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2 years ago

A pendulum completes 2 oscillation in 5s.

Physics

1 answer:

Otrada [13]2 years ago

6 0

Answer:

Explanation:

<h3>Given :-</h3><h2>Completes 2 oscillation in 5s.</h2><h2>A) Time period = ?</h2><h2 /><h2> $t \: = \frac{total \: time \: }{total \: no \: of \: oscillation}$ </h2><h2 /><h2> $\frac{5}{2} = 2.5s$ </h2>

<em>Thus 1 oscillation is completed in 2.5 </em><em>s.</em>

<h2>B) If g = 9.8 ms−², find its length.</h2><h2> $t \: = 2\pi \frac{ \sqrt{l} }{g }$ </h2><h3>= 2.5s </h3><h3> $2\pi \frac{ \sqrt{l} }{g}$ </h3><h3>Length = ? </h3><h2> $\frac{ {t}^{2} }{4\pi {}^{2} } \times g$ </h2><h3 /><h3> $\frac{2.5}{ {4}^{2} } \times 9.8 = 1.55m$ </h3>

Thus the length of g = 9.8 ms −² = 1.55m

<h2>Hope it helps u.......</h2><h2>STAY SAFE, STAY HEALTHY AND BLESSED .</h2><h2> HAVE A MARVELOUS DAY AHEAD !</h2><h2>THANK YOU !</h2>

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Two sounds are played for you. The first has a frequency of 196 Hz. The second has a frequency of 392 Hz. Which sound will be "h

kykrilka [37]

Answer:

The second one, the higher the hertz the higher the frequency we hear

Explanation:

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3 years ago

Explain how mirrors can produce images that are larger or smaller than life size, as well as upright or inverted

galina1969 [7]

Answer:

1) When $d_{o}$ < $d_{i}$ (hence $d_{o}$ < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

2) When $d_{o}$ > $d_{i}$ (and $d_{o}$ < f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted

3) When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror

Explanation:

The position of an object in front of a concave mirror of radius of curvature, R, determines the size and orientation of the image of the object as illustrated in the mirror equation

$\dfrac{1}{f}=\dfrac{1}{d_{o}} + \dfrac{1}{d_{i}}$

$Magnification, \, m = \dfrac{h_{i}}{h_{o}} = -\dfrac{d_{i}}{d_{o}}$

Where:

f = Focal length of the mirror = R/2

$d_{i}$ = Image distance from the mirror

$d_{o}$ = Object distance from the mirror

$h_{i}$ = Image height

$h_{o}$ = Object height

$d_{o}$ is positive for an object placed in front of the mirror and negative for an object placed behind the mirror

$d_{i}$ is positive for an image formed in front of the mirror and negative for an image formed behind the mirror

m is positive when the orientation of the image and the object is the same

m is negative when the orientation of the image and the object is inverted

f and R are positive in the situation where the center of curvature is located in front of the mirror (concave mirrors) and f and R are negative in the situation where the center of curvature is located behind the mirror (convex mirrors)

∴ When $d_{o}$ < $d_{i}$ (hence $d_{o}$ < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

When $d_{o}$ > $d_{i}$ (and $d_{o}$ < f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted

When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror.

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3 years ago

Using the law of conservation of angular momentum, estimate how fast a collapsed stellar core would spin if its initial spin rat

Nataly_w [17]

Answer:

$\omega_{f} = 1000000\,\frac{rev}{day}$

Explanation:

The law of conservation of angular momentum states that angular momentum remains constant when there is no external moment or forces applied to the system. Let assume that star can be modelled as an sphere, then:

$\frac{2}{5}\cdot M\cdot R_{o}^{2} \cdot \omega_{o} = \frac{2}{5}\cdot M\cdot R_{f}^{2} \cdot \omega_{f}$

The final angular speed is:

$\omega_{f} = \omega_{o}\cdot (\frac{R_{o}}{R_{f}})^{2}$

$\omega_{f} = (1\,\frac{rev}{day} )\cdot (\frac{10000\,km}{10\,km} )^{2}$

$\omega_{f} = 1000000\,\frac{rev}{day}$

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3 years ago

Trey races his bicycle for 192m. A wheel of his bicycle turns 48 times as the bicycle travels this distance. What is the diamete

chubhunter [2.5K]

Answer:

1.27 m

Explanation:

Distance = 192 m

number of rotations = 48

Distance traveled in one rotation = 2 x π x r

Where, r be the radius of wheel.

so, distance traveled in 48 rotations = 48 x 2 x 3.14 x r

It is equal to the distance traveled.

192 = 48 x 2 x 3.14 x r

r = 0.637 m

diameter of wheel = 2 x radius of wheel = 2 x 0.637 = 1.27 m

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2 years ago

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2 years ago