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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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In physics and chemistry, the law of conservation of energy states that the total energy of an isolated system remains constant; it is said to be conserved over time. This law means that energy can neither be created nor destroyed; rather, it can only be transformed or transferred from one form to another.
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3 years ago
The kinetic energy of a moving object is E=12mv2. A 61 kg runner is moving at 10kmh. However, her speedometer is only accurate t
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Answer:

e=3367.2J

%e=1.43%

Explanation:

From the exercise we know two information. The real speed and the experimental measured by the speedometer

v_{r}=10km/h=2.77m/s

Since the speedometer is only accurate to within 0.1km/h the experimental speed is

v_{e}=10km/h-0.1km/h=9.9km/h=2.75m/s

Knowing that we can calculate Kinetic energy for the real and experimental speed

E_{r}=\frac{1}{2}mv^2=\frac{1}{2}(61000g)(2.77m/s)^2=234023J

E_{e}=\frac{1}{2}mv^2=\frac{1}{2}(61000g)(2.75m/s)^2=230656J

Now, the potential error in her calculated kinetic energy is:

e=E_{r}-E_{e}=(234023-230656)J=3367.2J

%e=\frac{E_{r}-E_{e}}{E_{r}}x100=\frac{(234023-230656)J}{234023J}x100=1.43%

4 0
3 years ago
A man pushes on a trunk with a force of 250 newtons. The trunk does not move. How much positive work is done on the trunk?
Digiron [165]

Answer:

F is 250 N

d is 0 m

F x d

=250 x 0

=0

The answer is 0.0 J.

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3 years ago
You hold a metal block of mass 40 kg above your head at a height of 2 m.
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Answer:

The work done by gravity is 784 J.

Explanation:

Given:

Mass of the block is, m=40\ kg

Height to which it is raised is, h=2\ m

Acceleration due to gravity is, g=9.8\ m/s^2

Now, work done by gravity is equal to the product of force of gravity and the distance moved in the direction of gravity. So,

\textrm{Work by gravity}=F_g\times h

Force of gravity is given as the product of mass and acceleration due to gravity.

\therefore F_g=mg=40\times 9.8=392\ N. Now,

\textrm{Work by gravity}=F_g\times h=392\times 2=784\ J

Therefore, the work done by gravity is 784 J.

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In this lab you will use a cart and track to explore various aspects of motion. You will measure and record the time it takes th
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It is given that by using track and cart we can record the time and the distance travelled and also the speed of the cart can be recorded. With all this data we can solve questions on the laws of motion.  

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