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Nookie1986 [14]

3 years ago

7

15 PTS!!!!

2 answers:

faltersainse [42]3 years ago

7 0

B. group I think.
:))))))))

Anon25 [30]3 years ago

6 0

I thinks its period because they are on the last column to the right

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Explain how the planets in the solar system formed

Fed [463]

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3 years ago

A 0.144-kg baseball is moving toward home plate with a speed of 43 m/s when

algol [13]

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3 years ago

An apple is placed 20.0 cm in front of a diverging lens of focal length 10.0 cm. Find the image distance and the magnification o

jenyasd209 [6]

Answer:

Image distance of apple=-6.7 cm

Magnification of apple=0.33

Explanation:

We are given that an apple is placed 20.cm in front of a diverging lens.

Object distance=u=-20 cm

Focal length=f=-10 cm

Because focal length of diverging lens is negative.

We have to find the image distance and magnification of the apple.

Lens formula

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

Substitute the values then we get

$-\frac{1}{10}=\frac{1}{v}+\frac{1}{20}$

$\frac{1}{v}=-\frac{1}{10}-\frac{1}{20}$

$\frac{1}{v}=\frac{-2-1}{20}=-\frac{3}{20}$

$v=-\frac{20}{3}=-6.7 cm$

Image distance of apple=-6.7 cm

Magnification=m= $\frac{v}{u}=\frac{-\frac{20}{3}}{-20}$

Magnification of apple= $\frac{1}{3}=0.33$

Hence, the magnification of apple=0.33

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2 years ago

Peg P is driven by the forked link OA along the path described by r = eu, where r is in meters. When u = p4 rad, the link has an

8_murik_8 [283]

Answer:

The transverse component of acceleration is 26.32 $m/s^2$ where as radial the component of acceleration is 8.77 $m/s^2$

Explanation:

As per the given data

u=π/4 rad

ω=u'=2 rad/s

α=u''=4 rad/s

$r=e^u$

So the transverse component of acceleration are given as

$a_{\theta}=(ru''+2r'u')\\$

Here

$r=e^u\\r=e^{\pi/4}\\r=2.1932 m$

$r'=e^u.u'\\r'=2.1932 \times 2\\r'=4.3864 m$

So

$a_{\theta}=(ru''+2r'u')\\a_{\theta}=(2.1932\times 4+2\times 4.3864 \times 2)\\a_{\theta}=26.32 m/s\\$

The transverse component of acceleration is 26.32 $m/s^2$

The radial component is given as

$a_r=r''-r\theta'^2$

Here

$r''=e^u.u'^2+e^u u''\\r''=2.1932 \times (2)^2+2.1932\times 4\\r''=17.5456 m$

So

$a_r=r''-ru'^2\\a_r=17.5456-2.1932\times (2)^2\\a_r=8.7728 m/s^2$

The radial component of acceleration is 8.77 $m/s^2$

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3 years ago

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