1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
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

You might be interested in
Two charges are located in the x-y plane. If q1=-4.55 nC and is located at x=0.00 m, y=0.680 m and the second charge has magnitu
Elden [556K]

Answer:

Ex= -23.8 N/C  Ey = 74.3 N/C

Explanation:

As the  electric force is linear, and the electric field, by definition, is just this electric force per unit charge, we can use the superposition principle to get the electric field produced by both charges at any point, as the other charge were not present.

So, we can first the field due to q1, as follows:

Due  to q₁ is negative, and located on the y axis, the field due to this charge will be pointing upward, (like the attractive force between q1 and the positive test charge that gives the direction to the field), as follows:

E₁ = k*(4.55 nC) / r₁²

If we choose the upward direction as the positive one (+y), we can find both components of E₁ as follows:

E₁ₓ = 0   E₁y = 9*10⁹*4.55*10⁻⁹ / (0.68)²m² = 88.6 N/C (1)

For the field due to q₂, we need first to get the distance along a straight line, between q2 and the origin.

It will be just the pythagorean distance between the points located at the coordinates (1.00, 0.600 m) and (0,0), as follows:

r₂² = 1²m² + (0.6)²m² = 1.36 m²

The magnitude of the electric field due to  q2 can be found as follows:

E₂ = k*q₂ / r₂² = 9*10⁹*(4.2)*10⁹ / 1.36 = 27.8 N/C (2)

Due to q2 is positive, the force on the positive test charge will be repulsive, so E₂ will point away from q2, to the left and downwards.

In order to get the x and y components of E₂, we need to get the projections of E₂ over the x and y axis, as follows:

E₂ₓ = E₂* cosθ, E₂y = E₂*sin θ

the  cosine of  θ, is just, by definition, the opposite  of x/r₂:

⇒ cos θ =- (1.00 m / √1.36 m²) =- (1.00 / 1.17) = -0.855

By the same token, sin θ can be obtained as follows:

sin θ = - (0.6 m / 1.17 m) = -0.513

⇒E₂ₓ = 27.8 N/C * (-0.855) = -23.8 N/C (pointing to the left) (3)

⇒E₂y = 27.8 N/C * (-0.513) = -14.3 N/C (pointing downward) (4)

The total x and y components due to both charges are just the sum of the components of Ex and Ey:

Ex = E₁ₓ + E₂ₓ = 0 + (-23.8 N/C) = -23.8 N/C

From (1) and (4), we can get Ey:

Ey = E₁y + E₂y =  88.6 N/C + (-14.3 N/C) =74.3 N/C

7 0
3 years ago
Consider a single photon with a wavelength of lambda, a frequency of nu, and an energy of E. What is the wavelength, frequency,
BARSIC [14]

Answer: lambda \lambda, nu \nu, and 100E

Explanation:

The energy E of a photon is given by:

E=h\nu   (1)

Where:

h is the Planck constant

nu is the frequency

On the other hand, we have an expression that relates the frequency of the photn with its wavelength \lambda:

nu=\frac{c}{\lambda} (2) where c is the speed of light

Substituting (2) in (1):

E=h\frac{c}{\lambda}   (3) This is the energy for a single photon

For 100 photons, the energy is:

100E=100(h\frac{c}{\lambda})=100h\nu   (3)

Where the wavelength and the frequency of the light remains constant.

Therefore, the answer is:

\lambda, \nu, and 100E

5 0
3 years ago
3. Looking at the image below label the areas of highest and lowest kinetic energy of both gravitational potential
MArishka [77]

Explanation:

max potential your welcome i love helping outhers :)

8 0
3 years ago
Consulting your data table, choose the date that the Moon is closest to the Descending Ecliptic. Note that the Descending Node i
Alenkinab [10]
Found the choices that accompanied this question.

The date that the Moon is closest to the Descending Ecliptic based on the data table is JANUARY 1, 2017.

The date that the Moon is closest to the Ascending Ecliptic based on the data table is JANUARY 16, 2017.

 The Ecliptic Longitude oF the Moon when it is nearest<span>to the descending node is 322 degrees.</span>
6 0
3 years ago
A projectile is fired vertically with an initial velocity of 192 m/s. Calculate the maximum altitude h reached by the projectile
artcher [175]

Answer:

a) Maximum height reached = 1878.90 m

b) Time of flight = 39.14 seconds.

Explanation:

Projectile motion has two types of motion Horizontal and Vertical motion.  

Vertical motion:  

We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.  

Considering upward vertical motion of projectile.  

In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g m/s^2 and final velocity = 0 m/s.  

0 = u sin θ - gt  

t = u sin θ/g  

Total time for vertical motion is two times time taken for upward vertical motion of projectile.  

So total travel time of projectile , t=\frac{2usin\theta }{g}

Vertical motion (Maximum height reached, H) :

We have equation of motion, v^2=u^2+2as, where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.  

Initial velocity = vertical component of velocity = u sin θ, acceleration = -g, final velocity = 0 m/s at maximum height H

0^2=(usin\theta) ^2-2gH\\ \\ H=\frac{u^2sin^2\theta}{2g}

In the give problem we have u = 192 m/s, θ = 90° we need to find H and t.

a) H=\frac{u^2sin^2\theta}{2g}=\frac{192^2\times sin^290}{2\times 9.81}=1878.90m

Maximum height reached = 1878.90 m

b) t=\frac{2usin\theta }{g}=\frac{2\times 192\times sin90}{9.81}=39.14s

Time of flight = 39.14 seconds.

5 0
3 years ago
Other questions:
  • WILL GET BRAINLIEST!!
    6·2 answers
  • Which of the following is a physical change? burning a piece of wood rust forming on an iron fence sawing a piece of wood in hal
    6·1 answer
  • How many significant figures from 0,020170 kg ?<br> a. 3<br> b. 4<br> c. 5<br> d. 6<br> e. 7
    5·1 answer
  • In 1665 Sir Isaac Newton proposed the fundamental law of gravitation as a universal force of attraction between any two bodies.
    13·2 answers
  • What are the products of the light reactions that are subsequently used by the calvin cycle?
    15·1 answer
  • Communication satellite use__sent by a transmitting station to transmit signals over long distances A microwaves B polar waves C
    13·1 answer
  • A soap bubble, when illuminated at normal incidence with light of 463 nm, appears to be especially reflective. If the index of r
    5·1 answer
  • A small steel roulette ball rolls around the inside of a 30 cm diameter roulette wheel. It is spun at 150 rpm, but is slows to 6
    13·1 answer
  • How do meteorologists gather data
    13·1 answer
  • A source charge generates an electric field of 4286 N/C at a distance of 2. 5 m. What is the magnitude of the source charge? (Us
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!