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4 years ago

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A physical pendulum consists of a meter stick that is pivoted at a small hole drilled through the stick a distance d from the 50

.00 cm mark. (a) The period of oscillation is 2.50 s. Find d. (b) If you moved the pivot 5.00 cm closer to the 50.00 cm mark, what would the period of oscillation be

1 answer:

PSYCHO15rus [73]4 years ago

7 0

Answer:

(a). The value of d is 0.056 cm and 1.496 cm.

(b). The time period is 1.35 sec.

Explanation:

Given that,

Length = 50.00 cm

Time period = 2.50 s

Time period of pendulum is defined as the time for one complete cycle.

The period depends on the length of the pendulum.

Using formula of time period

$T=2\pi\sqrt{\dfrac{I}{mgh}}$

Where, I = moment of inertia

We need to calculate the value of d

Using parallel theorem of moment of inertia

$I=I_{cm}+md^2$

For a meter stick mass m , the rotational inertia about it's center of mass

$I_{cm}-\dfrac{mL^2}{12}$

Where, L = 1 m

Put the value into the formula of time period

$T=2\pi\sqrt{\dfrac{\dfrac{mL^2}{12}+md^2}{mgd}}$

$T=2\pi\sqrt{\dfrac{L^2}{12gd}+\dfrac{d}{g}}$

$T^2=4\pi^2(\dfrac{L^2}{12gd}+\dfrac{d}{g})$

Multiplying both sides by d

tex]T^2d=4\pi^2(\dfrac{L^2}{12g}+\dfrac{d^2}{g})[/tex]

$(\dfrac{4\pi^2}{g})d^2-T^2d+\dfrac{\pi^2L^2}{3g}=0$

Put the value of T, L and g into the formula

$4.028d^2-6.25d+0.336=0$

$d = 0.056\ m, 1.496\ m$

The value of d is 0.056 cm and 1.496 cm.

(b). Given that,

L = 50-5 = 45 cm

We need to calculate the time period

Using formula of period

$T=2\pi\sqrt{\dfrac{l}{g}}$

Put the value into the formula

$T=2\pi\sqrt{\dfrac{45\times10^{-2}}{9.8}}$

$T=1.35\ sec$

Hence, (a). The value of d is 0.056 cm and 1.496 cm.

(b). The time period is 1.35 sec.

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Estimate the kinetic energy of the earth with respect to the sun as the sum of two terms.

nekit [7.7K]

The definition of kinetic energy allows to find the result for the relationship between the energy of the sun and the Earth is:

The kinetic energy ratio is $\frac{K_{Sum} }{K_{Earth}} = 5.3 \ 10^2$

<h3 /><h3 /><h3> Kinetic enrgy.</h3>

Kinetic energy is the energy due to the movement of bodies, it is given by the relation

K = ½ m v²

where K is the kinetic energy, m the mass of the body and v the velocity of the body.

In a compound motion it is common to separate energy into parts to simplify calculations.

Translational kinetic energy. Due to the linear movement of the body

$K_{tras} =\frac{1}{2} m v^2$

Rotational kinetic energy. Due to the rotational movement of the body.

$K_{rot} = \frac{1}{2} I w^2$

Where I is the inrtia momentum and w the angular velocity.

They indicate that we compare the kinetic energy of the sun and the Earth.

The Earth has two movements, one of rotation about its axis with a period of T = 24 h and one of translation with respect to the Sun with a period of T= 365 days, therefore the kinetic energy of the Earth.

$K_{earth} = K_{tras} + K_{rot}$

Linear and rotational speed are related.

v = w r

The Earth is an almost spherical body therefore the moment of inertia of a solid sphere.

I = $\frac{2}{5 } m r^2$

Let's subatitute.

$K_{earth} = \frac{1}{2} \ m r^2_{tras} w^2_{tras} + \frac{1}{2} ( \frac{2}{5} m r^2_{earth}) w^2_{rot}$

The movement of the Earth around the sun is almost circular, therefore we can use the relations of the uniform circular movement, where the angle for one revolution is 2π radians and the time is called the period.

$w = \frac{2 \pi}{T}$

Let's substitute.

$K_{earth} = \frac{1}{2} m ( \frac{2\pi r^2_{tras}}{T_{tras}})^2 \ + \frac{1}{5} m (\frac{2\pi r^2_{earth} }{T^2_{rot}})^2$

$K_{earth} = 4 \pi^2 \ m \ ( \frac{1}{2} [ \frac{r_{tras}}{T_{tras}y} ]^2 + \frac{1}{5} [ \frac{r_{rot}}{T_{rot}}]^2)$

Data for Earth are tabulated:

Mass m = 5.98 1024 kg

Radius r = 6.37 10⁶ m

Radius orbits tras = 1.496 10¹¹ m

Rotation period $T_{rot}$ = 24 h ( $\frac{3600s}{1h}$ ) = 8.64 10⁴s

Translation period $T_{tras}$ = 365 d ( $\frac{24h}{1 d}$ ) ( $\frac{3600s}{1h}$ ) = 3.15 10⁷ s

Let's calculate.

$K_{earth} = 4 \pi^2 5.98 \ 10^{24} ( \frac{1}{2} ( \frac{1.496 \ 10^{11}}{3.15 \ 10^7 } )^2 \ + \frac{1}{5}( \frac{6.37 \ 10^6 }{8.64 \ 10^4})^2 )$

$K_{earth} = 2.36 \ 10^{26 } \ (1.128 \ 10^7 + 1.087 \ 10^3)$

$K_{earth}= 2.66 \ 10^{33} J$

Let's analyze the kinetic energy for the Sun, this is inside the solar system therefore it has no translation movement and is approximately a sphere with a rotation period of $T_{Sum}$ = 27 days.

The kinetic energy of the sun is;

$K_{sum} = K_{rot} = \frac{1}{2} I w^2$

$K_{sum} = \frac{1}{2} (\frac{2}{5} M R^2) (\frac{2\pi}{T_{sum}})^2$

$K_{sum} = \frac{4\pi^2 }{5} M (\frac{R}{T_{rot}})^2$

The tabulated data for the sun are:

Mass m = 1,991 1030 kg.

Radius R = 6.96 10⁸ m

Period T = 27 d ( $\frac{24h}{1 d}$ ) ( $\frac{3600s}{1h}$ ) = 2.33 10⁶ s

Let's calculate.

$K_{sum} = 1.40 \ 10^{36} J$

The relationship of the kinetic energy of the sun and the Earth is:

$\frac{K_{sum}}{K_{earth}} = \frac{1.40 \ 10^{36}}{2.66 \ 10^{33}}$

$\frac{K_{sum}}{K_{earth}} = 5.3 \ 10^2$

In conclusion using the definition of kinetic energy we can shorten the result for the relationship between the energy of the sun and the Earth is:

The kinetic energy ratio is: $\frac{K_{Sum}}{K_{Earth}} = 5 \ 10^2$

Learn more about kinetic energy here: brainly.com/question/25959744

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Answer: -the primary electron acceptor.

Explanation:

The photosystem II has a reaction center, in the reaction center the energy from sunlight is converted into high energy electrons. At the center of the reaction center, chlorophyll molecule is present which absorbs the light and one of its electron is promoted to the higher energy.

The high energy electron is hop downward and it is transferred to the plastoquinone A, which is a primary electron acceptor. Then the electron is transferred to the plastoquinone B. The plastoquinone B will receives enough electrons it delivers its electron to the electron transfer chain.

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