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Oduvanchick [21]
3 years ago
10

A parcel of land is in the shape of an isosceles triangle. The base has a length of 425 ft.; the other sides, which are of equal

length, meet at an angle of 39 degrees. How long are they?

Physics
1 answer:
kogti [31]3 years ago
4 0

Answer:

The answer to your question is 636.6 ft    

Explanation:

Data

base = 425 ft

angle = 39°

See the picture below

1.- Divide the triangle to get two right triangles.

    Now the superior angle will measure 19.5° and the opposite side will measure 212.5 ft

2.- Use the trigonometric function sine to find the hypotenuse

     sin 19.5 = 212.5/hyp

solve for hyp

    hyp = 212.5 / sin 19.5

Result

    hyp = 212.5/ 0.333

    hyp = 636.6 ft    

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alexdok [17]
Acceleration=9.81m/s^2
initial velocity=0m/s
time=.28s

We have to find final velocity.

The equation we use is

Final velocity=initial velocity+acceleration x time

Vf=0m/s+(9.81m/s^2)(.28s)
Vf=2.7468m/s

We would round this to:

Vf (final velocity)=2.7m/s
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3 years ago
An intravenous (IV) system is supplying saline solution to a patient at the rate of 0.06 cm3/s through a needle of radius 0.2 mm
horsena [70]

Answer:

Pressure applied to the needle is 7528 Pa

Explanation:

As we know by poiseuille's law of flow of liquid through a cylindrical pipe

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now we know that

Q = 0.06 \times 10^{-6} m^3/s

radius = 0.2 mm

Length = 6.32 cm

\eta = 1\times 10^{-3} Pa s

now we have

6 \times 10^{-8} = \frac{\Delta P \pi (0.2 \times 10^{-3})^4}{8(1 \times 10^{-3})6.32 \times 10^{-2}}

3.03 \times 10^{-11} = \Delta P 5.02 \times 10^{-15}

\Delta P = 6028 Pa

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8 0
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Which is the smallest unit of life?
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7 0
2 years ago
An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b
lys-0071 [83]

Answer:

a)Distance traveled during the first second = 4.905 m.

b)Final velocity at which the object hits the ground = 38.36 m/s

c)Distance traveled during the last second of motion before hitting the ground = 33.45 m

Explanation:

a) We have equation of motion

             S = ut + 0.5at²

     Here u = 0, and a = g

              S = 0.5gt²

    Distance traveled during the first second ( t =1 )

              S = 0.5 x 9.81 x 1² = 4.905 m

   Distance traveled during the first second = 4.905 m.

b)  We have equation of motion

            v² = u² + 2as

      Here u = 0, s= 75 m and a = g

           v² = 0² + 2 x g x 75 = 150 x 9.81

           v = 38.36 m/s

      Final velocity at which the object hits the ground = 38.36 m/s

c) We have S = 0.5gt²

                   75 = 0.5 x 9.81 x t²

                    t = 3.91 s

   We need to find distance traveled last second

   That is

          S = 0.5 x 9.81 x 3.91² - 0.5 x 9.81 x 2.91² = 33.45 m

   Distance traveled during the last second of motion before hitting the ground = 33.45 m

       

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3 years ago
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It’s a chemical change
3 0
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