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sveta [45]
3 years ago
12

TP-6 What should you do when fueling an outboard boat with a portable tank?

Engineering
1 answer:
nikklg [1K]3 years ago
4 0
Portable is the same one u only knew
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Lydia is the CEO for a large pharmaceutical manufacturer. Her company is in the final stages of FDA
weqwewe [10]
OSHA inspections are generally unannounced. In fact, except in four exceptional circumstances when advance notice may be given.


It is a criminal offense for any person to give unauthorized advance notice of an OSHA inspection.
5 0
2 years ago
**Please Help. ASAP**
natima [27]

Answer:

The answer is below

Explanation:

1)

\frac{v-u}{a} =t\\\\Making \ v\ the \ subject\ of\ formula:\\\\First \ cross-multiply:\\\\v-u=at\\\\add\ u\ to \ both\ sides:\\\\v-u+u=at+u\\\\v=u+at

2)

\frac{y-x^2}{x}=3z\\ \\Making\ y\ the\ subject\ of\ formula:\\\\First \ cross \ multiply:\\\\y-x^2=3xz\\\\y=3xz+x^2\\\\y=x(x+3z)

3)

x+xy=y\\\\Making\ x\ the\ subject\ of\ formula:\\\\x(1+y)=y\\\\Divide\ through\ by\ 1+y\\\\\frac{x(1+y)}{1+y} =\frac{y}{1+y} \\\\x=\frac{y}{1+y}

4)

x+y=xy\\\\Making\ x\ the\ subject\ of\ formula:\\\\Subtract\ x\ from \ both\ sides:\\\\x+y-x=xy-x\\\\y=xy-x\\\\y=x(y-1)\\\\Divide\ through\ by \ y-1\\\\\frac{y}{y-1} =\frac{x(y-1)}{y-1}\\ \\x=\frac{y}{y-1}

5)

x=y+xy\\\\Making\ x\ the\ subject\ of\ formula:\\\\Subtract\ xy\ from \ both\ sides:\\\\x-xy=y+xy-xy\\\\x-xy=y\\\\x(1-y)=y\\\\Divide\ through\ by \ 1-y\\\\\frac{x(1-y)}{1-y} =\frac{y}{1-y}\\ \\x=\frac{y}{1-y}

6)

E=\frac{1}{2}mv^2-\frac{1}{2}mu^2\\  \\Making\ u\ the\ subject \ of\ formula:\\\\Multiply \ through\ by \ 2\\\\2E=mv^2-mu^2\\\\mu^2=mv^2-2E\\\\Divide\ through\ by\ m:\\\\u^2=\frac{mv^2-2E}{m}\\ \\Take\ square\ root\ of \ both\ sides:\\\\u=\sqrt{\frac{mv^2-2E}{m}}

7)

\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\\  \\Making\ y\ the\ subject \ of\ formula:\\\\\frac{x^2}{a^2}-1=\frac{y^2}{b^2}\\\\Multiply\ through\ by\ b^2\\\\b^2(\frac{x^2}{a^2} -1)=y^2\\\\Take\ square\ root\ of\ both\ sides:\\\\y=\sqrt{b^2(\frac{x^2}{a^2} -1)}

8)

ay^2=x^3\\\\Make\ y\ the\ subject\ of\ formula:\\\\Divide\ through\ by\ a:\\\\y^2=\frac{x^3}{a}\\ \\Take\ square\ root\ of\ both\ sides:\\\\y=\sqrt{\frac{x^3}{a}} \\

4 0
3 years ago
Write a statement to print the data members of InventoryTag. End with newline. Ex: if itemID is 314 and quantityRemaining is 500
Advocard [28]

Answer:

#include <stdio.h>

typedef struct InventoryTag_struct {

int itemID;

int quantityRemaining;

} InventoryTag;

int main(void) {

InventoryTag redSweater;

redSweater.itemID = 314;

redSweater.quantityRemaining = 500;

/* Your solution goes here */

printf("Inventory ID: %d, Qty: %d\n",redSweater.itemID,redSweater.quantityRemaining);

getchar();

return 0;

}

Explanation:

7 0
3 years ago
(a) Differentiate between heat treatment of ferrous and non-ferrous alloys (b) With your understanding of material's thermal pro
liubo4ka [24]

Answer:

In ferrous metal iron present but on the other hand in the non ferrous material iron does not present.That is why there is a different heat treatment process for ferrous and nonferrous materials.

Ferrous materials contains iron is the main constitute.Like steel ,cast iron ,wrought iron .Steel and cast iron are  the alloy element of iron ans carbon.Wrought iron is the purest from of iron.

Heat treatment process for  ferrous materials :

1.Normalizing

2.Annealing

3.Quenching

4.Surface hardening

Heat treatment process for non ferrous materials :

Mostly annealing process is used for non ferrous materials.After annealing non ferrous will become soft.

When two metal plates are joined then they form a bimetallic structure.The bimetallic structure is used to find the relationship of thermal temperature and the mechanical displacement.

The use of bimetallic structure -In clock ,thermometers ,engines.

7 0
3 years ago
The convection coefficient for flow over a solid sphere may be determined by submerging the sphere, which is initially at 25 °C,
sashaice [31]

Answer:

t = 59.37 s

Explanation:

Given data:

thermal diffusivity = \alpha = \frac{k}{\rho c_p} =0.40\times 10^{-0.5}

theraml conductivity = k = 22 W/m.K

h = 300 W/ m^2.K

T_i = 25 degree C = 298 k

T_o = 60 degree C = 333 k

T_{\infty}= 75 degree C =  348 L

diameter d = 0.1 m

characteristics length Lc = r/3 = = 0.0166

Bi = \frac{hLc}{K} = \frac{300\times 0.0166}{22} = 0.226

\tau = \frac{\alpha t}{lc^2} = \frac{0.4\times 10^{-5}\times t}{0.0166^2}

\tau = 0.036 t

\frac{T_o -T_{\infty}}{T_i -T_{\infty}} = Ae^[\lambda^2 \tau}

at Bi = 0.226

Ai = 0.982

\lambda = 0.876

\frac{333348}{298-348} = 0.982e^{-0.879^2 0.036t}

0.3 = 0.982 e^{-0.2t}

0.305 = e^{-0.2t}

-1.187 = - 0.02t

t = 59.37 s

7 0
2 years ago
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