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Shkiper50 [21]
3 years ago
6

oscillating spring mass systems can be used to experimentally determine an unknown mass without using a mass balance. a student

observes that a particular spring-mass system has a frequency of oscillation of 10 Hz. the spring constant of the spring is 250 N/m. what is the mass?​
Physics
1 answer:
12345 [234]3 years ago
3 0

Answer:

Mass, m = 6.18 kg

Explanation:

Given the following data;

Frequency, F = 10 Hz

Spring constant, k = 250 N/m

We know that pie, π = 22/7

To find the mass, we would use the following formula;

F = 1/2π√(k/m)

Where;

F is the frequency of oscillation.

k is the spring constant.

m is the mass of the spring.

Substituting into the formula, we have;

10 = 1/2 * 22/7 * √250/m

10 = 22/14 * √250/m

Cross-multiplying, we have;

140 = 22 * √250/m

Dividing both sides by 22, we have;

140/22 = √250/m

6.36 = √250/m

Taking the square of both sides, we have;

6.36² = (√250/m)²

40.45 = 250/m

Cross-multiplying, we have;

40.45m = 250

Mass, m = 250/40.45

Mass, m = 6.18 kg

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tamaranim1 [39]

Answer:

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x = ½ aw t²

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x = d + (0) t + ½ (-az) t²

x = d − ½ az t²

Solving the system of equations for x will give us the location the cars meet (relative to car W's starting point).

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3 years ago
An object that is moving in a linear path with an acceleration in the direction of motion has a(n) ______________ velocity. cons
Kamila [148]
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v(t)=v_0 + at
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4 years ago
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the diagram below represents the orbits of earth, comet temple-tuttle, and planet x, another planet in out solar system. arrows
mart [117]

Answer:

the most elliptical orbit is that of COMETA

Explanation:

The eccentricity of a curve in defined as the ratio between lacia to the focus, called c and the value of the axis greater than

         ε = c / a

if we use Pythagoras' theorem

         c = \sqrt{a^2 - b^2}

   substituting

           ε = \sqrt{1 - (b/a)^2 }

if   ε = 0 we have a circumference

In the diagram presented the orbit of the comet is an ellipse a> b

          ε=\sqrt{1- x}  \\ x = (\frac{b}{a} )^2

if we expand in series

             ε = 1 - x/2  

             ε=  1 - \frac{1}{2}  \ (\frac{ b}{a} )^2

if we neglect the non-linear terms

            ε = 1

Earth's orbit is a small ellipse

             b / a = 149 10⁶ / 151 10⁶

             b / a = 0.98675

             ε = \sqrt{1- 0.98675^2}

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Planet X, despite not having data, it seems that the sun is in the scepter of the orbit, so b = a

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5 0
3 years ago
The New England Merchants Bank Building in Boston is 152 m high. On windy days it sways with a frequency of 0.17 Hz, and the acc
aksik [14]

Answer:

d = 8.4 cm

Explanation:

In order to calculate the amplitude of oscillation of the top of the building, you use the following formula for the max acceleration of as simple harmonic motion:

a_{max}=A\omega^2           (1)

A: amplitude of the oscillation

w: angular speed of the oscillation = 2\pif

f: frequency = 0.17Hz

The maximum acceleration of the top of the building is a 2.0% of the free-fall acceleration. Then, you have:

a_{max}=0.02(9.8m/s^2)=0.196\frac{m}{s^2}

Then, you solve for A in the equation (1) and replace the values of the parameters:

A=\frac{a_{max}}{\omega^2}=\frac{a_{max}}{4\p^2i f^2}\\\\A=\frac{0.196m/s^2}{16\pi^2(0.17Hz)^2}\\\\A=0.042m=4.2cm

The total distance, side to side, of the oscilation of the top of the building is twice the amplitude A. Then you obtain:

d = 2A = 2(4.2cm) = 8.4cm

4 0
3 years ago
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