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sdas [7]
3 years ago
6

The New England Merchants Bank Building in Boston is 152 mm high. On windy days it sways with a frequency of 0.20 HzHz , and the

acceleration of the top of the building can reach 2.5 %% of the free-fall acceleration, enough to cause discomfort for occupants.What is the total distance, side to side, that the top of the building moves during such an oscillation?
Physics
1 answer:
kkurt [141]3 years ago
3 0

Answer:

The total distance, side to side, that the top of the building moves during such an oscillation = 31 cm

Explanation:

Let the total side to side motion be 2A. Where A is maximum acceleration.

Now, we know know that equation for maximum acceleration is;

A = α(max) / [(2πf)^(2)]

So 2A = 2[α(max) / [(2πf)^(2)] ]

α(max) = (0.025 x 9.81) while frequency(f) from the question is 0.2Hz.

Therefore 2A = 2 [(0.025 x 9.81) / [((2π(0.2)) ^(2)] ] = 2( 0.245 / 1.58) = 0.31m or 31cm

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A 5 meter long ladder leans against a wall. The bottom of the ladder slides away from the wall at the constant rate of 1 3 m/s.
Oksana_A [137]

Answer:9.75 m/s

Explanation:

Given

Length of ladder (L)=5 m

Foot the ladder is moving away with speed of \frac{\mathrm{d} x}{\mathrm{d} t}=13 m/s

From diagram

x^2+y^2=L^2------1

at x=3

y^2=25-9=16

y=4 m

Now differentiating equation 1 w.r.t time

2x\frac{\mathrm{d} x}{\mathrm{d} t}+2y\frac{\mathrm{d} y}{\mathrm{d} t}=0

x\frac{\mathrm{d} x}{\mathrm{d} t}=-y\frac{\mathrm{d} y}{\mathrm{d} t}

3\times 13=-4\times \frac{\mathrm{d} y}{\mathrm{d} t}

\frac{\mathrm{d} y}{\mathrm{d} t}=-\frac{3\times 13}{4}=-9.75 m/s

negative indicates distance is decreasing with time

5 0
3 years ago
Which of the following is not an example of centripetal acceleration?
Amanda [17]
An apple falling to the ground is not an example of centripetal acceleration.
5 0
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Which substance is a combination of different atoms?
Lady_Fox [76]

Answer:

The answer is compound

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5 0
3 years ago
What is energy that comes from the movement of particles
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A 1.50 3 103 - kg car starts from rest and accelerates uniformly to 18.0 m/s in 12.0 s. Assume that air resistance remains const
nexus9112 [7]

Answer:

Explanation:

let force exerted by engine be F.Net force =( F-400)N, applying newton law

     F-400 = 1.5 x 10³x18 =27000 ,

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velocity after 12 s  = 0 + 18 x 12 = 216 m/s

Average velocity = (0 + 216 )/2 = 108 m/s

Average power = force x average velocity = 27400 x 108 = 29.6 10⁵ W .⁶

b) At 12 s , velocity = 216 m/s

Instantaneous power = velocity x force = 216 x 27400 = 59.2 x 10⁶ W.

8 0
3 years ago
Read 2 more answers
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