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Gekata [30.6K]
4 years ago
9

Stages of global warming

Physics
1 answer:
ddd [48]4 years ago
7 0

Answer:

- Glaciers melt

- The seas rise

- Temperature changes

- Humans add heat trapping greenhouse gasses

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A cylindrical metal specimen having an original diameter of 11.77 mm and gauge length of 46.1 mm is pulled in tension until frac
Marta_Voda [28]

Answer:

% reduction in area = 54.26 %

percentage elongation = 43.16 %

Explanation:

a) percentage reduction in area = \frac{(A_1 - A_2)}{A_1 } * 100

A_1  =\pi r^2 = \pi * 5.88^2=108.8 mm2

A_2=\pi * 3.98^2= 49.97 mm2

% reduction in area =\frac{(108.8 - 49.97)}{108.8} * 100 = 54.26 %

b)percentage elongation = \frac{(66 - 46.1)}{46.1} *100 = 43.16 %

5 0
3 years ago
Which tricks do you use to reduce pressure in everyday activities? Which tricks do you use to reduce pressure in everyday activi
Molodets [167]

Answer:

same problem is here also bro i cannot find this plz help anyone

8 0
3 years ago
Classify the properties as extensive or intensive: mass density; color volume; total energy; temperature; melting point
m_a_m_a [10]

Answer:

Intensive properties

Density

Color

temperature

Melting point

Extensive properties

Mass

Volume

Total Energy

Explanation:

Intensive properties:  In Physics, Intensive properties which are not depend of the amount of matter in a sample, It only depends of the type of matter, some examples of intensive properties are:

1. Density: It is a intensive property. It can explain better with a example:  the water density is 1000 kg/m3, So if we have 1 liter or 1000 liters of water  the density will be the same for the two samples.

2. Color: Solid sodium chloride is white. If you have 2 samples the first recipient  with 2 kilograms of NaCl and the second with 10 kilograms of NaCl. The color of the substance does not depend on the amount of the substance.

As was mentioned before the same theory is applied to temperature and melting point concepts.

On the other hand,

Extensive properties are properties of the matter which depend on the amount of matter that is present in the system or sample. some examples are:

1. Mass: It is a property that measures the amount of matter that an object contains. For example, 10 kilograms of solid Copper contains a higher mass than 2 kilograms of the same metal.  

2. Volume: It is a property which measures the space occupied by an object or a substance. For example, the space occupied by a glass of milk is lower than the space occupied by a bottle of milk, Then the volume of the glass of milk is lower than the volume of the bottle of milk.

3. Finally the total energy is contained in molecules and atoms that constituted systems  so, if the amount of matter increases the number of molecules too, then the total energy will increase.

I hope it helps you.

6 0
3 years ago
Visible light passes through a diffraction grating that has 900 slits per centimeter, and the interference pattern is observed o
kobusy [5.1K]

Answer:

\Delta \lambda=14.3\ nm

Explanation:

It is given that,

The number of lines per unit length, N = 900 slits per cm

Distance between the formed pattern and the grating, l = 2.3 m

n the first-order spectrum, maxima for two different wavelengths are separated on the screen by 2.98 mm, \Delta Y=2.98\ mm = 0.00298\ m

Let d is the slit width of the grating,

d=\dfrac{1}{N}

d=\dfrac{1}{900\ cm}

d=1.11\times 10^{-5}\ m

For the first wavelength, the position of maxima is given by :

y_1=\dfrac{L\lambda_1}{d}

For the other wavelength, the position of maxima is given by :

y_2=\dfrac{L\lambda_2}{d}

So,

\Delta \lambda=\dfrac{\Delta y d}{l}

\Delta \lambda=\dfrac{0.00298\times 1.11\times 10^{-5}}{2.3}

\Delta \lambda=1.43\times 10^{-8}\ m

or

\Delta \lambda=14.3\ nm

So, the difference between these wavelengths is 14.3 nm. Hence, this is the required solution.

3 0
3 years ago
Red light of wavelength 633 nmnm from a helium-neon laser passes through a slit 0.360 mmmm wide. The diffraction pattern is obse
777dan777 [17]

Answer:

      Δy= 5,075 10⁻⁶ m

Explanation:

The expression that describes the interference phenomenon is

      d sin θ = (m + ½) λ

As the observation is on a distant screen

     tan θ = y / x

     tan θ= sin θ/cos θ

As in ethanes I will experience the separation of the vines is small and the distance to the big screen

          tan θ = sin θ

Let's replace

     d y / x = (m + ½) λ

The width of a bright stripe at the difference in distance  

     y₁ = (m + ½) λ x / d

     m = 1

      y₁ = 3/2 λ x / d

Let's use m = 1, we look for the following interference,

             m = 2

             y₂ = (2+ ½) λ x / d

The distance to the screen is constant x₁ = x₂ = x₀

The width of the bright stripe is

           Δy = λ x / d (5/2 -3/2)

           Δy = 630 10⁻⁹ 2.90 /0.360 10⁻³ (1)

           Δy= 5,075 10⁻⁶ m

8 0
3 years ago
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