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Ber [7]
3 years ago
14

Research indicates that inattentional blindness often decreases when people work on tasks that require a great deal of attention

. Please select the best answer from the choices provided T F
Physics
2 answers:
Shalnov [3]3 years ago
7 0

Answer:

False

Explanation:

Inattentionla blindness is a condition that happens when you overview or do not notice an object that is not supposed to be in a certain place and that is strange for its context because you are paying attention to something else, this means that if you do a task that needs a great deal of attention you will actually increase the inattentional blindness.

QveST [7]3 years ago
6 0

The correct answer is F

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A race-car drives around a circular track of radius RRR. The race-car speeds around its first lap at linear speed v_iv i ​ v, st
OLEGan [10]

Answer:

The magnitude of the centripetal acceleration increases by 16 times when the linear speed increases by 4 times.

Explanation:

The initial centripetal acceleration, a of the race-car around the circular track of radius , R with a linear speed v is a = v²/R.

When the linear speed of the race-car increases to v' = 4v, the centripetal acceleration a' becomes a' = v'²/R = (4v)²/R = 16v²/R.

So the centripetal acceleration, a' = 16v²/R.

To know how much the magnitude of the car's centripetal acceleration changes, we take the ratio a'/a = 16v²/R ÷ v²/R = 16

a'/a = 16

a' = 16a.

So the magnitude of the centripetal acceleration increases by 16 times when the linear speed increases by 4 times.

8 0
4 years ago
To get copper from the solid phase to the liquid phase must of the copper sample
Elina [12.6K]
You would want to increase the temperature. 
8 0
3 years ago
Read 2 more answers
Imagine that an electron in an excited state in a nitrogen molecule decays to its ground state, emitting a photon with a frequen
mash [69]
Since energy cannot be created nor destroyed, the change in energy of the electron must be equal to the energy of the emitted photon.

The energy of the emitted photon is given by:
E=hf
where
h is the Planck constant
f is the photon frequency
Substituting f=8.88 \cdot 10^{14}Hz, we find
E=hf=(6.6 \cdot 10^{-34} Js)(8.88 \cdot 10^{14} Hz)=5.86 \cdot 10^{-19} J

This is the energy given to the emitted photon; it means this is also equal to the energy lost by the electron in the transition, so the variation of energy of the electron will have a negative sign (because the electron is losing energy by decaying from an excited state, with higher energy, to the ground state, with lower energy)
\Delta E= -5.86 \cdot 10^{-19} J
6 0
3 years ago
Read 2 more answers
Using Newton's third law of motion, explain what happens when you let an untied balloon go.
mash [69]

Answer:

balloon pushes you back

Explanation:

3rd Law: Every action has an equal and opposite reaction

So, when you let go of the balloon it's pushed forward so the balloon pushes you back

7 0
4 years ago
Read 2 more answers
Two cylindrical resistors are made from the same material. The shorter one has length L, diameter D, and resistance R1. The long
nordsb [41]

Answer:

the resistance of the longer one is twice as big as the resistance of the shorter one.

Explanation:

Given that :

For the shorter cylindrical resistor

Length = L

Diameter = D

Resistance = R1

For the longer cylindrical resistor

Length = 8L

Diameter = 4D

Resistance = R2

So;

We all know that the resistance of a given material can be determined by using the formula :

R = \dfrac{\rho L }{A}

where;

A = πr²

R = \dfrac{\rho L }{\pi r ^2}

For the shorter cylindrical resistor ; we have:

R = \dfrac{\rho L }{\pi r ^2}

since 2 r = D

R = \dfrac{\rho L }{\pi (\frac{2}{2 \ r}) ^2}

R = \dfrac{ 4 \rho L }{\pi \ D   ^2}

For the longer cylindrical resistor ; we have:

R = \dfrac{\rho L }{\pi r ^2}

since 2 r = D

R = \dfrac{ \rho (8 ) L }{\pi (\frac{2}{2 \ r}) ^2}

R = \dfrac{32\rho L }{\pi \ (4 D)   ^2}

R = \dfrac{2\rho L }{\pi \ (D)   ^2}

Sp;we can equate the shorter cylindrical resistor to the longer cylindrical resistor as shown below :

\dfrac{R_s}{R_L} = \dfrac{ \dfrac{ 4 \rho L }{\pi \ D   ^2}}{ \dfrac{2\rho L }{\pi \ (D)   ^2}}

\dfrac{R_s}{R_L} ={ \dfrac{ 4 \rho L }{\pi \ D   ^2}}* { \dfrac  {\pi \ (D)   ^2} {2\rho L}}

\dfrac{R_s}{R_L} =2

{R_s}=2{R_L}

Thus; the resistance of the longer one is twice as big as the resistance of the shorter one.

7 0
3 years ago
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