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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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The uncertainty Δp sets a lower bound on the average momentum of a particle in the nucleus. If a particle's average momentum wer
kirill [66]

Answer: The minimum kinetic energy Kmin is 1.3 × 10^-13 J

Explanation: Please see the attachments below

8 0
3 years ago
Consider a semi-spherical bowl machined out of wood (A drawing of a semi-spherical bowl can be found in UNIT- 4 Introduction, so
Mrrafil [7]

Answer:

The total surface are of the bowl is given by: 0.0532*pi m² (approximately 0.166533 m²)

Explanation:

The total surface area of the semi-spherical bowl can be decomposed in three different sections: 1) an outer semi-sphere of radius 12 cm, 2) an inner semi-sphere of radius 10 cm, and 3) the edge, which is a 2-dimensional ring with internal radius of 10 cm and external radius of 12 cm. We will compute the areas independently and then sum them all.

a) Outer semi-sphere:

A1 = 2*pi*r² = 2*pi*(12 cm)² = 288*pi cm² = 904.78 cm²

b) Inner semi-sphere:

A2 = 2*pi*(10 cm)² = 200*pi cm² = 628.32 cm²

c) Edge (Ring):

A3 = pi*(r1² - r2²) = pi*((12 cm)²-(10 cm)²) = pi*(144-100) cm² = 44*pi cm² = 138.23 cm²

Therefore, the total surface area of the bowl is given by:

A = A1 + A2 + A3 = 288*pi cm² + 200*pi cm² + 44*pi cm² = 532*pi cm² (approximately 1665.33 cm²)

Changing units to m², as required in the problem, we get:

A = 532*pi cm² * (1 m² / 10, 000 cm²) = 0.0532*pi m² (approximately 0.166533 m²)

5 0
3 years ago
Kawas 200 osebanywhes he remembered he had to return some books they​
tia_tia [17]
I think you need to add more.. but I may know where you are leading

Was he 200 m away and made the trip in 200 seconds?

If yes...
2 m/s was his speed and 0 velocity
6 0
3 years ago
Suppose the original segment of wire is stretched to 10 times its original length. How much charge must be added to the wire to
Debora [2.8K]

Here we want to study how the linear charge density changes as we change the measures of our body.

We will find that we need to add 9*Q of charge to keep the linear charge density unchanged.

<em>I will take two assumptions:</em>

The charge is homogeneous, so the density is constant all along the wire.

As we work with a linear charge density we work in one dimension, so the wire "has no radius"

Originally, the wire has a charge Q and a length L.

The linear charge density will be given by:

λ = Q/L

Now the length of the wire is stretched to 10 times the original length, so we have:

L' = 10*L

We want to find the value of Q' such that λ' (the <u>linear density of the stretched wire</u>) is still equal to λ.

Then we will have:

λ' = Q'/L' = Q'/(10*L) = λ = Q/L

Q'/(10*L) = Q/L

Q'/10 = Q

Q' = 10*Q

So the new <u>charge must be 10 times the original charge</u>, this means that we need to add 9*Q of charge to keep the linear charge density unchanged.

If you want to learn more, you can read:

brainly.com/question/14514975

6 0
2 years ago
A bus accelerates from 5.75 m/s at a rate of 1.25 m/s/s for 3.50
marysya [2.9K]

Answer:

10.125 meters?

Explanation:

Im taking 5.75m/s + 1.25 m/s/s (3.5) = my answer.

In those 3.5 seconds it travels 4.375.

I added that to 5.75 to get 10.125m

3 0
3 years ago
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