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3 years ago

The phase of matter which is exposed to normal atmospheric pressure is solely dependent upon temperature? true or false

1 answer:

6 0

The statement above is true. The phase of matter which is exposed to normal atmospheric pressure is indeed solely dependent upon temperature. If the matter is exposed to the normal atm pressure, its temperature depends on it.

You might be interested in

The velocity of a mass moving in a circular path is..

Alenkasestr [34]

Constantly changing.

5 0

3 years ago

Consider four atoms from the second period: lithium, beryllium, boron, carbon, and nitrogen. Which of these emblems has the lowe

xz_007 [3.2K]

If you go to the right along the periodic table, electronegativity increases.
So the larger the column number, the greater the electronegativity.

-Lithium has lowest as it is in the 1st column

-Beryllium (2nd column)
-Boron s (13th column)
-Nitrogen (15th column)

6 0

3 years ago

You are trying to overhear a most interesting conversation, but from your distance of 10.0 m , it sounds like only an average wh

Alexandra [31]

Answer:

r₂=0.1 m

Explanation:

Given that

r₁= 10 m , β₁ = 20 dB

At r₂ ,β₂= 60 dB

As we know that intensity level of sound given as

$\beta =10\ log\dfrac{I}{10^{-12}}$

$\beta _1=10\ log\dfrac{I_1}{10^{-12}}$

$20=10\ log\dfrac{I_1}{10^{-12}}$

10² x 10⁻¹² = I₁

I₁=10⁻¹⁰ W/m²

$\beta _2=10\ log\dfrac{I_2}{10^{-12}}$

$60=10\ log\dfrac{I_1}{10^{-12}}$

10⁶ x 10⁻¹² = I₂

I₂ = 10⁻⁶ W/m²

I₁=10⁻¹⁰ W/m²

P = I A

P=Power ,I =Intensity ,A=Area

$\dfrac{I_1}{I_2}=\dfrac{r^2_2}{r^2_1}$

$\dfrac{10^{-10}}{10^{-6}}=\dfrac{r^2_2}{10^2}$

r₂=0.1 m

4 0

3 years ago

If the weight of the ruler is one Newton ,Gc cannot have a value more than 25cm

nevsk [136]

If the length of the ruler is 50 cm, the center of gravity cannot be greater than 25 cm.

The given parameters:

Weight of the ruler = 1 N

<h3>What is center of gravity (CG)?</h3>

Center of gravity is the point at which the weight of an object is concentrated.

Let the length of the ruler = L

The center of the gravity of the ruler is calculated as follows;

$X_{CG} = \frac{W(L_0) + W(L -X_{CG})}{W} \\\\X_{CG} = \frac{1(0) + 1(L -X_{CG})}{1}\\\\X_{CG} = L - X_{CG}\\\\X_{CG } + X_{CG} = L\\\\2X_{CG} = L\\\\X_{CG} = \frac{L}{2} \\\\when , \ L = 50 \ cm\\\\X_{CG} = \frac{50}{2} \\\\X_{CG} = 25 \ cm$

Thus, if the length of the ruler is 50 cm, the center of gravity cannot be greater than 25 cm. This may change if the length of the ruler changes because the center of gravity of uniform ruler depends on the length of the ruler.

Learn more about center of gravity here: brainly.com/question/6765179

4 0

2 years ago

Water, with a density of 1000 kg/m3, flows out of a spigot, through a hose, and out a nozzle into the air. The hose has an inner

stepan [7]

Answer:

P₁ = 2.3506 10⁵ Pa

Explanation:

For this exercise we use Bernoulli's equation and continuity, where point 1 is in the hose and point 2 in the nozzle

P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

A₁ v₁ = A₂ v₂

Let's look for the areas

r₁ = d₁ / 2 = 2.25 / 2 = 1,125 cm

r₂ = d₂ / 2 = 0.2 / 2 = 0.100 cm

A₁ = π r₁²

A₁ = π 1.125²

A₁ = 3,976 cm²

A₂ = π r₂²

A₂ = π 0.1²

A₂ = 0.0452 cm²

Now with the continuity equation we can look for the speed of water inside the hose

v₁ = v₂ A₂ / A₁

v₁ = 11.2 0.0452 / 3.976

v₁ = 0.1273 m / s

Now we can use Bernoulli's equation, pa pressure at the nozzle is the air pressure (P₂ = Patm) the hose must be on the floor so the height is zero (y₁ = 0)

P₁ + ½ ρ v₁² = Patm + ½ ρ v₂² + ρ g y₂

P₁ = Patm + ½ ρ (v₂² - v₁²) + ρ g y₂

Let's calculate

P₁ = 1.013 10⁵ + ½ 1000 (11.2² - 0.1273²) + 1000 9.8 7.25

P₁ = 1.013 10⁵ + 6.271 10⁴ + 7.105 10⁴

P₁ = 2.3506 10⁵ Pa

7 0

3 years ago