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2 years ago

What is a table saw for

Engineering

2 answers:

bezimeni [28]2 years ago

7 0

Cutting the table because a table saw is for cutting tables.......

svet-max [94.6K]2 years ago

5 0

Answer:

a table

Explanation:

because you can saw the table

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In a much smaller model of the Gizmo apparatus, a 5 kg mass drops 86 mm (0.086 m) and raises the temperature of 1 gram of water

Orlov [11]

Answer:

The amount of energy transferred to the water is 4.214 J

Explanation:

The given parameters are;

The mass of the object that drops = 5 kg

The height from which it drops = 86 mm (0.086 m)

The potential energy P.E. is given by the following formula

P.E = m·g·h

Where;

m = The mass of the object = 5 kg

g = The acceleration de to gravity = 9.8 m/s²

h = The height from which the object is dropped = 0.086 m

Therefore;

P.E. = 5 kg × 9.8 m/s² × 0.086 m = 4.214 J

Given that the potential energy is converted into heat energy, that raises the 1 g of water by 1°C, we have;

The amount of energy transferred to the water = The potential energy, P.E. = 4.214 J.

6 0

2 years ago

A 03-series cylindrical roller bearing with inner ring rotating is required for an application in which the life requirement is

-BARSIC- [3]

Answer:

$\mathbf{C_{10} = 137.611 \ kN}$

Explanation:

From the information given:

Life requirement = 40 kh = 40 $40 \times 10^{3} \ h$

Speed (N) = 520 rev/min

Reliability goal $(R_D)$ = 0.9

Radial load $(F_D)$ = 2600 lbf

To find C10 value by using the formula:

$C_{10}=F_D\times \pmatrix \dfrac{x_D}{x_o +(\theta-x_o) \bigg(In(\dfrac{1}{R_o}) \bigg)^{\dfrac{1}{b}}} \end {pmatrix} ^{^{^{\dfrac{1}{a}}$

where;

$x_D = \text{bearing life in million revolution} \\ \\ x_D = \dfrac{60 \times L_h \times N}{10^6} \\ \\ x_D = \dfrac{60 \times 40 \times 10^3 \times 520}{10^6}\\ \\ x_D = 1248 \text{ million revolutions}$

$\text{The cyclindrical roller bearing (a)}= \dfrac{10}{3}$

The Weibull parameters include:

$x_o = 0.02$

$(\theta - x_o) = 4.439$

$b= 1.483$

∴

Using the above formula:

$C_{10}=1.4\times 2600 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{1}{\dfrac{10}{3}}}$

$C_{10}=3640 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{3}{10}}$

$C_{10} = 3640 \times \bigg[\dfrac{1248}{0.9933481582}\bigg]^{\dfrac{3}{10}}$

$C_{10} = 30962.449 \ lbf$

Recall that:

1 kN = 225 lbf

∴

$C_{10} = \dfrac{30962.449}{225}$

$\mathbf{C_{10} = 137.611 \ kN}$

7 0

2 years ago

The Energy Losses Associated with Valves and Fittings: a)- are generally associated with a K factor b)- are generally associated

madam [21]

Answer:

a)Are generally associated with factor.

Explanation:

We know that losses are two types

1.Major loss :Due to friction of pipe surface

2.Minor loss :Due to change in the direction of flow

As we know that when any hindrance is produced during the flow of fluid then it leads to generate the energy losses.If flow is along uniform diameter pipe then there will not be any loss but if any valve and fitting placed is the path of fluid flow due to this direction of fluid flow changes and it produce losses in the energy.

Lot' of experimental data tell us that loss in the energy due to valve and fitting are generally associated with K factor.These losses are given as

$Losses=K\dfrac{V^2}{2g}$

8 0

3 years ago

Consider a Carnot heat pump cycle executed in a steady-flow system in the saturated mixture region using R-134a flowing at a rat

attashe74 [19]

Answer:

7.15

Explanation:

Firstly, the COP of such heat pump must be measured that is,

$COP_{HP}=\frac{T_H}{T_H-T_L}$

Therefore, the temperature relationship, $T_H=1.15\;T_L$

Then, we should apply the values in the COP.

$=\frac{1.15\;T_L}{1.15-1}$

$=7.67$

The number of heat rejected by the heat pump must then be calculated.

$Q_H=COP_{HP}\times W_{nst}$

$=7.67\times5=38.35$

We must then calculate the refrigerant mass flow rate.

$m=0.264\;kg/s$

$q_H=\frac{Q_H}{m}$

$=\frac{38.35}{0.264}=145.27$

The $h_g$ value is 145.27 and therefore the hot reservoir temperature is 64° C.

The pressure at 64 ° C is thus 1849.36 kPa by interpolation.

And, the lowest reservoir temperature must be calculated.

$T_L=\frac{T_H}{1.15}$

$=\frac{64+273}{1.15}=293.04$

$=19.89\°C$

the lowest reservoir temperature = 258.703 kpa

So, the pressure ratio should be = 7.15

8 0

2 years ago

Water vapor at 1.0 MPa, 300°C enters a turbine operating at steady state and expands to 15 kPa. The work developed by the turbin

Andre45 [30]

Answer:

a) isentropic efficiency = 84.905%

b) rate of entropy generation = .341 kj/(kg.k)

Please kindly see explaination and attachment.

Explanation:

a) isentropic efficiency = 84.905%

b) rate of entropy generation = .341 kj/(kg.k)

The Isentropic efficiency of a turbine is a comparison of the actual power output with the Isentropic case.

Entropy can be defined as the thermodynamic quantity representing the unavailability of a system's thermal energy for conversion into mechanical work, often interpreted as the degree of disorder or randomness in the system.

Please refer to attachment for step by step solution of the question.

5 0

3 years ago