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3 years ago

Pls help me answer my module

Engineering

1 answer:

Otrada [13]3 years ago

4 0

Answer:

Hand tools based on job requirement and its importance and the classification of hand tools according to its function and its importance are discussed below in details.

Explanation:

Hand tools based on work requirement is essential because Every tool is specifically invented for a particular purpose, so picking the accurate tool will also reduce the amount of energy needed to get work done right without causing injury or harm to either the tools or the exterior being worked on.

classifying of hand tools: wrenches, screwdrivers, cutters, striking tools, hammer tool or struck, pliers, vise, clamps, snips, saws, drills, and knives.

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A rectangular open box, 25 ft by 10 ft in plan and 12 ft deep weighs 40 tons. Sufficient amount of stones is placed in the box a

dimulka [17.4K]

Answer:

44.95 tonnes

Explanation:

According to principle of buoyancy the object will just sink when it's weight is more than the weight of the liquid it displaces

It is given that empty weight of box = 40 tons

Let the mass of the stones to be placed be = M tonnes

Thus the combined mass of box and stones = (40+M) tonnes..........(i)

Since the box will displace water equal to it's volume V we have $volume of box = 25ft*10ft*12ft= 3000ft^{3}$

$Volume= 84.95m^{3}$

$Since 1ft^{3} =0.028m^{3}$

Now the weight of water displaced = $Weight =\rho \times Volumewhererho$ is density of water = 1000kg/ $m^{3}$

Thus weight of liquid displaced = $\frac{84.95X1000}{1000}tonnes=84.95 tonnes$ ..................(ii)

Equating i and ii we get

40 + M = 84.95

thus Mass of stones = 44.95 tonnes

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

Amiraneli [1.4K]

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 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

Rear axles are usually lubricated by the same gear oil that lubricates the differential. True or false

KIM [24]

Answer:

true

Explanation:

8 0

2 years ago

A rich industrialist was found murdered in his house. The police arrived at the scene at 11:00 PM. The temperature of the corpse

d1i1m1o1n [39]

Answer:

The dude was killed around 6:30PM

Explanation:

Newton's law of cooling states:

$T = T_m + (T_0-T_m)e^{kt}$

where,

$T_0$ = initial temp

$T_m$ = temp of room

$T$ = temp after t hours

$k$ = how fast the temp is changing

$t$ = time (hours)

$T_0 = 31$ because the body was initlally 31ºC when the police found it

$T_m = 22$ because that was the room temp

$T = 30$ because the body temp drop to 30ºC after 1 hour

t = 1 because that's the time it took for the body temp to drop to 30ºC

$k=???$ we don't know k so we must solve for this

rearrange the equation to solve for k

$T = T_m + (T_0-T_m)e^{kt}$

$T - T_m= (T_0-T_m)e^{kt}$

$\frac{T - T_m}{(T_0-T_m)}= e^{kt}$

$ln(\frac{T - T_m}{T_0-T_m})=kt$

$\frac{ln(\frac{T - T_m}{T_0-T_m})}{t}=k$

plug in the numbers to solve for k

$k = \frac{ln(\frac{T - T_m}{T_0-T_m})}{t}$

$k = \frac{ln(\frac{30 - 22}{31-22})}{1}$

$k=ln(\frac{8}{9})$

Now that we know the value for k, we can find the moment the murder occur. A crucial information that the question left out is the temperature of a human body when they're still alive. A living human body is about 37ºC. We can use that as out initial temperature to solve this problem because we can assume that the freshly killed body will be around 37ºC.

$T_0 = 37$ because the body was 37ºC right after being killed