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Diano4ka-milaya [45]

3 years ago

13

The New Horizons probe that passed by Pluto during July 2015 is one of the fastest spacecraft ever assembled. It was moving at a

bout 14 km/s when it went by Pluto. If it maintained this speed, how long would it take New Horizons to reach the nearest star, Proxima Centauri, which is about 4.3 light-years away? (Note: It isn’t headed in that direction, but you can pretend that it is.)

1 answer:

LenKa [72]3 years ago

7 0

Answer:

2.91 x 10¹² sec

Explanation:

d = distance of nearest star, Proxima Centauri = 4.3 ly = 4.3 x 9.46 x 10¹⁵ m

v = speed of new horizon probe = 14 km/hr = 14000 m/s

t = time taken for the new horizon probe to reach nearest star, Proxima Centauri = ?

Using the equation

d = v t

Inserting the values given

4.3 x 9.46 x 10¹⁵ = (14000) t

t = 2.91 x 10¹² sec

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Its o.16 m/s2....really

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Where is the near point of an eye for which a contact lens with a power of +2.55 diopters is prescribed?Where is the far point o

tankabanditka [31]

Answer:

(a). The eye's near point is 68.98 cm from the eye.

(b). The eye's far point is 33.33 cm from the eye.

Explanation:

Given that,

Power = 2.55 D

Object distance = 25 cm for near point

Object distance = ∞ for far point

Suppose where is the far point of an eye for which a contact lens with a power of -3.00 D is prescribed for distant vision?

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$f =\dfrac{1}{P}$

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$f=\dfrac{100}{2.55}$

$f=39.21\ cm$

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Using formula of lens

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

Put the value into the formula

$\dfrac{1}{39.21}=\dfrac{1}{v}-\dfrac{1}{-25}$

$\dfrac{1}{39.21}-\dfrac{1}{25}=\dfrac{1}{v}$

$\dfrac{1}{v}=-\dfrac{1421}{98025}$

$v=-68.98\ cm$

The eye's near point is 68.98 cm from the eye.

(b). We need to calculate the focal length

Using formula of power

$f =\dfrac{1}{P}$

Put the value into the formula

$f=-\dfrac{100}{3.00}$

$f=-33.33\ cm$

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Using formula of lens

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

Put the value into the formula

$\dfrac{1}{-33.33}=\dfrac{1}{v}-\dfrac{1}{\infty}$

$-\dfrac{1}{33.33}+\dfrac{1}{\infty}=\dfrac{1}{v}$

$\dfrac{1}{v}=-\dfrac{1}{33.33}$

$v=-33.33\ cm$

The eye's far point is 33.33 cm from the eye.

Hence, (a). The eye's near point is 68.98 cm from the eye.

(b). The eye's far point is 33.33 cm from the eye.

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3 years ago

An electron is released from rest on the axis of a uniform positively charged ring, 0.200 m from the ring's center. If the linea

melisa1 [442]

Answer:

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<u>Given:</u>

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$V=\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\$

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If two equal forces act on an object in opposite directions, what is the net

labwork [276]

Answer:

Net Force = 0

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3 years ago