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kozerog [31]
3 years ago
13

Uniform circular motion can best be described as A) movement with right-angle turns. B) the motion of an object orbiting at vary

ing speeds. C) acceleration parallel to the direction of movement. D) the motion of an object in a circle at a constant speed.
Physics
1 answer:
murzikaleks [220]3 years ago
5 0
D
The motion of an object in a circle at a constant speed.
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Air enters a nozzle steadily at 2.21 kg/m3 and 20 m/s and leaves at 0.762 kg/m3 and 150 m/s. If the inlet area of the nozzle is
saveliy_v [14]

Answer:

a) The mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle is 23.6 cm².

Explanation:

a) The mass flow rate through the nozzle can be calculated with the following equation:

\dot{m_{i}} = \rho_{i} v_{i}A_{i}

Where:

v_{i}: is the initial velocity = 20 m/s

A_{i}: is the inlet area of the nozzle = 60 cm²  

\rho_{i}: is the density of entrance = 2.21 kg/m³

\dot{m} = \rho_{i} v_{i}A_{i} = 2.21 \frac{kg}{m^{3}}*20 \frac{m}{s}*60 cm^{2}*\frac{1 m^{2}}{(100 cm)^{2}} = 0.27 kg/s  

Hence, the mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle can be found with the Continuity equation:

\rho_{i} v_{i}A_{i} = \rho_{f} v_{f}A_{f}

0.27 kg/s = 0.762 kg/m^{3}*150 m/s*A_{f}

A_{f} = \frac{0.27 kg/s}{0.762 kg/m^{3}*150 m/s} = 0.00236 m^{2}*\frac{(100 cm)^{2}}{1 m^{2}} = 23.6 cm^{2}

Therefore, the exit area of the nozzle is 23.6 cm².

I hope it helps you!                                                                  

5 0
3 years ago
Read 2 more answers
Elements that typically give up electrons CHECK ALL THAT APPLY
lys-0071 [83]

Answer:

B. have a lower ionization energy

D. are metals

Explanation:

An atom can be defined as the smallest unit comprising of matter that forms all chemical elements. Thus, atoms are basically the building blocks of matters and as such determines or defines the structure of a chemical element.

Generally, atoms are typically made up of three distinct particles and these are protons, neutrons and electrons.

In Chemistry, electrons can be defined as subatomic particles that are negatively charged and as such has a magnitude of -1.

Valence electrons can be defined as the number of electrons present in the outermost shell of an atom. Valence electrons are used to determine whether an atom or group of elements found in a periodic table can bond with others. Thus, this property is typically used to determine the chemical properties of elements.

Valency can be defined as a measure of the combining power of a chemical element with other atoms to form a molecule or chemical compound.

Typically, valency is measured by the amount of hydrogen atoms that a chemical element can combine with or displace to form a molecule or chemical compound.

Ionization energy can be defined as the minimum energy required to remove or detach an electron from a neutral atom in a gaseous state.

Generally, the ionization energy of chemical elements tend to increase from left to right across a period on the periodic table. This increase is due to the fact that the atomic radius of chemical elements generally decreases across the periodic table, typically from alkali metals (group one elements such as hydrogen, lithium and sodium) to noble gases (group eight elements such as argon, helium and neon) i.e from left to the right of the periodic table. Also, the atomic radius of a chemical element increases down each group of the periodic table, typically from top to bottom (column).

This ultimately implies that, atoms with relatively large atomic radii tend to have a low electron affinity and a low ionization energy.

In conclusion, chemical elements that typically give up electrons are metals because their outermost shell contains excess electrons and have a lower ionization energy.

4 0
2 years ago
A wave is a:
liraira [26]

Answer:

wave

Explanation:

5 0
3 years ago
To apply Problem-Solving Strategy 12.2 Sound intensity. You are trying to overhear a most interesting conversation, but from you
Ivenika [448]

Answer:

r₂ = 0.316 m

Explanation:

The sound level is expressed in decibels, therefore let's find the intensity for the new location

            β = 10 log \frac{I}{I_o}

let's write this expression for our case

           β₁ = 10 log \frac{I_1}{I_o}

           β₂ = 10 log \frac{I_2}{I_o}

           

          β₂ -β₁ = 10 ( log \frac{I_2}{I_o} - log \frac{I_1}{I_o})

          β₂ - β₁ = 10 log \frac{I_2}{I_1}

          log \frac{I_2}{I_1} = \frac{60 - 20}{10} = 3

           \frac{I_2}{I_1} = 10³

           I₂ = 10³ I₁

having the relationship between the intensities, we can use the definition of intensity which is the power per unit area

           I = P / A

           P = I A

the area is of a sphere

          A = 4π r²

           

the power of the sound does not change, so we can write it for the two points

          P =  I₁ A₁ =  I₂ A₂

          I₁ r₁² = I₂ r₂²

we substitute the ratio of intensities

          I₁ r₁² = (10³ I₁ ) r₂²

         r₁² = 10³ r₂²

         

         r₂ = r₁ / √10³

         

we calculate

          r₂ = \frac{10.0}{\sqrt{10^3} }

          r₂ = 0.316 m

8 0
3 years ago
How many hydrogen and carbon atoms in a diamond
Wewaii [24]

Answer:

Explanation:

Thus, total 4+4=8 C atoms are present per unit cell of diamond. Carbon has an electronic arrangement of 2,4. In diamond, each carbon shares electrons with four other carbon atoms - forming four single bonds.

4 0
2 years ago
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