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Hoochie [10]
3 years ago
5

The height of a circular tower is 179 meters and its diameter is 48 meters. (Assume that the building is a perfect cylinder). Wh

at is the volume of a 1: 166 scale model in cm^3?
Physics
1 answer:
EastWind [94]3 years ago
6 0

The definition of a scale of 1: 166 will mean that the scale of 1 in the model will be equivalent to 166 times the measurement in the real model, therefore we will have that the height would be 166 times smaller than the 179m given:

h = \frac{179m}{166} = 1.078m = 107.8cm

The same for the diameter,

\phi = \frac{48m}{166}= 0.2891m = 28.91cm

The volume of a cylinder is given as

V = (\pi r^2)(h)

V = (\pi (\frac{d}{2})^2)(h)

V = (\pi (\frac{ 28.91}{2})^2)(107.8)

V = 70762.8cm^3

Therefore the volume would be V = 70762.8cm^3

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A 220 kg crate hangs from the end of a rope of length L = 14.0 m. You push horizontally on the crate with a varying force F to m
kifflom [539]

<u>Answer</u>:

(a) magnitude of F = 797 N

(b)the total work done  W = 0

(c)work done by the gravitational force =  -1.55 kJ

(d)the work done by the pull  = 0

(e) work your force F does on the crate = 1.55 kJ

<u>Explanation</u>:

<u>Given</u>:

Mass of the crate, m =  220 kg

Length of the rope, L = 14.0m

Distance, d =  4.00m

<u>(a) What is the magnitude of F when the crate is in this final position</u>

Let us first determine vertical angle as follows

=>Sin \theta = \frac{d }{L}

=> \theta = Sin^{-1} \frac{d}{L} =

Now substituting thje values

=> \theta = Sin^{-1} \frac{4}{12} =

=> \theta = Sin^{-1} \frac{1}{3}

=> \theta = Sin^{-1}(0.333)

=> \theta = 19.5^{\circ}

Now the tension in the string resolve into components

The vertical component supports the weight

=>Tcos\theta = mg

=>T = \frac{mg}{cos\theta}

=>T = \frac{230 \times 9.8 }{cos(19.5)}

=>T = \frac{2254 }{cos(19.5)}

=>T = \frac{2254 }{0.9426}

=>T =2391N

Therefore the horizontal force

F = TSin(19.5)

F = 797 N

b) The total work done on it

As there is no change in Kinetic energy

The total work done W = 0

<u>c) The work done by the gravitational force on the crate</u>

The work done by gravity

Wg = Fs.d = - mgh

Wg = - mgL ( 1 - Cosθ )

Substituting the values                                                            

= -230 \times 9.8\times 12 ( 1 - cos(19.5) )

= -230 \times 9.8\times 12 ( 1 - 0.9426) )

= -230 \times 9.8\times 12 (0.0574)

= -230 \times 9.8\times 0.6888

=  -230 \times 6.750

= -1552.55 J

The work done by gravity = -1.55 kJ

<u>d) the work done by the pull on the crate from the rope</u>

Since the pull  is perpendicular to the direction of motion,

The work done = 0

e)Find the work your force F does on the crate.

Work done by the Force on the crate

WF = - Wg  

WF = -(-1.55)

WF = 1.55 kJ

<u>(f) Why is the work of your force not equal to the product of the horizontal displacement and the answer to (a)</u>

Here the work done by force is not equal to F*d  

and it is equal to product of the cos angle and F*d

So, it is not equal to the product of the horizontal displacement and the answer to (a)      

7 0
3 years ago
Two parallel plates of area 0.155 m2<br> are separated by 0.00100 m. What<br> is their capacitance?
KonstantinChe [14]

Answer:

1.37 x 10^-9

Explanation:

Trust me bro.

4 0
3 years ago
Calculate the required rate of return for Climax Inc., assuming that (1) investors expect a 4.0% rate of inflation in the future
Westkost [7]

Answer:

Required rate of return = 18.5 %

Explanation:

given,                            

rate of inflection = 4 %

risk free rate = 3 %                      

market risk premium = 5 %                    

firm has a beta  = 2.30                                  

rate of return has averaged 15.0% over the last 5 years

now,                                                                      

Nominal risk free rate = risk free rate  + inflation

                                    = 3%  +  4%

                                   = 7%

Required rate of return = Nominal risk free rate + β (RPM)

                                       = 7%       +          2.3 x 5.0%

Required rate of return = 18.5 %

5 0
3 years ago
Which of the following is an example if harmonic motion?
german

Answer:

"A pendulum swinging back and forth" is an example of harmonic motion

X = Xo cos ω t

Explains the back and forth motion of the pendulum

3 0
2 years ago
If the speedometer of your car reads a constant speedometer of your car reads a constant speed of 40km/hr , can you say the car
Harrizon [31]
I think u can say thats a constant velocity, but remember if ur turning, or going around a curve, that is also changing velocity. Hope this helps have a great day! 
4 0
3 years ago
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