which is =50 watt</em>

<em>Power =∅ΔP</em>

<em>50 = 1273885.3 ∅</em>

5 0

3 years ago

Koch traded Machine 1 for Machine 2 when the fair market value of both machines was $60,000. Koch originally purchased Machine 1

Mariana [72]

Answer:

Koch's adjusted basis in machine 2 after the exchange is $60,000

Explanation:

given data

fair market value = $60,000

originally purchased Machine 1 = $76,900

Machine 1 adjusted basis = $40,950

Machine 2 seller purchase = $64,050

Machine 2 adjusted basis = $55,950

solution

As he exchanged machine for another at $60,000

and this exchanged in fair market

so adjusted basis = $50,000

Adjusted basis is the price of the item that affects the factors that are considered price. These factors usually include taxes, depreciation value, and other costs of acquiring and maintaining a given item. Adjusted basis is important so the right amount to sell

Adjusted basis increases when a person deducts expenses from factor taxes and operating statements

so Koch's adjusted basis in machine 2 after the exchange is $60,000

3 0

3 years ago

Water flows through a pipe at an average temperature of T[infinity] = 70°C. The inner and outer radii of the pipe are r1 = 6 cm

Paul [167]

Answer:

The differential equation and the boundary conditions are;

A) -kdT(r1)/dr = h[T∞ - T(r1)]

B) -kdT(r2)/dr = q'_s = 734.56 W/m²

Explanation:

We are given;

T∞ = 70°C.

Inner radii pipe; r1 = 6cm = 0.06 m

Outer radii of pipe;r2 = 6.5cm=0.065 m

Electrical heat power; Q'_s = 300 W

Since power is 300 W per metre length, then; L = 1 m

Now, to the heat flux at the surface of the wire is given by the formula;

q'_s = Q'_s/A

Where A is area = 2πrL

We'll use r2 = 0.065 m

A = 2π(0.065) × 1 = 0.13π

Thus;

q'_s = 300/0.13π

q'_s = 734.56 W/m²

The differential equation and the boundary conditions are;

A) -kdT(r1)/dr = h[T∞ - T(r1)]

B) -kdT(r2)/dr = q'_s = 734.56 W/m²

6 0

3 years ago