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hoa [83]
3 years ago
9

The mass of an empty container is 3/10kg. When the container is half filled with water, the mass becomes 1 1/10kg. What is the m

ass of the container when it is completely filled with water
Mathematics
1 answer:
Nastasia [14]3 years ago
6 0

Answer:

1.9 kg

Step-by-step explanation:

Empty container: 3/10 kg = 0.3 kg

Container half filled: 1 1/10 kg = 1.1 kg

Mass of water that fills it halfway: 1.1 kg - 0.3 kg = 0.8 kg

Mass of water that fills it completely: 2 * 0.8 kg = 1.6 kg

Mass of container completely filled with water: 0.3 kg + 1.6 kg = 1.9 kg

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Answer:

1) \frac{dy}{dt}=2.5-\frac{3y}{2t+100}

2) y(t)=(50+t)- \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }}

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4) The salt concentration will increase without bound.

Step-by-step explanation:

1) Let y represent the amount of salt in the tank at time t, where t is given in minutes.

Recall that: \frac{dy}{dt}=rate\:in-rate\:out

The amount coming in is 0.5\frac{lb}{gal}\times 5\frac{gal}{min}=2.5\frac{lb}{min}

The rate going out depends on the concentration of salt in the tank at time t.

If there is y(t) pounds of  salt and there are 100+2t gallons at time t, then the concentration is: \frac{y(t)}{2t+100}

The rate of liquid leaving is is 3gal\min, so rate out is =\frac{3y(t)}{2t+100}

The required differential equation becomes:

\frac{dy}{dt}=2.5-\frac{3y}{2t+100}

2) We rewrite to obtain:

\frac{dy}{dt}+\frac{3}{2t+100}y=2.5

We multiply through by the integrating factor: e^{\int \frac{3}{2t+100}dt }=e^{\frac{3}{2} \int \frac{1}{t+50}dt }=(50+t)^{\frac{3}{2} }

to get:

(50+t)^{\frac{3}{2} }\frac{dy}{dt}+(50+t)^{\frac{3}{2} }\cdot \frac{3}{2t+100}y=2.5(50+t)^{\frac{3}{2} }

This gives us:

((50+t)^{\frac{3}{2} }y)'=2.5(50+t)^{\frac{3}{2} }

We integrate both sides with respect to t to get:

(50+t)^{\frac{3}{2} }y=(50+t)^{\frac{5}{2} }+ C

Multiply through by: (50+t)^{-\frac{3}{2}} to get:

y=(50+t)^{\frac{5}{2} }(50+t)^{-\frac{3}{2} }+ C(50+t)^{-\frac{3}{2} }

y(t)=(50+t)+ \frac{C}{(50+t)^{\frac{3}{2} }}

We apply the initial condition: y(0)=0

0=(50+0)+ \frac{C}{(50+0)^{\frac{3}{2} }}

C=-12500\sqrt{2}

The amount of salt in the tank at time t is:

y(t)=(50+t)- \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }}

3) The tank will be full after 50 mins.

We put t=50 to find how pounds of salt it will contain:

y(50)=(50+50)- \frac{12500\sqrt{2} }{(50+50)^{\frac{3}{2} }}

y(50)=98.23

There will be 98.23 pounds of salt.

4) The limiting concentration of salt is given by:

\lim_{t \to \infty}y(t)={ \lim_{t \to \infty} ( (50+t)- \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }})

As t\to \infty, 50+t\to \infty and \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }}\to 0

This implies that:

\lim_{t \to \infty}y(t)=\infty- 0=\infty

If the tank had infinity capacity, there will be absolutely high(infinite) concentration of salt.

The salt concentration will increase without bound.

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