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9966 [12]
3 years ago
9

How many cranial nerves do we have

Physics
1 answer:
maxonik [38]3 years ago
8 0
12.


The twelve cranial nerves are: olfactory nerve, optic nerve, oculomotor nerve, trochlear nerve, trigeminal nerve, abducens nerve, facial nerve, vestibulocochlear nerve, glossopharengeal nerve, vagus nerve, spinal accessory nerve, and hypoglossal nerve.
:)
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An open cooking pot containing 0.5 liter of water at 20°C, 1 bar sits on a stove burner. Once the burner is turned on, the water
sesenic [268]

Answer:

a. the time required for the onset of evaporation is: 196.1 seconds and b. the time required for all of the water to evaporate is: 1328.3 seconds.

Explanation:

We need to stablish that there is 3 states at this problem. At the firts one, water is compressed liquid and the conditions for this state are: P1=100KPa,T1=20°C,V1=0.5m^3. From the compressed liquid chart and using extrapolation, we can get: v1=vf1=0.0010018 (m^3/Kg) and u1=uf1=83.95(KJ/kg). Now we can find the mass of water at the state 1 as: m=\frac{V_{1} }{v_{1} } =\frac{0.5*10^{-3} }{0.0010018}=0.5(Kg) Then the liquid water is heated at a rate of 0.85KW, and its volume increase, while work is done by the system at the boundary, we can assume that the pressure remains constant throughout the entire process. At the second state the water is saturated liquid and the conditions are: P2=100KPa, T2=Tsat=99.63°C, v2=vf2=0.001043(m^3/Kg) and u2=uf2=417.36(KJ/Kg). Now we can find the work as:W=mP(v_{2} -v_{1} )=0.5*100*(0.001043-0.0010018)=0.00207(KJ). (a) After that we need to do an energy balance for process 1-2 and get: U=Q-W or m(u_{2} -u_{1} )= Q*t-W, solving for t we get the time required for the onset of evaporation:t=\frac{0.5*(417.36-83.95)+0.00207}{0.85}=196.1(s).(b) Then continue heat transfer to the cooking pot and results in phase change getting vapor at 99.63°C. At the final state or third state the mass is zero because all liquid was evaporated and the initial mass at this state is the same for the second state: 0.5 (Kg) and doing an energy balances results in:(m_{3} u_{3} -m_{2} u_{2})=Q*t-W+( m_{3}-m_{2})h_{e}, but m3=0, now solving for t we can get the time required for all of the water to evaporate as:t=\frac{m_{2}(h_{e}-u_{2})+W}{Q}. We can get from the saturated liquid chart the enthalpy he=hge=2675.5(KJ/Kg) @P=100KPa. Now we need to calculate the work related with the volume decreases as vapor exits the control volume or process 2-3 work boundary as: W=\int\limits^3_2 {p} \, dV= p*(V_{3} -V_{2} )=-m_{2} P_{2} v_{2} =-(0.5)*100*0.001043=-0.0522(KJ). Now replacing every value in the time equation we get:t=\frac{0.5(2675.5-417.36)+(-0.0522)}{0.85}=1328.3(s)

6 0
3 years ago
What happens when condensation causes clouds to become very heavy?
nordsb [41]

Answer:

C. precipitation

Explanation:

if clouds store too much water from evaporation, it will rain which is precipitation

8 0
3 years ago
If a critical mass of fissionable material in a spherical shape is flattened like a hamburger, it will be?
lawyer [7]

A subcritical state will result by flattening a critical mass of fissionable material into the shape of a hamburger.

To find the answer, we need to know about the critical mass.

<h3>What is critical mass?</h3>
  • A material's nuclear characteristics (particularly, its nuclear fission cross-section), density, shape, enrichment, purity, temperature, and environment all affect its critical mass.
  • A mass of fissile material is considered to be in a critical state when a nuclear chain reaction in the mass is self-sustaining and there is no change in power, temperature, or neutron population.
  • At a specific temperature, a mass might be precisely critical.
  • A subcritical state will result by flattening a critical mass of fissionable material into the shape of a hamburger.
  • The cross sections for fission and absorption grow as the relative neutron velocity drops.

Thus, we can conclude that, a subcritical state will result by flattening a critical mass of fissionable material into the shape of a hamburger.

Learn more about the critical mass here:

brainly.com/question/12545809

#SPJ4

8 0
1 year ago
If you were standing on the moon and you had a radio signal transmitter, you could send the radio signal to Earth in 1.28 second
Hatshy [7]

Answer:

Distance between moon and earth = 3.84*10^{5}km

Explanation:

 Distance traveled by radio waves = Velocity of radio waves * Time taken by radio waves

 Velocity of radio waves = 3*10^{8}m/s

 Time taken by radio waves = 1.28 seconds

 So distance traveled by radio wave = 3*10^{8}*1.28=3.84*10^{8}m

 Distance between moon and earth = 3.84*10^{8}m=3.84*10^{5}km

7 0
3 years ago
Molybdenum has the BCC crystal structure, has a density of 10.22 g cm-3 and an atomic mass of 95.94 g mol-1. What is the atomic
puteri [66]

Answer:

atomic concentration = 2 atoms/unit cell

lattice parameter: a= 3.22 x 10⁻¹⁰ m

atomic radius: r= 1.39 x 10⁻¹⁰m

Explanation:

The atomic concentration is the number of atoms that can fit into a unit cell. It is a known number for each unit cell crystal structure. For a BCC (body-centered cube) crystal structure, atomic concentration is 2 atoms/unit cell because there are a 1/8 part of an atom in each corner of the cube (1/8 x 8= 1 atom) and 1 central atom in the central position of the cube ⇒ n= 1 atom + 1 atom= 2 atoms/unit cell

In order to calculate the lattice parameter a, we introduce the atomic mass 95.94 g/mol and the density 10.22 g/cm³ in the expression for the volume of the cube:

Vc= a³= \frac{(95.94 g/mol) x (2 atoms/unit cell)}{(10.2 g/cm^{3}) x (6.023 x 10^{23} atoms/mol)  }

a³= 3.12 x 10⁻²³ m³

⇒ a = ∛(3.12 x 10⁻²³ m³) = 3.22 x 10⁻¹⁰m

Once we know the lattice parameter a, we can calculate the atomic radius r by using the expression of a for a BCC structure:

a= \frac{4r}{\sqrt{3}}

⇒ r= a x √3/4= (3.22 x 10⁻¹⁰ m) x √3/4 = 1.39 x 10⁻¹⁰ m

5 0
3 years ago
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