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Rus_ich [418]
3 years ago
7

See the picture below

Mathematics
1 answer:
svetlana [45]3 years ago
6 0

Answer:

a) 204.20 square centimeters

b) 282.74 square centimeters

c) 314.16 cubic centimeters

Step-by-step explanation:

a)

The curved surface area of a cone is given by:

Curved Surface Area of Cone =  \pi rl

Where r is the radius (given as 5) and l is the slant height, the hypotenuse of the right triangle formed (right side) with 5 as base and 12 as height.

We can find the slant height by using pythagorean theorem on the triangle:

12^2 + 5 ^2 = l^2

144 + 25 = l^2

169 = l^2

l = 13

Now,

Curved Surface Area of Cone = \pi rl=\pi (5)(13) = 204.20 square centimeters

b)

Total surface area is curved surface area PLUS the area of the bottom, which is a circle with area  \pi r^2

So the bottom has an area of  \pi r^2 = \pi (5)^2 = 78.54

Hence, The total surface area = 204.20 + 78.54 = 282.74 square centimeters

c)

The volume of cone is given by  \frac{1}{3}\pi r^2 h

Where radius is 5 and height is 12, so we have:

Volume of Cone = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (5)^2 (12) = 314.16 cubic centimeters

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Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains liters of a dye solution with a
Alja [10]

Answer:

t = 460.52 min

Step-by-step explanation:

Here is the complete question

Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 liters of a dye solution with a concentration of 1 g/liter. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 liters/min, the well-stirred solution flowing out at the same rate.Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value.

Solution

Let Q(t) represent the amount of dye at any time t. Q' represent the net rate of change of amount of dye in the tank. Q' = inflow - outflow.

inflow = 0 (since the incoming water contains no dye)

outflow = concentration × rate of water inflow

Concentration = Quantity/volume = Q/200

outflow = concentration × rate of water inflow = Q/200 g/liter × 2 liters/min = Q/100 g/min.

So, Q' = inflow - outflow = 0 - Q/100

Q' = -Q/100 This is our differential equation. We solve it as follows

Q'/Q = -1/100

∫Q'/Q = ∫-1/100

㏑Q  = -t/100 + c

Q(t) = e^{(-t/100 + c)} = e^{(-t/100)}e^{c}  = Ae^{(-t/100)}\\Q(t) = Ae^{(-t/100)}

when t = 0, Q = 200 L × 1 g/L = 200 g

Q(0) = 200 = Ae^{(-0/100)} = Ae^{(0)} = A\\A = 200.\\So, Q(t) = 200e^{(-t/100)}

We are to find t when Q = 1% of its original value. 1% of 200 g = 0.01 × 200 = 2

2 = 200e^{(-t/100)}\\\frac{2}{200} =  e^{(-t/100)}

㏑0.01 = -t/100

t = -100㏑0.01

t = 460.52 min

6 0
3 years ago
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