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Yakvenalex [24]
4 years ago
15

Near the top of the Citigroup Center building in New York City, there is an object with mass of 4.00 ✕ 105 kg on springs that ha

ve adjustable force constants. Its function is to dampen wind-driven oscillations of the building by oscillating at the same frequency as the building is being driven—the driving force is transferred to the object, which oscillates instead of the entire building.
(a) What effective force constant should the springs have to make the object oscillate with a period of 2.40 s?

(b) What energy is stored in the springs for a 2.20-m displacement from equilibrium?
Physics
1 answer:
jasenka [17]4 years ago
7 0

Explanation:

It is given that,

Mass of an object, m=4\times 10^5\ kg

(a) Time period of oscillation, T = 2.4 s

The formula for the time period of spring is given by :

T=2\pi \sqrt{\dfrac{m}{k}}

Where

k is the force constant

k=\dfrac{4\pi ^2 m}{T^2}

k=\dfrac{4\pi ^2 \times 4\times 10^5}{(2.4)^2}

k=2.74\times 10^6\ N/m

(b) Displacement in the spring, x = 2.2 m

Energy stored in the spring is given by :

U=\dfrac{1}{2}kx^2

U=\dfrac{1}{2}\times 2.74\times 10^6\ N/m\times (2.2\ m)^2

U=6.63\times 10^6\ J

Hence, this is the required solution.

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Also the last items on the list are recalled by the effect known as recency,

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Learn more here:

brainly.com/question/15180534

<em>The possible question options as obtained from a similar question online, are;</em>

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5 0
2 years ago
A spring whose stiffness is 3500 N/m is used to launch a 4 kg block straight up in the classroom. The spring is initially compre
jarptica [38.1K]

Answer:

the speed of the block at the given position is 21.33 m/s.

Explanation:

Given;

spring constant, k = 3500 N/m

mass of the block, m = 4 kg

extension of the spring, x = 0.2 m

initial velocity of the block, u = 0

displacement of the block, d =1.3 m

The force applied to the block by the spring is calculated as;

F = ma = kx

where;

a is the acceleration of the block

a = \frac{kx}{m} \\\\a = \frac{(3500) \times (0.2)}{4} \\\\a = 175 \ m/s^2

The final velocity of the block at 1.3 m is calculated as;

v² = u² + 2ad

v² = 0 + 2ad

v² = 2ad

v = √2ad

v = √(2 x 175 x 1.3)

v = 21.33 m/s

Therefore, the speed of the block at the given position is 21.33 m/s.

5 0
3 years ago
A 24.8 tall object is placed in front of a lens, which creates a -3.09 cm tall image. What is the magnification?
OleMash [197]

Answer: 0.125

Explanation:

Given that the

Object height U = 24.8 cm

Image height V = 3.09 cm

Magnification M is the ratio of the image height to the object height. That is, divide the given image height by the given object height.

M = V/U

Substitute V and U into the formula above.

M = 3.09/24.8

M = 0.125

Magnification is therefore equal to 0.125 approximately

6 0
3 years ago
The physician has prescribed an antiemetic for your 9-year-old patient to control nausea and vomiting. Which drug is the physici
FromTheMoon [43]

If the patient need a medicine that would control his or her nausea and vomiting, the best antiemetic drug that would reduce and would provide control of the patient’s symptoms of nausea and vomiting is the phenothiazines. This drug has the ability of blocking the dopamine, where in it will help control nausea and vomiting with the help of its ability of being a neurotransmitter. The doctor would likely recommend this drug for the patient in order to relieve the symptoms and to control the nausea and vomiting. The drug also helps in providing a patient calmness, where in it could also relieve or diminish stress or anxiety that the patient has.

8 0
3 years ago
When a certain air-filled parallel-plate capacitor is connected cross a battery, it acquires a charge (on each plate) of 160 µC.
ozzi

Answer:

k = 2.45

Explanation:

It is given that,

Charge acquired on each plate of the capacitor due to a battery, q=160\ \mu C

If V is the potential difference across the plates, capacitance in air is given by :

C_a=\dfrac{q}{V}=\dfrac{160}{V}.............(1)

Let k is the dielectric constant of the dielectric slab. While the battery connection is maintained, a dielectric slab is inserted into and fills the region between the plates. This results in the accumulation of an additional charge of 220 µC on each plate.

New charge becomes, Q=220\ \mu C+220\ \mu C=440\ \mu C

C_d=\dfrac{Q}{V}=\dfrac{440}{V}.............(2)

The dielectric constant of the dielectric slab is given by :

k=\dfrac{C_d}{C_a}

k=\dfrac{440}{160}

k = 2.45

So, the dielectric constant of the dielectric slab is 2.45. Hence, this is the required solution.

7 0
3 years ago
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