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Andrews [41]
3 years ago
14

The period of a simple pendulum, defined as the time necessary for one complete oscillation, is measured in time units and is gi

ven by
T = 2π×square root of l/g

where ℓ is the length of the pendulum and g is the acceleration due to gravity, in units of length divided by time squared. Show that this equation is dimensionally consistent.
Physics
1 answer:
Oksana_A [137]3 years ago
7 0

The task is to show that the right side of the equation has units of [Time], just like the left side has.

The right side of the equation is . . . 2 π √(L/G) .

We can completely ignore the  2π since it has no units at all, so it has no effect on the units of the right side of the equation.  Now the task is simply to find the units of  √(L/G) .

L . . . meters

G . . . meters/sec²

(L/G) = (meters) / (meters/sec²)

(L/G) = (meters) · (sec²/meters)

(L/G) = (meters · sec²) / (meters)

(L/G) = sec²

So  √(L/G) = seconds = [Time]

THAT's what we were hoping to prove, and we did it !

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3 years ago
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What would happen to the moon if there was no sun and earth ?
iragen [17]

Answer:

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4 0
3 years ago
A 53-N force is needed to keep a 50.0-kg box sliding across a flat surface at a constant velocity. What is the coefficient of ki
earnstyle [38]

The weight of the box is (mass) x (gravity) = (50 kg) x (9.8m/s²) = 490 newtons.

If the box is sliding at constant speed, and not speeding up or slowing down,
that means that the horizontal forces on it add up to zero. 

Since you're pushing on it with 53N in <em><u>that</u></em> direction, friction must be pulling
on it with 53N in the <u><em>other</em></u> direction.

 The 53N of friction is (the weight) x (the coefficient of kinetic friction).

                                                  53N  =  (490N) x (coefficient).

Divide each side by  490N :  Coefficient = (53N) / (490N)  =  0.1082 .

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5 0
3 years ago
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A sound source is moving at 80 m/s toward a stationary listener that is standing in still air (a) Find the wavelength of the sou
Setler [38]

Answer:

a. wavelength of the sound, \vartheta = 1.315\vartheta_{o}

b. observed frequecy, \lambda = 0.7604\lambda_{o}

Given:

speed of sound source, v_{s} = 80 m/s

speed of sound in air or vacuum, v_{a} = 343 m/s

speed of sound observed, v_{o} = 0 m/s

Solution:

From the relation:

v = \vartheta \lambda        (1)

where

v = velocity of sound

\vartheta = observed frequency of sound

\lambda = wavelength

(a) The wavelength of the sound between source and the listener is given by:

\lambda = \frac{v_{a}}{\vartheta }         (2)

(b) The observed frequency is given by:

\vartheta = \frac{v_{a}}{v_{a} - v_{s}}\vartheta_{o}

\vartheta = \frac{334}{334 - 80}\vartheta_{o}

\vartheta = 1.315\vartheta_{o}                (3)

Using eqn (2) and (3):

\lambda = \frac{334}{1.315} = \frac{1}{1.315}\frac{v_{a}}{\vartheta_{o}}

\lambda = 0.7604\lambda_{o}

4 0
3 years ago
What scientific observation did Edwin Hubble use to determine distances between galaxies?
Cerrena [4.2K]

Answer: the expanding universe

Explanation:

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