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rewona [7]
3 years ago
11

Scientists in a test lab are testing the hardness of a surface before constructing a building. Calculations indicate that the en

tire structure would sink by a certain amount for every additional floor that is added. If the maximum permissible limit for depression of the structure is 20 centimeters, how many floors can be safely added to the building?
A.14
B.15
C.18
D.23
Physics
2 answers:
Alex17521 [72]3 years ago
8 0

Answer:

C. 18

Explanation:

Hi, you haven't provided the depression per floor so I'll explain to you how to do it for a given depression/floor and you can extend it to your problem by applying the same steps. Since each additional floor generates a constant depression the depression generated for one floor (depression/floor) multiply by the number of extra floors will give you the total depression:

(depression per floor)*(number of floors added) = Total depression

Because the maximum depression accepted is 20 cm:

(depression per floor)*(number of floors added) < 20 cm

If the depression per floor is one centimeter:

1 cm * (number of floors added) < 20 cm

number of floors added < 20 cm

For this example, the answer would be 18 floors because is the maximum number of floors, in the list of options, that doesn't contradict the inequality (18 cm < 20 cm)

Artemon [7]3 years ago
5 0
<h2>Number of Floors</h2>

Thus each supplementary level will obtain the construction of tub with x so make the unit "cm".  It can be x cm. Presently we understand that the highest cavity is 20 cm, and it is also x*y, where y is the number of floors. Consequently, we possess x*y=20 since the number of floors is:

20/x  =

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A 0.750 kg block is attached to a spring with spring constant 17.5 N/m. While the block is sitting at rest, a student hits it wi
Dmitriy789 [7]

Answer:

a

 A =  0.081 \  m

b

The value is  u =  0.2569 \  m/s

Explanation:

From the question we are told that

   The mass is  m  =  0.750 \ kg

   The spring constant is  k  =  17.5 \  N/m

    The instantaneous speed is  v  =  39.0 \  cm/s= 0.39 \  m/s

    The position consider is  x =  0.750A  meters from equilibrium point

   

Generally from the law of  energy conservation we have that

        The kinetic energy induced by the hammer  =  The energy stored in the spring

So

          \frac{1}{2} *  m * v^2  =  \frac{1}{2}  *  k  *  A^2

Here a is the amplitude of the subsequent oscillations

=>      A =  \sqrt{\frac{m *  v^ 2 }{ k} }

=>      A =  \sqrt{\frac{0.750 *  0.39 ^ 2 }{17.5} }

=>       A =  0.081 \  m

Generally from the law of  energy conservation we have that

The kinetic energy  by the hammer  =  The energy stored in the spring at the point considered   +   The kinetic energy at the considered point

             \frac{1}{2}  * m *  v^2 = \frac{1}{2}  * k x^2 + \frac{1}{2}  * m *  u^2

=>          \frac{1}{2}  * 0.750 *  0.39^2 = \frac{1}{2}  * 17.5* 0.750(0.081 )^2 + \frac{1}{2}  * 0.750 *  u^2

=>          u =  0.2569 \  m/s

3 0
3 years ago
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