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4vir4ik [10]
3 years ago
10

How much water can be held by a cylindrical tank with a radius of 12 feet and a height of 30 feet? A. 2,260.8 cubic feet B. 13,5

64.8 cubic feet C. 33,912 cubic feet D. 54,259.20 cubic feet
Physics
1 answer:
MakcuM [25]3 years ago
6 0

Answer:

Volume, V = 13564.8  cubic feet

Explanation:

It is given that,

Radius of the cylindrical tank, r = 12 feet

Height of the tank, h = 30 feet

We need to find the water that can be held by a cylindrical tank i.e. we need to find the volume of the tank. It is given by :

V=\pi\times r^2\times h

V=3.14\times (12)^2\times 30

V = 13564.8  cubic feet

So, the water held by the tank is 13564.8  cubic feet. Hence, this is the required solution.

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Temka [501]

Answer:

F=32.24N

Explanation:

From the question we are told that:

Height h= 2.75 m

Lengthl = 5.25 m

Mass m=45kg

Final speed v_f=6.81

Generally the equation for Potential Energy P.E is mathematically given by

P.E=mgh

Therefore

Initial potential energy

P.E_1=45*9.8*2.75 \\\\P.E_1= 1212.75 J

Generally the equation for Kinetic Energy K.E is mathematically given by

K.E=0.5mv^2

Therefore

Final kinetic energy

K.E_2= 1/2*45*6.81*6.81 \\\\K.E_2= 1043.46J

Generally the equation for Work_done is mathematically given by

W=P.E_1-K.E_2\\\\W=169.3

Therefore

F=\frac{W}{d}\\\\F=\frac{169.3}{5.25}

F=32.24N

8 0
3 years ago
Which is an electromagnetic wave A. The waves that heat a cup of water in a microwave oven B. A flag waving in the wind C. Turni
viva [34]

Answer:

The answer would be A. The waves that heat a cup of water in a microwave oven

7 0
3 years ago
Photorefractive keratectomy (PRK) is a laser-based surgery process that corrects near- and farsightedness by removing part of th
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Answer:c

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Compare and Contrast the two main branches of physical science. How are they the same? How are they different?
nignag [31]
Physics
Chemistry


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Hope this helps you!!!
8 0
3 years ago
20 J of energy was used to
Leto [7]

Answer:

0.3m

Explanation:

Given parameters;

Elastic energy  = 20J

Spring constant = 445N/m

Unknown

Final extension of the spring  = ?

Solution:

The elastic potential energy of a stretched spring can be determined using the expression below;

         

           EP  = \frac{1}{2} k e²

k is the spring constant

e is the extension

  Insert the parameters and solve;

          20  = \frac{1}{2} x 445 x e²

        multiply those sides by 2;

         2(20) = 445e²

            40  = 445e²

              e² = \frac{40}{445}   = 0.09

              e  = √0.09 = 0.3m

6 0
3 years ago
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