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Eva8 [605]
3 years ago
12

A tank with some water in it begins to drain. The function v ( t ) = 46 − 3.5 t determines the volume of the water in the tank (

in gallons) given a number of minutes t since the water began draining. What is the vertical intercept of f ? What does your answer to part (a) represent? Select all that apply. The weight of the tank when it is empty. How many minutes it takes for all of the water to drain from the tank. The number of gallons of water in the tank when it starts draining. List all horizontal intercepts (or "roots") of f . What does your answer to part (b) represent? Select all that apply. The weight of the tank when it is empty. The number of gallons of water in the tank when it starts draining. How many minutes it takes for all of the water to drain from the tank.

Engineering
1 answer:
olchik [2.2K]3 years ago
7 0

Answer with Explanation:

Part a)

The volume of water in the tank as a function of time is plotted in the below attached figure.

The vertical intercept of the graph is 46.

Part b)

The vertical intercept represents the volume of water that is initially present in the tank before draining begins.

Part c)

To find the time required to completely drain the tank we calculate the volume of the water in the tank to zero.

0=46-3.5t\\\\3.5=46\\\\\therefore t=\frac{46}{3.5}=13.143minutes

Part d)

The horizontal intercept represents the time it takes to empty the tank which as calculated above is 13.143 minutes.

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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

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3 0
2 years ago
A single fixed pulley is used to lift a load of 400N by the application of an effort of 480N in 10s through a vertical height of
Allushta [10]

Answer:

(a) the velocity ratio of the machine (V.R) = 1

(b) The mechanical advantage of the machine (M.A) = 0.833

(c) The efficiency of the machine (E) = 83.3 %

Explanation:

Given;

load lifted by the pulley, L = 400 N

effort applied in lifting the, E = 480 N

distance moved by the effort, d = 5 m

(a) the velocity ratio of the machine (V.R);

since the effort applied moved downwards through a distance of d, the load will also move upwards through an equal distance 'd'.

V.R = distance moved by effort / distance moved by the load

V.R = 5/5 = 1

(b) The mechanical advantage of the machine (M.A);

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E = \frac{M.A}{V.R} \times 100\%\\\\E = 0.833 \ \times \ 100\%\\\\ E = 83.3 \ \%

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2 years ago
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