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4 years ago

14

Cutting and abrasive machining are the two major material processes. List the differences between Cutting tool and Abrasive mach

ining tool.

1 answer:

STatiana [176]4 years ago

7 0

Answer:

Explained

Explanation:

Cutting tools:

1. Cutting tools can either be single point or multi point.

2. Cutting tools can have variety of material depending on use like ceramics, diamonds, metals, CBN, etc.

3.Cutting tools have definite shapes and geometry.

Abrasive machining tools

1. Abrasive tools are always multi point tools.

2. Abrasive tools composed of abrasives bounded in medium of resin or metal.

3. They do not have definite geometry of shape

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Answer:thx

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3 years ago

Read 2 more answers

Plz answer all of these questions!

statuscvo [17]

Answer: all you need to to is go to help me with this question.cm

Explanation:

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3 years ago

Technician A states that air tools generally produce more noise than electric tools, so wear ear protection when using air tools

dusya [7]

Answer:

Both Technician A and Technician B are correct

Explanation:

Air tools and electric tools are both power tools as they are used to make work easier. Air tools generally use an air compressor that powers the motor of the tool making it possible to use it while electric tools as the name implies are powered by an electric source which in this case is batteries. An example of an air tool is the nail gun which can be used by furniture makers to drive nails and they are often louder than electric tools because of vibrations caused by the compressor making it necessary to use ear protection when using the tool for ear safety.

Technician B is also correct because it is always advisable to use impact sockets while using impact guns due to the ability of the impact sockets to withstand the force caused by operating impact guns and make work neater when nuts and bolts are being loosened or tightened.

5 0

3 years ago

The 30-kg gear is subjected to a force of P=(20t)N where t is in seconds. Determine the angular velocity of the gear at t=4s sta

tatyana61 [14]

Answer:

$\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}$

Explanation:

Previous concepts

Angular momentum. If we consider a particle of mass m, with velocity v, moving under the influence of a force F. The angular momentum about point O is defined as the “moment” of the particle’s linear momentum, L, about O. And the correct formula is:

$H_o =r x mv=rxL$

Applying Newton’s second law to the right hand side of the above equation, we have that r ×ma = r ×F =

MO, where MO is the moment of the force F about point O. The equation expressing the rate of change of angular momentum is this one:

MO = H˙ O

Principle of Angular Impulse and Momentum

The equation MO = H˙ O gives us the instantaneous relation between the moment and the time rate of change of angular momentum. Imagine now that the force considered acts on a particle between time t1 and time t2. The equation MO = H˙ O can then be integrated in time to obtain this:

$\int_{t_1}^{t_2}M_O dt = \int_{t_1}^{t_2}H_O dt=H_0t2 -H_0t1$

Solution to the problem

For this case we can use the principle of angular impulse and momentum that states "The mass moment of inertia of a gear about its mass center is $I_o =mK^2_o =30kg(0.125m)^2 =0.46875 kgm^2$ ".

If we analyze the staritning point we see that the initial velocity can be founded like this:

$v_o =\omega r_{OIC}=\omega (0.15m)$

And if we look the figure attached we can use the point A as a reference to calculate the angular impulse and momentum equation, like this:

$H_Ai +\sum \int_{t_i}^{t_f} M_A dt =H_Af$

$0+\sum \int_{0}^{4} 20t (0.15m) dt =0.46875 \omega + 30kg[\omega(0.15m)](0.15m)$

And if we integrate the left part and we simplify the right part we have

$1.5(4^2)-1.5(0^2) = 0.46875\omega +0.675\omega=1.14375\omega$

And if we solve for $\omega$ we got:

$\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}$

8 0

3 years ago

Cool water at 15°C is throttled from 5(atm) to 1(atm), as in a kitchen faucet. What is the temperature change of the water? What

Tresset [83]

Answer:

the lost work per kilogram of water for this everyday household happening = 0.413 kJ/kg

Explanation:

Given that:

Initial Temperature $T_1$ = 15°C

Initial Pressure $P_1$ = 5 atm

Final Pressure $P_2$ = 1 atm

Data obtain from steam tables of saturated water at 15°C are as follows:

Specific volume v = 1.001 cm³/gm

The change in temperature = 2°C

Specific heat of water = 4.19 J/gm.K

volume expansivity β = 1.5 × 10⁻⁴ K⁻¹

The expression to determine the change in temperature can be given as :

$\delta \ T = \frac{-V (1- \beta \ T}{C_p} * \delta \ P ( \frac{1}{9.87} \ \frac{J}{cm^3/atm})$ $\delta \ T = \frac{-1.001 \frac{cm^3}{gm} (1- 1.5*10^{-4} \ K^{-1} )*2}{4.19 \ \frac{J}{gm.K}} *(5-1)atm ( \frac{1}{9.87} \ \frac{J}{cm^3/atm})$

Δ T = 0.093 K

Now; we can calculate the lost work bt the formula:

$W_{lost} = T_{surr} *S$

where ;

$T_{surr}$ is the temperature of the surrounding. = 20°C = (20+273.15)K = 293.15 K

From above the change in entropy is:

$\delta \ S = C_p \ In (\frac{T+ \delta \ T }{T}) * \beta V \delta P$

$\delta \ S = 4.19* \ In (\frac{288.15+0.093 }{288.15}) - 1.5*10^{-4} * 1.001 (5-1)* (\frac{1}{9.87})$

$\delta \ S =1.408*10^{-3} \ J/gm.K$

$W_{lost} = T_{surr} *S$

$W_{lost} = 293.15* 1.408*10^{-3} \ J/gm.K$

$W_{lost} = 0.413 \ kJ/kg$

Thus, the lost work per kilogram of water for this everyday household happening = 0.413 kJ/kg

6 0

4 years ago