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katrin [286]
3 years ago
13

What is the total distance, side to side, that the top of the building moves during such an oscillation? The New England Merchan

ts Bank Building in Boston is 152 m high. On windy days it sways with a frequency of 0.22 Hz, and the acceleration of the top of the building can reach 1.5% of the free-fall acceleration, enough to cause discomfort for occupants. What is the total distance, side to side, that the top of the...
Physics
1 answer:
kramer3 years ago
5 0
I think the question should be the below:

<span>What is the total distance, side to side, that the top of the building moves during such an oscillation?
</span>
Answer is the below:

 <span>Acceleration .. a = (-) ω² x </span>
<span>(ω = equivalent ang. vel. = 2π.f) (x = displacement from equilibrium position) </span>

<span>x (max) = a(max) /ω² </span>

<span>x = (0.015 x 9.8m/s²) / (2π.f)² .. .. (0.147) / (2π*0.22)² .. .. ►x(max) = 0.077m .. (7.70cm)</span>
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You are removing branches from your roof after a big storm. You throw a branch horizontally from your roof, which is a height 3.
mart [117]

Answer:

The initial velocity in the x-direction with which the branch was thrown is approximately 10.224 m/s

Explanation:

The given parameters of the motion of the branch are;

The height from which the branch is thrown = 3.00 m

The horizontal distance the branch lands from where it was thrown, x = 8.00 m

The direction in which the branch is thrown = Horizontally

Therefore, the initial vertical velocity of the branch, u_y = 0 m/s

The time it takes an object in free fall (zero initial downward vertical velocity) to reach the ground is given as follows;

s = u_y·t + 1/2·g·t²

Where;

u_y = 0 m/s

s = The initial height of the object = 3.00 m

g = The acceleration due to gravity = 9.8 m/s²

∴ s = 0·t + 1/2·g·t² = 0 × t + 1/2·g·t² = 1/2·g·t²

t = √(2·s/g) = √(2 × 3/9.8) = (√30)/7 ≈ 0.78246

The horizontal distance covered before the branch touches the ground, x = 8.00 m

Therefore, the initial velocity in the horizontal, x-direction with which the branch was thrown, 'uₓ', is given as follows;

uₓ = x/t = 8.00 m/((√30)/7 s)

Using a graphing calculator, we get;

uₓ = 8.00 m/((√30)/7 s) = (28/15)·√30 m/s ≈ 10.224 m/s

The initial velocity in the horizontal, x-direction with which the branch was thrown, uₓ ≈ 10.224 m/s.

3 0
3 years ago
The table shows the diameters of the planets in our solar system. Assume that a basketball whose diameter is 25
tino4ka555 [31]

Answer:

Mercury:

.85

pea

Venus:

2.1

gumball

Earth:

2.2

gumball

Mars:

1.2

marble

Uranus:

9

grapefruit

Neptune:

8.6

softball

Explanation:

I have no clue if I'm right but hopefully, I am

3 0
2 years ago
What is not a raquet sport
uysha [10]

Answer:

what

Explanation:

Racket sports include tennis, badminton, squash or any other sport where you use rackets to hit a ball or shuttlecock to play. They can be played competitively or just for fun and are a great form of physical activity.

7 0
3 years ago
a student measures the speed of yellow light in water to be 2.00 x 10(to the 8th power) meters per second, 1.87 x 10(to the 8th
mamaluj [8]
<span>3) Neither precise or accurate. This is because of the deviation between the measurements, they vary and are not within a good range. And they are not close to the accepted value. In order to be precise the measurements have to be relatively close to each other, and to be accurate they have to be close to the accepted value.</span>
8 0
3 years ago
Suppose that a comet that was seen in 550 A.D. by Chinese astronomers was spotted again in year 1941. Assume the time between ob
pickupchik [31]

To solve the problem it is necessary to apply the concepts related to Kepler's third law as well as the calculation of distances in orbits with eccentricities.

Kepler's third law tells us that

T^2 = \frac{4\pi^2}{GM}a^3

Where

T= Period

G= Gravitational constant

M = Mass of the sun

a= The semimajor axis of the comet's orbit

The period in years would be given by

T= 1941-550\\T= 1391y(\frac{31536000s}{1y})\\T=4.3866*10^{10}s

PART A) Replacing the values to find a, we have

a^3= \frac{T^2 GM}{4\pi^2}

a^3 = \frac{(4.3866*10^{10})^2(6.67*10^{-11})(1.989*10^{30})}{4\pi^2}

a^3 = 6.46632*10^{39}

a = 1.86303*10^{13}m

Therefore the semimajor axis is 1.86303*10^{13}m

PART B) If the semi-major axis a and the eccentricity e of an orbit are known, then the periapsis and apoapsis distances can be calculated by

R = a(1-e)

R = 1.86303*10^{13}(1-0.997)

R= 5.58*10^{10}m

7 0
3 years ago
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