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Bess [88]
3 years ago
6

The frequency of a person's pendulum is 0.3204 Hz when at a location where g is known to be exactly 9.800 m/s^2.

Physics
1 answer:
wel3 years ago
7 0

Answer:

A) g = 9.751 m/s², B)       h = 2.573 10⁴ m

Explanation:

The angular velocity of a pendulum is

           w = √ g / L

Angular velocity and frequency are related.

          w = 2π f

          f = 1 / 2π √ g / L

A) with the initial data we can look for the pendulum length

        L = 1 /4π²   g / f²

        L = 1 /4π²   9,800 / 0.3204²

        L = 2.4181 m

The length of the pendulum does not change, let's look for the value of g for the new location

        g = 4π² f²  L

        g = 4π² 0.3196²  2.4181

        g = 9.75096 m / s²

         g = 9.751 m/s²

B) The value of the acceleration of gravity can be found with the law of universal gravitation

           F = G m M / R_{e}²

And Newton's second law

           W = m g

           W = F

         G m M / R_{e}² = mg

         g = G M / R_{e}²

         R_{e}² = G M / g

Let's calculate

          R_{e}² = 6.67 10⁻¹¹ 5.98 10²⁴ /9.75096

          R = √ 4.0905 10¹³ = √ 40.9053 10¹²

          R = 6.395726 10⁶ m

The height above sea level is

         h = R - [tex]R_{e}[/tex

         h = (6.395726 -6.37) 10⁶

         h = 0.0257256 106

         h = 2.573 10⁴ m

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* The third shell would be empty, so the eight electrons on the second level would be the outermost after the atom lost one electron

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The French high-speed train travels at 300 km/h. How long
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Suppose that an object is moving along a vertical line. Its vertical position is given by the equation L(t) = 2t3 + t2-5t + 1, w
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Answer:

The average velocity is

266\frac{m}{s},274\frac{m}{s} and 117\frac{m}{s} respectively.

Explanation:

Let's start writing the vertical position equation :

L(t)=2t^{3}+t^{2}-5t+1

Where distance is measured in meters and time in seconds.

The average velocity is equal to the position variation divided by the time variation.

V_{avg}=\frac{Displacement}{Time} = Δx / Δt = \frac{x2-x1}{t2-t1}

For the first time interval :

t1 = 5 s → t2 = 8 s

The time variation is :

t2-t1=8s-5s=3s

For the position variation we use the vertical position equation :

x2=L(8s)=2.(8)^{3}+8^{2}-5.8+1=1049m

x1=L(5s)=2.(5)^{3}+5^{2}-5.5+1=251m

Δx = x2 - x1 = 1049 m - 251 m = 798 m

The average velocity for this interval is

\frac{798m}{3s}=266\frac{m}{s}

For the second time interval :

t1 = 4 s → t2 = 9 s

x2=L(9s)=2.(9)^{3}+9^{2}-5.9+1=1495m

x1=L(4s)=2.(4)^{3}+4^{2}-5.4+1=125m

Δx = x2 - x1 = 1495 m - 125 m = 1370 m

And the time variation is t2 - t1 = 9 s - 4 s = 5 s

The average velocity for this interval is :

\frac{1370m}{5s}=274\frac{m}{s}

Finally for the third time interval :

t1 = 1 s → t2 = 7 s

The time variation is t2 - t1 = 7 s - 1 s = 6 s

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The position variation is x2 - x1 = 701 m - (-1 m) = 702 m

The average velocity is

\frac{702m}{6s}=117\frac{m}{s}

5 0
3 years ago
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