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mash [69]
3 years ago
5

When a substance goes directly from a gaseous state to a solid state as dry ice does

Physics
2 answers:
Naddik [55]3 years ago
6 0

Answer:  Sublimation I think.

Explanation:

wlad13 [49]3 years ago
4 0

Answer:

Sublimation

Explanation:

not entirly sure if I'm correct

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PLEASE HELP!!!!!
Arte-miy333 [17]

Answer:

it states that energy can either be gained or lost but it only changes its form.

Explanation:

for example:as a ball is still on the table it posses a potential energy of 100j and a k.e of 0j,as it falls it gains k.e so the midpoint the p.e is equal to the k.e (50j equally) as it approches the ground it completely gains k.e (100j) and the p.e is 0j.

total energy is 100j so it has been converted from p.e to k.e.

hope u have understood.

5 0
3 years ago
The atomic mass of an atom do not contain electrons because
Dimas [21]

Answer:

Electrons are so small that it does not affect the mass of atom .

Explanation:

Electrons are much smaller in mass than protons, weighing only 9.11 × 10^-28 grams, or about 1/1800 of an atomic mass unit. Therefore, they do not contribute much to an element's overall atomic mass.

7 0
3 years ago
How do fossils provide evidence that evolution has taken place
zzz [600]
<span>Fossils provide solid evidence that organisms from the past are not the same as those found today; they show a progression of evolution. Scientists calculate the age of fossils and categorize them to determine when the organisms lived relative to each other. Hope this helps</span>
7 0
3 years ago
Consider an object with s=12cm that produces an image with s′=15cm. Note that whenever you are working with a physical object, t
Leni [432]

A. 6.67 cm

The focal length of the lens can be found by using the lens equation:

\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}

where we have

f = focal length

s = 12 cm is the distance of the object from the lens

s' = 15 cm is the distance of the image from the lens

Solving the equation for f, we find

\frac{1}{f}=\frac{1}{12 cm}+\frac{1}{15 cm}=0.15 cm^{-1}\\f=\frac{1}{0.15 cm^{-1}}=6.67 cm

B. Converging

According to sign convention for lenses, we have:

- Converging (convex) lenses have focal length with positive sign

- Diverging (concave) lenses have focal length with negative sign

In this case, the focal length of the lens is positive, so the lens is a converging lens.

C. -1.25

The magnification of the lens is given by

M=-\frac{s'}{s}

where

s' = 15 cm is the distance of the image from the lens

s = 12 cm is the distance of the object from the lens

Substituting into the equation, we find

M=-\frac{15 cm}{12 cm}=-1.25

D. Real and inverted

The magnification equation can be also rewritten as

M=\frac{y'}{y}

where

y' is the size of the image

y is the size of the object

Re-arranging it, we have

y'=My

Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.

Also, the sign of s' tells us if the image is real of virtual. In fact:

- s' is positive: image is real

- s' is negative: image is virtual

In this case, s' is positive, so the image is real.

E. Virtual

In this case, the magnification is 5/9, so we have

M=\frac{5}{9}=-\frac{s'}{s}

which can be rewritten as

s'=-M s = -\frac{5}{9}s

which means that s' has opposite sign than s: therefore, the image is virtual.

F. 12.0 cm

From the magnification equation, we can write

s'=-Ms

and then we can substitute it into the lens equation:

\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}\\\frac{1}{f}=\frac{1}{s}+\frac{1}{-Ms}

and we can solve for s:

\frac{1}{f}=\frac{M-1}{Ms}\\f=\frac{Ms}{M-1}\\s=\frac{f(M-1)}{M}=\frac{(-15 cm)(\frac{5}{9}-1}{\frac{5}{9}}=12.0 cm

G. -6.67 cm

Now the image distance can be directly found by using again the magnification equation:

s'=-Ms=-\frac{5}{9}(12.0 cm)=-6.67 cm

And the sign of s' (negative) also tells us that the image is virtual.

H. -24.0 cm

In this case, the image is twice as tall as the object, so the magnification is

M = 2

and the distance of the image from the lens is

s' = -24 cm

The problem is asking us for the image distance: however, this is already given by the problem,

s' = -24 cm

so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.

3 0
4 years ago
HELP PLS<br> NO EXPLANATION NEEDED
GalinKa [24]

Explanation:

v ^2 - u ^2 = 2 a s

- u ^2 = 2 * - 9.8 * 0.411

- u ^2 = - 8.0556

u = √8.0556

u = 2.83 m / sec

3 0
3 years ago
Read 2 more answers
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