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3 years ago

7

Relative to the distance of an object in front of a plane mirror, how far behind the mirror is the image?

1 answer:

koban [17]3 years ago

3 0

It actually should be relatively close

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In which situation is the gravitational force between two objects hard to detect? (Options)

svlad2 [7]

A - the objects are too small

GRAVITATIONAL FORCE IS EXPERIENCED BY ALL OBJECTS IN THE UNIVERSE ALL THE TIME. BUT THE ORDINARY OBJECTS YOU SEE EVERY DAY HAVE MASSES SO SMALL THAT THEIR ATTRACTION TOWARD EACH OTHER IS HARD TO DETECT. -https://www.ftsd.org/cms/lib6/MT01001165/Centricity/ModuleInstance/630/CHAPTER_2_NOTES_FOR_EIGHTH_GRADE_PHYSICAL_SCIENCE.pdf

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3 years ago

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A wheel rotates with a constant angular acceleration of  rad/s2. During a certain time interval its angular displacement is  r

Hatshy [7]

Answer:

The angular velocity at the beginning of the interval is $\pi\sqrt{2}\ rad/s$ .

Explanation:

Given that,

Angular acceleration $\alpha=\pi\ rad/s^2$

Angular displacement $\theta=\pi\ rad$

Angular velocity $\omega =2\pi\ rad/s$

We need to calculate the angular velocity at the beginning

Using formula of angular velocity

$\alpha =\dfrac{\omega^2-\omega_{0}^2}{2\theta}$

$\omega_{0}^2=\omega^2-2\alpha\theta$

Where, $\alpha$ = angular acceleration

$\omega$ = angular velocity

Put the value into the formula

$\omega_{0}^2=(2\pi)^2-2\times\pi\times\pi$

$\omega=\sqrt{2\pi^2}$

$\omega_{0}=\pi\sqrt{2}\ rad/s$

Hence, The angular velocity at the beginning of the interval is $\pi\sqrt{2}\ rad/s$ .

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3 years ago

A 4.0-kg object has 72 J of kinetic energy.<br> Determine its speed.

Svetlanka [38]

Answer:

6m/s

Explanation:

v = sqrt of 2KE/m

Where:

KE = kinetic energy

m = mass of a body

v = velocity of a body

= sqrt of 2(72)/4

= sqrt of 144/4

= sqrt of 36

= 6m/s

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2 years ago

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Aliun [14]

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Zolol [24]

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