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Ray Of Light [21]
3 years ago
6

If the mercury barometer is 75.58 cm at the base of the mountain and 66.37 cm at the summit. What is the height of the mountain?

Physics
1 answer:
Licemer1 [7]3 years ago
3 0

Answer:

5016.2446

Explanation:

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What happens inside when an air-filled, sealed can, is heated?.
g100num [7]

Answer:

It would explode

Explanation:

you're welcome

5 0
2 years ago
Read 2 more answers
Please help me whit this question
mariarad [96]

Answer:

Gypsum

Explanation:

Gypsum is a soft sulfate mineral composed of Calcium Sulfate di-hydrate. It is widely used in composition of fertilizer. It is also known as land plaster. On adding gypsum to soil, the quality of soil is improved. It conditions the soil and adds nutrients. The properties of soil is improved. Hence, the correct answer is Gypsum.

Hope this helped:)

pls mark brainlist

6 0
3 years ago
PLEASE HELP ILL MARK BRAINLIEST
ikadub [295]
I think the answer is A?
7 0
3 years ago
Using Mirror equation A, Calculate The Frequency Of The Long Stand And The Shortest Wave Length Of That An Object Is Placed Of A
Leya [2.2K]

The Image distance and Magnification of The Image ​will be 30 cm and 3.

<h3>What is focal length?</h3>

The focal length of the lens, which is often expressed in millimeters, is the distance between the lens and the image sensor when the subject is in focus.

Given data;

Focal length,f=?

Image distance,v=?

Object distance,u= 10 cm

Magnification,m= 2.85

The focal length is half of the radius;

f=R/2

f=30 Cm/2

f= 15 Cm

The mirror equation is found as;

\rm \frac{1}{f} =\frac{1}{v} +\frac{1}{u} \\\\\ \frac{1}{15}=\frac{1}{v}+\frac{1}{10} }  \\\\\ v= -30 \ cm

The magnification of the lens is found as;

\rm m=\frac{30}{10}\\\\ m=3

Hence, the image distance and magnification of The image ​will be 30 cm and 3.

To learn more about the focal length refer;

brainly.com/question/16188698

#SPJ1

6 0
2 years ago
What will be the final temperature if a 4.00 g silver ring at 41.0◦C if it gives off 18.0 J of heat to the surroundings? The spe
Eduardwww [97]

Answer:

Final temperature, T_f=21.85^{\circ}

Explanation:

Given that,

Mass of silver ring, m = 4 g

Initial temperature, T_i=41^{\circ}C

Heat released, Q = -18 J (as heat is released)

Specific heat capacity of silver, c=0.235\ J/g\ C

To find,

Final temperature

Solution,

The expression for the specific heat is given by :

Q=mc\Delta T

Q=mc(T_f-T_i)

T_f=\dfrac{-Q}{mc}+T_i

T_f=\dfrac{-18}{4\times 0.235}+41

T_f=21.85^{\circ}

So, the final temperature of silver is 21.85 degrees Celsius.

5 0
3 years ago
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