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tangare [24]
3 years ago
7

Following a gunshot wound to his head, Jack became more uninhibited, irritable, and profane. It is likely that his personality c

hange was the result of injury to his:
a.parietal lobe.
b.temporal lobe.
c.occipital lobe.
d.frontal lobe.
e.endocrine system.
Physics
1 answer:
Olegator [25]3 years ago
7 0

Answer:

d.Frontal lobe.

Explanation:

Frontal lobe :

This is our brain portion that controls important human mental abilities like the expression of emotion, solving problems, memory, judgment, language, and sexual behavior. Essentially, this is our personality's "control panel" and our ability to interact.

The capacity of a person to make good choices and weigh consequences is often impaired after a frontal lobe injury. Frontal lobe injury can result in increased irritability, which can include a change in attitude and an inability to control behavior.

Total volume of cerebral cortex (human): frontal lobe = 41%; occipital lobe = 18% ; parietal lobe = 19%;temporal lobe = 22%.

Therefore the answer is d.

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G a magnetic field perpendicular to the plane of a wire loop is uniform in space but changes with time t in the region of the lo
Luden [163]

Answer:

e = Δφ / Δt     induced emf is proportional to enclosed flux

Also φ  = B * A      flux is proportional to area and enclosed field

If the induced emf e increases with time than the flux and hence the magnetic field is increasing with time  (replace B with G)

Since e = ΔG * A / Δt    if e is linear then G must also be linear and be proportional to the time

6 0
2 years ago
2.0 mol of monatomic gas A initially has 5000 J of thermal energy. It interacts with 3.0 mol of monatomic gas B, which initially
Naddika [18.5K]

Answer:

 E_particle = 1,129 10⁻²⁰ J / particle

  T= 817.5 K

Explanation:

Energy is a scalar quantity so it is additive, let's look for the total energy of each gas

Gas a

         E_a = 2 5000 = 10000 J

Gas b

         E_b = 3 8000 = 24000 J

When the total system energy is mixed it is

          E_total = E_a + E_b

          E_total = 10000 + 24000 = 34000

The total mass is

           M = m_a + m_b

           M = 2 +3 = 5

The average energy among the entire mass is

           E_averge = E_total / M

            E_averago = 34000/5

            E_average = 6800 J

One mole of matter has Avogadro's number of atoms 6,022 10²³ particles

Therefore, each particle has an energy of

                E_particle = E_averag / 6.022 10²³ = 6800 /6.022 10²³

                E_particle = 1,129 10⁻²⁰ J / particle

For  find the temperature let's use equation

               E = kT

               T = E / k

     

               T = 1,129 10⁻²⁰ / 1,381 10⁻²³

               T = 8.175 102 K

               T= 817.5 K

5 0
4 years ago
A modern compact fluorescent lamp contains 1.4 mg of mercury (Hg). If each mercury atom in the lamp were to emit a single photon
Reika [66]

Answer:

A. 1.64 J

Explanation:

First of all, we need to find how many moles correspond to 1.4 mg of mercury. We have:

n=\frac{m}{M_m}

where

n is the number of moles

m = 1.4 mg = 0.0014 g is the mass of mercury

Mm = 200.6 g/mol is the molar mass of mercury

Substituting, we find

n=\frac{0.0014 g}{200.6 g/mol}=7.0\cdot 10^{-6} mol

Now we have to find the number of atoms contained in this sample of mercury, which is given by:

N=n N_A

where

n is the number of moles

N_A=6.022\cdot 10^{23} mol^{-1} is the Avogadro number

Substituting,

N=(7.0\cdot 10^{-6} mol)(6.022\cdot 10^{23} mol^{-1})=4.22\cdot 10^{18} atoms

The energy emitted by each atom (the energy of one photon) is

E_1 = \frac{hc}{\lambda}

where

h is the Planck constant

c is the speed of light

\lambda=508 nm=5.08\cdot 10^{-7}nm is the wavelength

Substituting,

E_1 = \frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{5.08\cdot 10^{-7} m}=3.92\cdot 10^{-19} J

And so, the total energy emitted by the sample is

E=nE_1 = (4.22\cdot 10^{18} )(3.92\cdot 10^{-19}J)=1.64 J

4 0
3 years ago
An ant does 1 N-m of work in dragging a 0.0020-N grain of sugar. How far does the ant drag the sugar.
algol13
The ant would drag the sugar 500m


Good luck!
8 0
3 years ago
Calculate a rate of cooling down of air from 80 C to 5C Show calculation. Give an answer in cubic meters per minute and cfm.
antoniya [11.8K]

Explanation:

Given that,

Rate of cooling of air

Initial temperature= 80°C

Final temperature = 5°C

We need to calculate

Using newton's law of cooling

\dfrac{dT}{dt}=c(T-T_{0})

\dfrac{dT}{dt}=c(\dfrac{T_{1}+T_{2}}{2}-T_{0})

Where, dT=T_{1}-T_{2}

Here, T =\dfrac{T_{1}+T_{2}}{2}

T_{0} = 25°C  (surrounding temperature)

dt = 1 minute

\dfrac{dT}{dt}=c(\dfrac{T_{1}+T_{2}}{2}-T_{0})

Put the value into the formula

\dfrac{80-5}{1}=c(\dfrac{85}{2}-25)

c=\dfrac{75}{17.5}

c=4.285\ cubic\ meter/minute

Hence, This is the required answer.

3 0
3 years ago
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