In a study of the effect of lighting on work productivity, the workers were told
that the lights would be changed on Monday, and researchers would measure
their response. The fact that the workers produced more than ever might not
be due to the lighting, but to the:
A. placebo effect
B. Hawthorne effect.
C. hypothesis theory,
D. informed consent theory.

Answers

Answer 1

Answer:

B. Hawthorne effect.

Explanation:

APEX

Answer 2

Answer:

b

Explanation:

Apex as always :)


Related Questions

The amount of space between two points is measure in unit?

Answers

I think it’s meter Hope this helps :)

The measurement of distance is the length between two places. To measure is to establish the distance between two geometric objects. With a ruler, you can measure distance the most frequently. Ordinarily, eighth-inch (or 0.125 in) segments make up inch rulers.

What space between two points is the measure unit?

When calculating the distance to stars, many astronomers choose to use parsecs (abbreviated pc). This is due to the definition's tight connection to a technique for determining the separation between stars. The distance at which 1 AU subtends an angle of 1 arc sec is measured in parsecs.

A unit of measurement is a specific magnitude of a quantity that is established and used as a standard for measuring other quantities of the same kind.

Therefore, Any additional amount of that type can be stated as a multiple of the measurement.

Learn more about measure here:

https://brainly.com/question/4725561

#SPJ2

What is the difference between mass and weight?

A) Mass measures the force of gravity and weight measures volume.
B) Mass is measured in metric units and weight is measured in US Standard System.
C) Mass measures amount of matter in an object and weight measures the force of gravity.
D) Mass measures volume and weight measures the force of gravity.

Answers

Answer:

C.  Mass is independent of the force acting on the mass.

Weight depends on the gravitational force acting on the mass.

Example:  A mass of 1 kg on the earth still has a mass of 1 kg on the moon;

whereas the weight of that mass on earth is 9.8 N and only 1/6 of that amount on the moon.

Which of the following is NOT an important part of exercise clothing? A belts B. shoes C. shirts D. socks​

Answers

Answer:

BELTS

Explanation:

why would you need belts for running and stuff

Belts are not an important part of exercise clothing.

What is exercise?

An exercise is a physical activity performed to increase strength and agility of the muscles as well as to enhance fitness.

During an exercise, it is important to wear clothes that will not hinder movement or cause injury.

Some items of clothing needed in exercise include:

shoes shirts socks

Therefore, belts are not important during exercise.

Learn more about exercise at: https://brainly.com/question/13490156

What forces are acting on a book sitting on a table? Are action - reaction force involved in this situation? Newton's third law

Answers

Answer:Newton’s law also states that larger bodies with heavier masses exert more gravitational pull, which is why those who walked on the much smaller moon experienced a sense of weightlessness, as it had a smaller gravitational pull. To help explain his theories of gravity and motion, Newton helped create a new, specialized form of mathematics.

Explanation: I think I'm very stupid

Convert 47 meters to gigameters

Answers

Answer: 4.7× 10-8 gigameter.

Exgiga meter. Hope this helps :)

A 28-g rifle bullet traveling 190 m/s embeds itself in a 3.1-kg pendulum hanging swing upward l an arc. Determine the vertical and horizontal components of the pendulum's maximum displacement.

Answers

Answer:

V = 0.15m

H = 0.85m

Explanation:

We will be using the conservation of momentum to solve the problem.

m(i).v(i) = m(f).v(f) where

0.028 * 190 = 3.1 * v(f)

v(f) = 5.32/3.1

v(f) = 1.72 m/s

next, we use the conservation of energy to find the vertical displacement.

½mv² = mgh

½v² = gh, making h the subject of formula, we have

h = v²/2g

h = 1.72²/ 2 * 9.81

h = 2.9584 / 19.62

h = 0.15 m

The height, h is the vertical displacement.

Next, we find the angle of the pendulum at the top of the swing.

Φ = cos^-1 [(2.5 - 0.15)/2.5] *the height at which the pendulum is hanging is not given, so I assumed it to be 2.5m, yo can substitute that for whatever it is

Φ = cos^-1(2.35/2.5)

Φ = cos^-1 (0.94)

Φ = 19.95°

This gotten angle is used to find the horizontal displacement

x = 2.5 sinΦ

x = 2.5 sin 19.95

x = 2.5 * 0.34

x = 0.85 m

What is the weight in newtons of lily’s convertible, which has a mass of 1800 kg?

Answers

Answer:

Explanation:

W=?

Mass=1800kg

g=(9.8m/s^2 constant)

W=mg

W=1800*9.8

W=17649N

You walk to the corner store at 2 m/s for a total time of 76 seconds. What is the distance to the store?

Answers

Answer:

152.

Explanation:

2 x 76 = 152

bruh its not that hard

Answer:

152 m

Explanation:

Distance = Speed x Time

[tex]2 \times 76 \\ 152[/tex]

Thus the distance is equal to 152 m.

What is the velocity of a plane that traveled 1500 miles from New York City to Orlando in 10.0 hours?

Answers

Answer:

v = (1500/10) = 150 [miles/h]

Explanation:

The velocity can be easily calculated using the following kinematics expression:

v = x/t

where:

v = velocity [miles/h]

t = time = 10 [hr]

x = distance = 1500 [miles]

v = (1500/10) = 150 [miles/h]

A train accelerated from 10km/hr to 40km/hr in 2 minutes. How much distance does it cover in this period? Assume that the tracks are straight?

Answers

Distance s=829.586 m
Explanation:
The initial velocity of the train is u=10 km/hr=
The final velocity of the train is v=40 km/hr=
The time taken by the train is t=2 min=120 sec
Let the distance traveled by the train during acceleration be s
The acceleration of the train is,

From the equation of motion

A piston-cylinder device containing a fluid is fitted with a paddle wheel stirring device operated by the fall of an external weight of mass 51kg. As the mass drops by a height of 5.6m, the paddle wheel makes 10100 revolutions. Meanwhile the free moving piston (frictionless and weightless) of 0.51m diameter moves out by a distance of 0.71m. Determine the net work for the system if atmospheric pressure is 101 kPa.

Answers

Answer:

The value is [tex]W_N = 11849 \ J [/tex]

Explanation:

From the question we are told that

The mass of the external weight is [tex]m = 51 \ kg[/tex]

The height through which the mass drops is [tex]h = 5.6 \ m[/tex]

     The  number of revolution made is   [tex]N  = 10100 \  kg[/tex]

    The  diameter of the free moving piston is     [tex]d  = 0.51 \  m \  kg[/tex]

     The distance moved by the free moving piston is   [tex]s  = 0.71 \  m \  kg[/tex]

     The  atmospheric pressure is  [tex]P  = 101 \ kPa = 101*10^{3}\ Pa [/tex]

Generally the workdone by the  external weight is mathematically represented as

      [tex]W_w  =m *  g   *  h[/tex]

=     [tex]  51 *  9.8  *  5.6 [/tex]

=     [tex]2799 N  [/tex]

Generally the workdone by the free moving piston  is mathematically represented as  

        [tex]W_p  =P *  A  *  s[/tex]

Here   A is the cross-sectional area with value  

        [tex] A  =  \pi *  \frac{d^2}{4}[/tex]

        [tex] A  =  3.142 *  \frac{0.51^2}{4}[/tex]

So

       [tex]W_p  =101*10^{3} * 3.142 *  \frac{0.51^2}{4}  *  0.71[/tex]

=>       [tex]W_p  = 14651 [/tex]

So

   The net workdone is mathematically evaluated as

        [tex]W_N   = -W_w+W_p  [/tex]

The negative sign shows that it is acting in opposite direction to [tex]W_N[/tex]

So

       [tex]W_N   = -2799+14651  [/tex]

      [tex]W_N   = 11849 \ J   [/tex]

A cruise ship sails due south at 2.50 m/s while a coast guard patrol boat heads 19.0° north of west at 4.80 m/s. What are the x-component and y-component of the velocity of the cruise ship relative to the patrol boat? (Assume that the +x-axis is east and the +y-axis is north. Enter your answers in m/s.)

Answers

Answer:

Explanation:

We shall represent the velocity of cruise ship and coast guard petrol boat in vector form .

velocity of cruise ship

Vcs = - 2.5 j

Vpb =  - 4.8 cos 19 i + 4.8 sin 19 j = - 4.54 i + 1.56 j

velocity of the cruise ship relative to the patrol boat

= Vcs - Vpb

=  - 2.5 j - ( - 4.54 i + 1.56 j  )

=  - 2.5 j  +  4.54 i -   1.56 j  

= 2.04 i - 1.56 j .

x-component of the velocity of the cruise ship relative to the patrol boat

= 2.04 m /s

y-component of the velocity of the cruise ship relative to the patrol boat

= - 1.56 m /s .

Consider a thin circular disk that has been heated up to 400 °C and then left inside a chamber to cool down. The chamber surface is kept at 40 °C and the air inside is maintained at 25 °C. Assume that the disk is held at the center of the chamber, and we can ignore heat conduction effects in this problem. The disk surface property is known as ε=0.65. For the diameter of D=200 mm, what is the total rate of heat transferfrom the disk?

Answers

Answer:

hello your question lacks the required diagram attached below is the complete question with the required diagram

answer : Qtotal = 807.4 Mw

Explanation:

Given Data :

disk properties :

∈ = 0.65

D = 200 mm

Ts = 400⁰c

attached below is the detailed solution

The total rate of Heat transferred from the disk

Qtotal = 807.4 Mw

A truck moving at 13.3 m/s hits a concrete wall. As a result of the collision, a 6-kg wrench moves forwards and strikes the wall of the tool compartment. If the wrench stops after being in contact with the wall for 0.07 s, what is the average force exerted on the wrench by the wall

Answers

Answer:

Explanation:

The velocity of the wrench must be equal to the velocity of the truck . So momentum of the wrench before it hits the wall

= mv = 6 x 13.3 = 79.8 kg m /s

If resisting force of wall be F , impulse on the wrench = F x time

= F x .07

Impulse = change in momentum of the wrench = mv - 0 = mv = 79.8 kgm/s

So F x .07 = 79.8

F = 1140 N .

Galileo's telescopes were not of high quality by modern standards. He was able to see the moons of Jupiter, but he never reported seeing features on Mars. Use the small-angle formula to find the angular diameter of Mars when it is closest to Earth. How does that compare with the maximum angular diameter of Jupiter? (Assume circular orbits with radii equal to the average distance from the sun.)

Answers

Answer:

Angular diameter of Mars = 15.80 * 10^5 arc seconds

The Angular diameter of Mars is 3 times the angular diameter of Jupiter

Explanation:

Average distance of the earth from sun = 150.67 * 10^6 km

assuming the radius of Mars ( average distance from sun) = 209.33 * 10^6 km

assuming the radius of Jupiter(average distance from sun) = 768.71 * 10^6 km

The small-angle formula for mars

angular diameter = ( linear diameter / distance ) * (2.06 * 10^5 )

distance between earth and mars = 54.6 * 10^6 km

linear diameter = 2 * radius = 418.66 * 10^6 km

angular diameter = ( 418.66 / 54.6 ) * 2.06 * 10^5

                             = 15.80 * 10^5 arc seconds

small angel formula for Jupiter

Angular diameter = ( linear diameter / distance ) * (2.06 * 10^5)

distance between Jupiter and earth = 588 * 10^6 km

linear diameter = 2 * radius = 1537.42 * 10^6 km

Angular diameter = ( 1537.42 / 588) * 2.06*10^5

                             = 5.39 * 10^5 arc seconds

comparing the angular diameter of the Mars and that of Jupiter :

The angular diameter of mars / angular diameter of Jupiter

= 15.80 / 5.39 = 2.931 ≈ 3

How long would it take for Sofia to walk 300 meters if she is walking at a velocity of 2.5 m/s?

Answers

Answer:

Time=120seconds

Explanation:

S=300m

V=2.5m/s

t=?

V=S/t

t=S/V

t=300/2.5

t=120 second

Answer:

120 seconds

Explanation:

.....

...

.....

A model airplane is flying North with a velocity of 15 m/s. A strong wind is blowing East at 12 m/s

Answers

Answer:

19.21 m/s^2

Explanation:

Given that a model airplane is flying North with a velocity of 15 m/s. A strong wind is blowing East at 12 m/s.

The resultant velocity can be calculated by using pythagorean theorem.

Velocity = sqrt( 15^2 + 12^2 )

Velocity = sqrt ( 225 + 144 )

Velocity = sqrt ( 369 )

Velocity = 19.21 m/s^2

Therefore, the plane will be travelling at velocity of 19.21 m/s^2

A certain ultrasound device can measure a fetal heart rate as low as 50 beats per minute. This corresponds to the surface of the heart moving at about 4.0 x 10-4 m/s. If the probe generates ultrasound that has a frequency of 2.0 MHz (1 MHz = 1 megahertz = 106 Hz), what frequency shift must the machine be able to detect

Answers

Answer:

The. Machine must detect a shift of

1 Hz

Explanation:

Frequency shift is given as

={ ( Vsound +V/ V sound -V) -1}f emitted

So by substitution we have

= { 1540+4E-4/1540-4E)-1) 2*10^6

= 1Hz

a car speeds up from 8.50 m/s to 22.2 m/s in 3.84 s. what is the acceleration of the car

Answers

Answer:

3.57

Explanation:

22.2-8.50=13.7                                                                                                                      

13.7/3.84=3.57

Acceleration (a) is the change in velocity (Δv) over the change in time (Δt), represented by the equation a = Δv/Δt.

a light bulb is a kind of lever is the statement true or false​

Answers

Answer:

False

Explanation:

It would be a screw

50 points =) The morning after a massive snowstorm, Michaela gets into her car to drive to work. The storm caused her windows to freeze, so she first needs to defrost the car. While the engine is running, she checks the thermometer. It shows the air inside of her car has a temperature of 0 °C. Does this mean the air inside of her car has no kinetic energy? Explain your answer.

Answers

Answer:

YES!

THE AIR INSIDE A CAR THE AIR RESISTANCE WILL CONVERT SOME OF THE KINETIC ENERGY FROM THE CAR INTO THE KINETIC ENERGY IN THE AIR AS IT'S PUSHED AWAY FROM THE FRONT OF THE CAR AND SUCKED BACK INTO THE GAP BEHIND IT.this TOO DEGRADES INTO HEAT.

PLEASE I NEED A BRAINLIEST

Why are chemical processes unable to produce the same amount of energy flowing out of the sun as nuclear fusion?

Answers

Answer:

Explanation:

One of the major differences between nuclear reactions and chemical reactions is that nuclear reactions involve larger amount of energy than chemical energy. This is because the force between the protons and neutrons in the nucleus of an atom is much higher than the force of attraction between electrons and the positively charged nucleus, hence nuclear reactions involves/requires a larger amount of energy (because it's reactions involve the nucleus) than chemical reactions (because it's reactions involve the electrons).

Thus, during nuclear fusion, two light nuclei are bombarded against one another to produce a larger/heavier nuclei with the release of large amount of energy (because the forces between the protons and neutrons are much higher) unlike when two atoms/molecules are chemically combined together to form a new molecule with the rearrangement of electrons in the valence shells of the participating molecules.

If the distance d (in meters) traveled by an object in time t (in seconds) is given by the formula d = A + Bt^2, the SI units of A and B must be:_______

Answers

Answer:

The SI units of the “A” is m (meters)

The SI units of the “B” is m/s^2

Explanation:

Given the distance = d meters.

Time taken to travel = t (seconds)

Function of the distance, d = A + Bt^2

Now we have given the above information and from the given distance function, we have to find the SI units of the A and B. Here, below are the SI units.

Thus, the SI units of the “A” is = m (meters)

The SI units of the “B” is = m/s^2

An observer in frame S sees lightning simultaneously strike two points 100 m apart. The first strike occurs at xx1 = yy1 = zz1 = tt1 = 0 and the second at xx2 = 100 mm, yy2 = zz2 = tt2 = 0. (a) What are the coordinates of these two events in a frame S’ moving in the standard configuration (motion along the common xx − xx’ axis) at 0.70c relative to S? (b) How far apart are the events in S’? (c) Are the events simultaneous in S’? If not, what is the difference in time between the events, and which event occurs first?

Answers

Answer:

a) 0, = -0.33 us

b) 140m

c) No, The event are not simultaneous i.e they did not occur at the same time, the second even (-0.33 usec) occurs 0.33 usec earlier than the first event.

Explanation:

a)

the lorentz factor expression is written as;

y = 1₀ / √(1 - (v²/c²))

where v  is the relative speed of an observer and c is the speed of light

so we were given that relative speed to be o.7c

therefore

y = 1 / √(1 - ((0.7c)² / c²))

y = 1 / √(1 - (0.49c² / c²))

y = 1 / √(1 - 0.49)

y = 1 / 0.7141

y = 1.4

1 - the coordinates  of the first event, the s' frame of reference is,

x1 ' = y(x1 - vt1) = 0

y1 ' = y1, z1' = z1 and

t1 ' = y [t1 - v/c²x1]

= 0

2 - the coordinates of the second event, the s ' frame of reference is'

x2 ' = y(x2-vt2)

= 1.4(100m - 0)

= 140m

y2 ' = y2, z2 ' = z2

t2 ' = y [ t2 - v/c²x2 ]

= 1.4 [ 0 - 0.7c/c²(100) ]

using speed of light c as 3*10^8

1.4 [ 0 - (0.7*3*10^8) / (3*10^8)²(100) ]

= -0.33 us

b)

distance between

delltaX' = X2' - X1'

= 140m - 0

= 140m

c)

No, The event are not simultaneous i.e they did not occur at the same time.

the second even (-0.33 us) occurs 0.33 us earlier than the first event.

sovle it please help me​

Answers

Explanation:

(a)

F = (GM1M2)/r²

F = (6.7 X 10^-11 x 30.3 x 40.17)/0.5²

F = 326.197 x 10^-⁹N

(b)

F =750N

G = 6.7 x 10^-11

M1 = 2050kg

r = 70.3m

M2 = ?

F = (GM1M2)/r²

F x r² = GM1M2

(F x r²)/GM1 = M2

M2 = (750 x 70.3²)/(6.7 x 10^-11 x 2050)

M2 = 26.99kg

A force vector points due east and has a magnitude of 140 newtons. A second force is added to . The resultant of the two vectors has a magnitude of 420 newtons and points along the (a) east/ (b) west line. Find the magnitude and direction of . Note that there are two answers. (a) Number newtons (b) Number newtons

Answers

Answer:

(a) When the resultant force is pointing along east line, the magnitude and direction of the second force is 280 N East

(b)  When the resultant force is pointing along west line, the magnitude and direction of the second force is 560 N West

Explanation:

Given;

a force vector points due east, [tex]F_1[/tex] = 140 N

let the second force = [tex]F_2[/tex]

let the resultant of the two vectors = F

(a) When the resultant force is pointing along east line

the second force must be pointing due east

[tex]F = F_1 + F_2\\\\F_2 = F - F_1\\\\F_2 = 420 \ N - 140 \ N\\\\F_2 = 280 \ N[/tex]

[tex]F_2 = 280 \ East[/tex]

(b) When the resultant force is pointing along west line

the second force must be pointing due west and it must have a greater magnitude compared to the first force in order to have a resultant in west line.

[tex]F = F_2 - F_1\\\\F_2 = F + F_1\\\\F_2 = 420 \ N + 140 \ N\\\\F_2 = 560 \ N[/tex]

[tex]F_2 = 560 \ West[/tex]

Suppose you wish to whirl a pail full of water in a vertical circle without spilling any of its contents. If your arm is 0.82 m long (from shoulder to fist) and the distance from the handle to the surface of the water is 18.5 cm, what minimum speed is required?

Answers

Answer:

Approximately [tex]3.1\; \rm m \cdot s^{-1}[/tex].

Explanation:

The content of this pail is in a centripetal motion because its path forms part of a vertical circle. Let [tex]m[/tex] denote the mass of the contents of this pail, let [tex]v[/tex] denote the (linear) velocity of the content, and let [tex]r[/tex] denote the radius of this circle. The net force on the contents of this pail will thus be:

[tex]\displaystyle F(\text{net}) = \frac{m\, v^2}{r}[/tex] towards the center of the circle.

Assume that there is no friction between the content and walls of the pail. The only two possible forces on the contents pail towards the center would be:

The downwards gravitational pull from the earth, Normal force between walls of the pail and the contents (except at the top and bottom of the circle,) andIf the rotation is fast enough, the normal force from the bottom of the pail, which also points downwards.

Note that at the top of the circle, both the gravitational pull and the normal force from the bottom point towards the center of the circle. On the other hand, the normal force from the walls of the pail would be perpendicular to the line towards the center of the circle. At that point in the circle, there's no upward force to support the content of the pail. The uniform rotation will be sufficiently fast if it could allow the content to stay in the pail at the top of the circle.

Let [tex]g[/tex] denote the gravitational field strength at the top of this circle. The size of the gravitational pull on the content would be [tex]m\cdot g[/tex]. Let [tex]F(\text{normal})[/tex] denote the normal force from the bottom of the pail on the contents. The sum of these two forces should be equal to the vertical net force on the contents of this pail. That is:

[tex]F(\text{net}) = m\cdot g + F(\text{normal})[/tex].

From the centripetal motion of the content:

[tex]\displaystyle \frac{m\, v^2}{r} = m\cdot g + F(\text{normal})[/tex].

Rearrange to obtain an expression for the normal force:

[tex]\displaystyle F(\text{normal}) = \frac{m\, v^2}{r} - m\cdot g[/tex].

Note, that the normal force the bottom of the pail exerts on the contents should be greater than or equal to zero. While the pail is at the top of the circle, the normal force from the bottom of the pail cannot pull the contents upwards. Hence:

[tex]\displaystyle \frac{m\, v^2}{r} - m\cdot g = F(\text{normal}) \ge 0[/tex].

[tex]\displaystyle \frac{m\, v^2}{r} - m\cdot g \ge 0[/tex].

Rearrange and simplify to obtain:

[tex]\displaystyle \frac{v^2}{r} - g \ge 0[/tex].

[tex]v^2 \ge g\cdot r[/tex].

[tex]v \ge \sqrt{g \cdot r}[/tex].

In other words, if the gravitational field strength is [tex]g[/tex] and the radius of the circle is [tex]r[/tex], the minimum linear velocity required to keep the content in the pail at the top of the circle is [tex]\sqrt{g \cdot r}[/tex].

If [tex]g = 9.81\; \rm N \cdot kg^{-1} = 9.81\; \rm m \cdot s^{-2}[/tex] and [tex]r = 0.82 \; \rm m + 0.185\; \rm m \approx 1.005\; \rm m[/tex], then the minimum value of [tex]v[/tex] would be approximately:

[tex]\sqrt{9.81 \; \rm m \cdot s^{-1} \times 1.005\; \rm m} \approx 3.1\; \rm m \cdot s^{-1}[/tex].

A charged object is immersed in a 17 N/C E-field. Then the total charge on the object increases by a factor of 22 over 10 s. During this time, the electrostatic force on the object remains constant. Determine the magnitude of the E-field after the total charge has changed.

Answers

Answer:

Eu = 0.65 N/C (Approx)

Explanation:

Given:

Time taken = 10 sec

E-field (Initial)(Ei) = 17 N/C

Find:

E-field (Final)

Assume;

Initial charge = Qi

Initial immersed force (Fi) = QiE

Fi = Qi(17)

After 10 sec

Fu = (26)(Qi)(Eu)

So

Fi = Fu

So,

Qi(17) = (26)(Qi)(Eu)

Eu = 17 / 26

Eu = 0.65 N/C (Approx)

Given :

Initial electric field , [tex]E_i=17\ N/C[/tex] .

The total charge on the object increases by a factor of 22 .

To Find :

The magnitude of the E-field after the total charge has changed .

Solution :

Let , initial charge is q .

Therefore , final charge is 22q .

Electrostatic force F is given by :

[tex]F=qE[/tex]

It is also given that the force remains constant .

Therefore ,

[tex]F_i=F_f\\\\qE_i=22qE_f\\\\E_f=\dfrac{E_i}{22}\\\\E_f=\dfrac{17}{22}\ N/C\\\\E_f=0.78\ N/C[/tex]

Hence , this is the required solution .

Lucy took 3 hours to cover 2/3 of a journey. She covered the remaining 60 miles in 2 hours. What was the average speed for the whole journey?

Answers

The correct answer is 36 miles per hour

Explanation:

To find the average speed, the first step is to find the total time and the total distance Lucy covered because the formula for average speed is the total distance divided into the total time.

Total Distance

It is known Lucy covered 2/3 of the total distance first, and then she completed the remaining distance (60 miles). According to this, the last distance or 60 miles represents 1/3 of the total distance.

[tex]\frac{2}{3} + \frac{1}{3} = \frac{3}{3}[/tex]  (Total distance)

Also, if 60 miles is equivalent to 1/3 this number can be used to find the total distance. In this case, just multiply 60 by 3 = 180 miles as 60 is one-third.

This means the total distance is 180 miles

Total Time

It is known the first part of this journey took 3 hours and the second took 2 hours. This means the total time was 5 hours (3 + 2 = 5)

Find the average speed

[tex]Average speed =\frac{Distance}{time}[/tex]

[tex]Average speed = \frac{180 m}{5 h}[/tex]

[tex]Average speed = 36m/h[/tex]

Specialized visual receptors that assist mostly in nighttime vision are __________. A. cones B. rods C. fovea D. optic disks Please select the best answer from the choices provided A B C D

Answers

Answer:

B

Explanation:

EDGE 2020

The rod photoreceptor cell type, which is found in the retina, transmits low-light vision and is primarily in charge of the neural transmission of night vision. Thus, option B  is correct.

What visual receptors that assist mostly in nighttime vision?

We can see at night thanks to rods, and we can see color during the day thanks to cones. Although daylight vision is far more important for us, the topic of why there are roughly 20 times more rods than cones in a human retina has typically been met with a shrug.

Rods function in extremely dim lighting. Because just some light particles (photons) may trigger a rod, we employ these for night vision. Because rods hinder color vision, we experience a grayscale world at night. More than 100 million rod cells make up the human eye.

Therefore, The retina's rod photoreceptor cell type conveys vision in low light and is principally in charge of the neuronal activity of night vision.

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