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Brilliant_brown [7]
3 years ago
10

These two cubes represent a small cell and large cell .

Physics
1 answer:
Elena L [17]3 years ago
7 0

Answer:

Idk

Explanation:

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1/3 the weight than it is on earth, duh
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You need to build a prototype with machined parts that withstand a saline corrosive environment and temperatures above 200 degre
kondor19780726 [428]

Answer:

Stainless steel

Explanation:

I will try to order the solutions from the least correct to the most correct.

Since a temperature greater than 200 ° F is required, that is to say approximately 93 ° c, <em>Polycaprolactone</em> is the least indicated. Its melting point is approximately 60 ° C, so it would not serve the required application.

On the other hand we have<em> Untreated aluminum</em>, which although it has a melting point higher than the required one, without a zinc and magnesium treatment it will easily oxidize in a salty environment, so it cannot be used in this choice either.

We have to compare the two steels.

The<em> Mild Steel </em>has a better corrosion resistance than the previous ones, but in a long-term cycle it will end up full of corrosion and therefore its properties will be highly affected.

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In addition its melting point is above 1500 ° c.

The best choice is stainless steel.

5 0
3 years ago
If the president vetoes a bill, it can still become law If congress overrides the presidents veto with ___ majority
vovangra [49]
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THIS MARCIN
nekit [7.7K]

Answer:

The image is formed at a ‘distance of 16.66 cm’ away from the lens as a diminished image of height 3.332 cm. The image formed is a real image.

Solution:

The given quantities are

Height of the object h = 5 cm

Object distance u = -25 cm

Focal length f = 10 cm

The object distance is the distance between the object position and the lens position. In order to find the position, size and nature of the image formed, we need to find the ‘image distance’ and ‘image height’.

The image distance is the distance between the position of convex lens and the position where the image is formed.

We know that the ‘focal length’ of a convex lens can be found using the below formula

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

f

1

=

v

1

−

u

1

Here f is the focal length, v is the image distance which is known to us and u is the object distance.

The image height can be derived from the magnification equation, we know that

Magnification=h′h=vu\text {Magnification}=\frac{h^{\prime}}{h}=\frac{v}{u}Magnification=

h

h

′

=

u

v

Thus,

h′h=vu\frac{h^{\prime}}{h}=\frac{v}{u}

h

h

′

=

u

v

First consider the focal length equation to find the image distance and then we can find the image height from magnification relation. So,

1f=1v−1(−25)\frac{1}{f}=\frac{1}{v}-\frac{1}{(-25)}

f

1

=

v

1

−

(−25)

1

1v=1f+1(−25)=110−125\frac{1}{v}=\frac{1}{f}+\frac{1}{(-25)}=\frac{1}{10}-\frac{1}{25}

v

1

=

f

1

+

(−25)

1

=

10

1

−

25

1

1v=25−10250=15250\frac{1}{v}=\frac{25-10}{250}=\frac{15}{250}

v

1

=

250

25−10

=

250

15

v=25015=503=16.66 cmv=\frac{250}{15}=\frac{50}{3}=16.66\ \mathrm{cm}v=

15

250

=

3

50

=16.66 cm

Then using the magnification relation, we can get the image height as follows

h′5=−16.6625\frac{h^{\prime}}{5}=-\frac{16.66}{25}

5

h

′

=−

25

16.66

So, the image height will be

h′=−5×16.6625=−3.332 cmh^{\prime}=-5 \times \frac{16.66}{25}=-3.332\ \mathrm{cm}h

′

=−5×

25

16.66

=−3.332 cm

Thus the image is formed at a distance of 16.66 cm away from the lens as a diminished image of height 3.332 cm. The image formed is a ‘real image’.

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Around 70-72% of earth’s surface is covered in water (most of it is salt water).
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