A satellite travels around the earth at 37,000 km/hr. How far will it travel after 13 hours?

Answers

Answer 1

Explanation:

Distance = speed × time

d = (37,000 km/hr) (13 hr)

d = 481,000 km


Related Questions

Match the story events on the left to the correct element of plot structure on the right


Victor pretends he can speak French


climax


Victor gets his school schedule.


resolution


Victor tries to get Teresa's attention


after homeroom and at lunch.


exposition


Teresa asks Victor if he will help her in


French


rising action


Victor checks out books to learn French


and help Teresa


falling action

Answers

Explanation:

- Victor pretends he can speak French > Rising action.

- Victor gets his school schedule > Exposition.

- Victor tries to get Teresa's attention after homeroom and at lunch > Rising action.

- Teresa asks Victor if he will help her in French > Falling action.

- Victor checks out books to learn French and help Teresa > Climax

Answer:

In "Seventh Grade" by Gary Soto, the story reaches its climax when Mr. Bueller stays quiet about Victor not knowing French. When Mr. Bueller asks if anyone in the class knows French and then Victor raises his hand, although he doesn't speak the language, Mr. Bueller decides not to make fun of it, and instead, he continues with the class normally. This action had a positive effect on Victor, who considers Mr. Bueller to be a good person and motivates him to do well in French, despite of his previous attempt to impress Teresa. Regarding the other options, although they occur at the beginning (Teresa sees Victor in the lunch area and smiles at him and Victor raises his hand in French to impress Teresa) and at the end (Victor assures Teresa that helping her will not be a bother), they aren't considered to be the highest point of the conflict in the story

he cans have essentially the same size, shape, and mass. Which can has more energy at the bottom of the ramp

Answers

Answer:

c. both have same energy

Explanation:

The complete question is

suppose you have two cans, one with milk, and the other with refried beans. The cans have essentially the same size, shape, and mass. If you release both cans at the same time, on a downhill ramp, which can has more energy at the bottom of the ramp? ignore friction and air resistance..

a. can with beans

b. can with milk

c. both have same energy

please explain your answer

Since both cans have the same size, shape, and mass, and they are released at the same height above the ramp, they'll possess the same amount of mechanical energy. This is because their mechanical energy, which is the combination of their potential and kinetic energy are both dependent on their mass. Also, having the same physical quantities like their size and shape means that they will experience the same environmental or physical factors, which will be balanced for both.

If VF=Vi+AT and Vi=0,A=3,T=4 find Vf?

Answers

Answer: 12

Explanation:

Given: VF=Vi+AT

-------------------------

In this case, substitute all the given values into the equation

VF=Vi+AT

VF=0+(3)(4)

VF=0+12

VF=12

Hope this helps!! :)

Answer:

[tex]\huge \boxed{V_f=12}[/tex]

Explanation:

[tex]V_f=V_i+AT[/tex]

This is the formula for final velocity.

The values are given for initial velocity, acceleration, and time elapsed.

[tex]V_i=0, \ A=3, \ T=4[/tex]

Solve for [tex]V_f[/tex].

[tex]V_f=0+(3)(4)[/tex]

Evaluate.

[tex]V_f=12[/tex]

When the k. E of
the object
object is increases
by 100% the momentin
the body is
increased by
how to solve plz​

Answers

[tex]\sqrt{2}[/tex]Answer:

KE2 = 2 KE1

1/2 M V2^2 = 2 * (1/2 M V1^2)

V2^2 = 2 V1^2

V2 = [tex]\sqrt{2}[/tex] V1

Since momentum = M V  the momentum increases by [tex]\sqrt{2}[/tex]

A motorboat starting from rest travels in a straight line on a lake. If the boat achieves a speed of 9.0 m/s in 13 s, what is the boat's average acceleration?

Answers

Answer:

Acceleration, [tex]a=0.69\ m/s^2[/tex]

Explanation:

Given that,

Initial speed of the motorboat, u = 0

Final speed off the motorboat, v = 9 m/s

Time, t = 13 s

We need to find the boat's average acceleration. It is equal to the change in velocity divided by time taken. SO,

[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{9-0}{13}\\\\a=0.69\ m/s^2[/tex]

So, the acceleration of the boat is [tex]0.69\ m/s^2[/tex].

Gravel is __ than clay.

Answers

Answer:

more permeable

Explanation:

no idea, i just remember learning this in school lol.

... noisier when it's in a paper bag ...

A box is sitting on a board. The coefficient of static friction between the box and the board is 0.830216. The coefficient of kinetic friction between the box and the board is 0.326245. One side of the board is raised until the box starts sliding. Give a variable legend for this problem.
a) What is the angle at which the box starts sliding? The model for this problem:
θ=__________________________________ Answer________________________________

b) What is the magnitude of its acceleration after it starts sliding? The model for this problem:
a=__________________________________ Answer________________________________

Answers

Answer:

Explanation:

Coefficient of static friction μs = .830216

Coefficient of kinetic friction μk = .326245

a ) The angle at which the box starts sliding depends upon coefficient of static friction . If θ be the required angle

tanθ = μs

tanθ = .830216

θ = 39.7°

b )

When the box starts sliding , kinetic friction will be acting on it .

frictional force on the box = μk mg cos 39.7

net force on the box

= mg sin39.7 -  μk mg cos 39.7

Applying Newton's law of motion

mg sin39.7 -  μk mg cos 39.7  = m a

a = g sin39.7 -  μk g cos 39.7

= 9.8 x sin 39.7 - .326245 x 9.8 x cos 39.7

= 6.26 - 2.46

= 3.8 m /s² .

(I) A car slows down from 28 m????s to rest in a distance of 88 m. What was its acceleration, assumed constant?

Answers

Answer:

The  value is  [tex]a = - 4.45 m/s^2[/tex]

Explanation:

From the question we are told that  

       The  initial speed is  [tex]u = 28 \ m/s[/tex] at a distance of  [tex]s_1 = 0 \ m[/tex]

        The  final speed is  [tex]v = 0 \ m/s[/tex]    at a distance of  [tex]s_2 = 88 \ m[/tex]

Generally  from the  kinematic equation we have that

       [tex]v^2 = u^2 +2as[/tex]

=>   [tex]a = \frac{v^2 - u^2 }{ 2(s_2 - s_1 )}[/tex]

=>  [tex]a = \frac{0 - 28^2 }{ 2(88 - 0 )}[/tex]

=>   [tex]a = - 4.45 m/s^2[/tex]

The negative sign shows that it is decelerating

A boy who exerts a 300-N force on the ice of a skating rink is pulled by his friend with a force of 75 N, causing the boy to accelerate across the ice. If drag and the friction from the ice apply a force of 5 N on the boy, what is the magnitude of the net force acting on him? (a) 70 N (b) 370 N (c) 80 N (d) 380 N

Answers

Answer:

Choice a. [tex]70\; \rm N[/tex], assuming that the skating rink is level.

Explanation:

Net force in the horizontal direction

There are two horizontal forces acting on the boy:

The pull of his friend, andFrictions.

The boy should be moving in the direction of the pull of his friend. The frictions on this boy should oppose that motion. Therefore, the frictions on the boy would be in the opposite direction of the pull of his friend.

The net force in the horizontal direction should then be the difference between the pull of the friend, and the friction on this boy.

[tex]\text{Net force, horizontal} = 75\; \rm N - 5\; \rm N = 70\; \rm N[/tex].

Net force in the vertical direction

The net force on this boy should be zero in the vertical direction. Consider Newton's Second Law of motion. The net force on an object is proportional to its acceleration. In this question, the net force on this boy in the vertical direction should be proportional to the vertical acceleration of this boy.

However, because (by assumption) the ice rink is level, the boy has no motion in the vertical direction. His vertical acceleration will be zero. As a result, the net force on him should also be zero in the vertical direction.

Net force

Therefore, the (combined) net force on this boy would be:

[tex]\sqrt{(70\; \rm N)^2 + (0\; \rm N)^2} = 70\; \rm N[/tex].

The answer is a hope it helps

A 4.8-kg block attached to a spring executes simple harmonic motion on a frictionless horizontal surface. At time t=0 s, the block has a displacement of -0.50m, a velocity of -0.80m/s and an acceleration of +8.3m/s2 The force constant of the spring is closest to:______.
A) 62 N/m
B) 67 N/m
C) 56 N/m
D) 73 N/m
E) 80 N/m

Answers

Answer:

E) 80 N/m

Explanation:

Given;

mass of the block, m = 4.8 kg

displacement of the block, x = -0.5 m

velocity of the block, v = -0.8 m/s

acceleration of the block, a = 8.3 m/s²

From Newton's second law of motion;

F = ma

Also, from Hook's law;

F = -Kx

where;

k is the force constant

Thus, ma = -kx

k = -ma/x

k = -(4.8 x 8.3) / (-0.5)

k = 79.7 N/m

k ≅ 80 N/m

Therefore, the force constant of the spring is closest to 80 N/m

A helicopter goes straight up 500m from a landing pad. It then goes north 20m. Then it goes down 452m. a) What is the displacement of the helicopter?
Express as components of a vector.
x-component_____________________
y-component_____________________

b) What is the displacement of the helicopter? Express as a vector (magnitude and direction).

Answer_____________________

Answers

Answer:

a

  x-component             [tex]20 \ m[/tex]

y-component     [tex]500 - 452 = 48 \ m[/tex]

b

 Magnitude [tex]d = 52 \ m[/tex]

direction is  [tex]\theta = 67.4^o[/tex]

Explanation:

From the question we are told that

   The first  vertical distance is  [tex]y_1 = 500 \ m[/tex]

    The  first horizontal distance  is  [tex]x = 20 \ m[/tex]

    The  second vertical distance is  [tex]y_2 = 452 \ m[/tex]

Generally the displacement is  

x-component             [tex]20 \ m[/tex]

y-component     [tex]500 - 452 = 48 \ m[/tex]

Generally the helicopters displacement is mathematically evaluated as  

       [tex]d = \sqrt{ x- component ^2 + y- component ^2 }[/tex]

      [tex]d = \sqrt{ 20t ^2 + 48 ^2 }[/tex]

      [tex]d = 52 \ m[/tex]

The  direction is the angle the displacement of the helicopter makes with the horizontal which is mathematically evaluated as

         [tex]\theta = tan ^{-1}[ \frac{48}{20}][/tex]

=>       [tex]\theta = tan ^{-1}[ 2.4 ][/tex]

=>      [tex]\theta = 67.4^o[/tex]

   

Base of wall of water dam is made wider.Give reason.

Answers

dams are made broader at the base than at the top to withstand the pressure of the water behind it. As the pressure is greatest at the bottom, the dam must be made thickest at it base

A student submits the following work on reference frames and centripetal force, but she has made a few mistakes. Select all sentences that contain mistakes. All non-inertial reference frames exhibit “fictitious forces.” One of these fictitious forces is the centripetal force. For example, consider a car moving in a straight line. When the car turns to the right, the passengers experience a “force” to the right. However, there is no actual force applied. The passenger is merely continuing in a straight direction. When the car is turning, the reference frame of the car is an inertial reference frame. Hence, the passenger experiences this fictitious force, even though there is no actual force there.

Answers

Answer:

"However, there is no actual force applied."

"The passenger is merely continuing in a straight direction."

Explanation:

Am AP Phys student

As a professional teacher who has been assigned to teach science in an elementary school class design activities to teach source of energy to your learners

Answers

Answer:

For solar energy, I would show them how a magnifying glass works when exposed to the sun.

For wind energy, I would teach them how to make a paper windmill and explain how it works.

For the hydroelectric energy, I would have them make a plastic turbine and explain to them how to use it in rivers or streams.

For electromagnetic energy, I would tell them to rub a balloon until their hairs stand on end.

And for electricity, I would teach them how the other energy sources create electricity and what electricity works for in these times.

Explanation:

To explain something so complicated to a child is not as easy as it would be with a teenager or an adult.

To make the children learn about the forms of energy, I would use the nemotechnique rule, using short and easy-to-remember sentences and explaining with many examples about how to get each type of energy and its use, in addition to adding didactic, visual and auditory content, which are the most common types of learning in children.

An electron from a Ti ^ + 2 hydrogen ion leaps from one orbit with radius 13.25 angstrom to another orbit with radius 2.12 angstrom. determine the energy (Joule) e produced in said transition and the wavelength (in cm)

Answers

Answer:

ΔE = 59.75 A,

Explanation:

Titanium has 3 electrons in its last shell, as it is doubly ionized, it is left with a single electron in this shell, which is why it behaves like a hydrogen-type atom, consequently we can use Bohr's atomic theory

                  rₙ = a₀ /Z     n²

                 Eₙ = 1k e² / 2a₀ (Z² / n²)

Where a₀ is Bohrd's atomic radius so  = 0.529 núm

Let's find out what quantum number n has each orbit

rn = 13.25 A = 1.325 nm

for Titanium with atomic number 22

            n² = Z rₙ / a₀

           n = √ (22 (1.325 / 0.529))

           n = 7.4

since N is an entry we take

           n = 7

rn = 2.12 A = 0.212 nm

           n = √ (22 / 0.529) 0.212

           n = 3

With these values ​​we can calculate the energy of the transition from level ne = 7 to level no = 3

         ΔE = ka e2 Z2 / 2ao (1n02 - 1 / nf2)

          ΔE = 9 10⁹ 1.6 10⁻¹⁹ 22² (2 0.529 10⁻⁹) (1/3² - 1/7²)

          ΔED = 6.5875 10² (0.111 - 0.0204)

          ΔE = 59.75 A

let us be the Planck relation between energy and frequency

          E = h f

the frequency is related to the speed of light

           c = λ f

            f = c / λ

we substitute

           E = h c /y

           E = ΔE

           h c /λ = E

           λ  = 6.63 10-34 3 108 / 59.75

           λ= 3.01939 10⁻²⁴ m

          λ = 3.01939 10⁻²² cm

you are working in a physics lab where you have made a simple circuit with a battery and bulb in which part of your circuit is the current flow maximum through the bulb filament or through the battery if you reverse the polarity would there be any difference in the intensity of the bulb​

Answers

Answer:

The current moves in the terminal.

An electrical cable consists of 125 strands of fine wire, each having 2.65 m0 resistance. The same potential difference is applied between the ends of all the strands and results in a total current of 0.750 A. (a) What is the current in each strand

Answers

Answer:

I = 6 mA

Explanation:

Given that,

Number of strands are 125

Resistance of each strand is 2.65 mΩ

The same potential difference is applied between the ends of all the strands and results in a total current of 0.750 A.

We need to find the current in each strand.

Total current is 0.75 A

Number of strands are 125

So, current in each strand :

[tex]I=\dfrac{0.75}{125}\\\\I=0.006\ A\\\\I=6\ mA[/tex]

So, 6 mA of current flows in each strand.

Dry air is primarily composed of nitrogen. In a classroom demonstration, a physics instructor pours 3.6 L of liquid nitrogen into a beaker. After the nitrogen evaporates, how much volume does it occupy if its density is equal to that of the dry air at sea level

Answers

Answer:

The  value is  [tex]V_n = 2.2498 \ m^3[/tex]

Explanation:

From the question we are told that

   The volume of  liquid nitrogen is  [tex]V_n = 3.6 \ L= 3.6 *10^{-3} \ m^3[/tex]

   The  density of  nitrogen at gaseous form   is  [tex]\rho_n = 1.2929 \ kg/m^3[/tex]  =  The dry air at sea level

   

Generally the density of nitrogen at liquid form is  

         [tex]\rho _l = 808 \ kg/m^3[/tex]

And this is mathematically represented as

      [tex]\rho_l = \frac{m}{V_l }[/tex]

=>   [tex]m = \rho_l * V_l[/tex]

Now the density of  gaseous nitrogen is

       [tex]\rho_n = \frac{m}{V_n }[/tex]

=>   [tex]m = \rho_n * V_n[/tex]

Given that the mass is constant

       [tex]\rho_n * V_n = \rho_l * V_l[/tex]

        [tex]1.2929* V_n = 808 * 3.6*10^{-3}[/tex]

=>   [tex]V_n = 2.2498 \ m^3[/tex]

       

For a certain experiment, Juan must measure the concentration of a certain substance in a solution over time. He needs to collect a measurement every 0.05 seconds. He then needs to display his data in a graph and place that graph in a text document. Select the best tools to use for this experiment. Check all that apply.

Answers

Answer:

Probeware and computer

Explanation:

Computers are more powerful and better than a graphing calculator for this situation.

Probeware and Computer

are the tools he must use.

A small spherical body is tied to a string of length 1 m and revolved in a vertical circle such that the tension in the string is zero at the highest point . Find the linear speed of the body in the 1) lowest position & 2) highest position






Answers

Explanation:

At the highest point, the tension force is 0, so the only force acting on the sphere is gravity.  Sum of forces on the sphere in the centripetal direction:

∑F = ma

mg = mv²/r

v = √(gr)

v = √(9.8 m/s² × 1 m)

v = 3.13 m/s

If the speed is constant, then the linear speed at the lowest point is also 3.13 m/s.  Otherwise, we would need to know the tension in the string at that point.

A long solenoid that has 1 200 turns uniformly distributed over a length of 0.420 m produces a magnetic field of magnitude 1.00 10-4 T at its center. What current is required in the windings for that to occur

Answers

Answer:

The current required  winding is  [tex]2.65*10^-^2 mA[/tex]

Explanation:

We can use the expression B=μ₀*n*I-------1 for the magnetic field that enters a coil  and

n= N/L (number of turns per unit length)

Given data

The number of turns n= 1200 turns

length L= 0.42 m

magnetic field B= 1*10^-4 T

μ₀= [tex]4\pi*10^-^7 T.m/A[/tex]

Applying the equation  B=μ₀*n*I

I= B/μ₀*n

I= B*L/μ₀*n

[tex]I= \frac{1*10^-^4*0.42}{4\pi*10^-^7*1.2*10^3 }[/tex]

[tex]I= 2.65*10^-^2 mA[/tex]

How fast must a meter stick be moving if its length is observed to shrink to 0.57 m?

Answers

Answer:

0.8216c

Explanation:

Using the relationship

L' = L√(1 - v²/c²)

where

L = original length,

L' = observed length,

v = velocity,

c =speed.

L'/L = 0.57

Then

0.57 = √(1 - v²/c²)

1 - v²/c² = 0.57² = 0.3249

v²/c² = 1 - 0.3249 = 0.6751

v² = 0.6751c²

v = c√0.6751 = 0.8216c

Explanation:

The astronomical unit (AU) is defined as the mean center-to-center distance from Earth to the Sun, namely 1.496x10^(11) m. The parsec is the radius of a circle for which a central angle of 1 s intercepts an arc of length 1 AU. The light-year is the distance that light travels in 1 y.
(a) How many parsecs are there in one astronomical unit?
(b) How many meters are in a parsec?
(c) How many meters in a light-year? (d) How many astronomical units in a light-year? (e) How many light-years in a parsec?

Answers

Answer:

a) How many parsecs are there in one astronomical unit?

[tex]4.85x10^{-6}pc[/tex]

(b) How many meters are in a parsec?

[tex]3.081x10^{16}m[/tex]

(c) How many meters in a light-year?

[tex]9.46x10^{15}m[/tex]

(d) How many astronomical units in a light-year?

[tex]63325AU[/tex]

(e) How many light-years in a parsec?

3.26ly

Explanation:

The parallax angle can be used to find out the distance using triangulation. Making a triangle between the nearby star, the Sun and the Earth, knowing that the distance between the Earth and the Sun ([tex]1.496x10^{11} m[/tex]) is defined as 1 astronomical unit:

[tex]\tan{p} = \frac{1AU}{d}[/tex]

Where d is the distance to the star.

Since p is small it can be represent as:

[tex]p(rad) = \frac{1AU}{d}[/tex]  (1)

Where p(rad) is the value of in radians

However, it is better to express small angles in arcseconds

[tex]p('') = p(rad)\frac{180^\circ}{\pi rad}.\frac{60'}{1^\circ}.\frac{60''}{1'}[/tex]

[tex]p('') = 2.06x10^5 p(rad)[/tex]

[tex]p(rad) = \frac{p('')}{2.06x10^5}[/tex] (2)

Then, equation 2 can be replace in equation 1:

[tex]\frac{p('')}{2.06x10^5} = \frac{1AU}{d}[/tex]  

[tex]\frac{d}{1AU} = \frac{2.06x10^5}{p('')}[/tex]  (3)

From equation 3 it can be see that [tex]1pc = 2.06x10^5 AU[/tex]

a) How many parsecs are there in one astronomical unit?

[tex]1AU . \frac{1pc}{2.06x10^5AU}[/tex] ⇒ [tex]4.85x10^{-6}pc[/tex]

(b) How many meters are in a parsec?

[tex]2.06x10^{5}AU . \frac{1.496x10^{11}m}{1AU}[/tex] ⇒ [tex]3.081x10^{16}m[/tex]

(c) How many meters in a light-year?

To determine the number of meters in a light-year it is necessary to use the next equation:

[tex]x = c.t[/tex]

Where c is the speed of light ([tex]c = 3x10^{8}m/s[/tex]) and x is the distance that light travels in 1 year.

In 1 year they are 31536000 seconds

[tex]x = (3x10^{8}m/s)(31536000s)[/tex]

[tex]x = 9.46x10^{15}m[/tex]

(d) How many astronomical units in a light-year?

[tex]9.46x10^{15}m . \frac{1AU}{1.496x10^{11}m}[/tex] ⇒ [tex]63325AU[/tex]

(e) How many light-years in a parsec?

[tex]2.06x10^{5}AU . \frac{1ly}{63235AU}[/tex] ⇒ [tex]3.26ly[/tex]

Tech A says voltage drops can be measured as long as current is flowing. Tech B says voltage drops can be measured across components, connectors, or cables. Who is correct?
A. Tech A
B. Tech B
C. Both Techs A and B
D. Neither Tech A nor B

Answers

Answer:

C. Both Techs A and B

Explanation:

For voltage drop to be measured in the circuit, then  there must be a voltage in the circuit. Once there is a voltage across the circuit, there will be current flowing through the the circuit, hence technician A is correct. Voltage drop is usually measured across components in the circuit. Components in a circuit are consumptive in the circuit, hence their is usually a voltage drop when current flows through them in a circuit.  Technician B is correct.

Answer:

C

Explanation:

A drag racer can reach a top speed of 98 m/s. How long will it take the racer to travel 1500 m?

Answers

Answer:

[tex]t=15.3s[/tex]

Explanation:

Hello,

In this case, since the speed is defined in terms of the distance over time:

[tex]V=\frac{x}{t}[/tex]

We can easily solve for the time with the given speed and distance:

[tex]t=\frac{x}{V}=\frac{1500m}{98m/s}\\ \\t=15.3s[/tex]

Regards.


Which statement is true regarding the waves shown?
A)
Doubling the frequency of the bottom waves by will cause it to match the
top waves.
B)
Cutting the frequency of the bottom waves in half will cause it to match
the top waves
C)
There is no way to match the bottom and top waves to the same frequency.
D)
Decreasing the frequency of the top waves by half will cause it to match
the bottom waves.

Answers

Answer:

The correct answer is B)

Cutting the frequency of the bottom waves in half will cause it to match top waves.

Explanation:

USATESTPREP

Cutting the frequency of the bottom waves will cause it to match the top waves. So the correct option is B.

What are waves?

A propagation of disturbance, from one point to another point is called a wave. Waves are either mechanical or non-mechanical. Electromagnetic waves are non-mechanical waves.

The mechanical waves cannot travel without a medium e.g sound waves. Non-mechanical waves do not require a medium to travel, this means that they can even travel through a vacuum.

Waves are of two types. Transverse waves and longitudinal waves.

If the direction of propagation of wave is perpendicular to the direction of movement of particles of the medium, it results in a transverse wave.

If the direction of propagation of wave is parallel to the direction of movement of particles in a medium, it results in a longitudinal wave.

Five properties of waves are amplitude, frequency, wavelength, time period and speed.

Maximum displacement from the mean position is the amplitude. The number of vibrations in a fixed point in unit time is the frequency. The distance between two identical points is the wavelength. Time taken by a wave to pass through a point is the time period and the distance travelled between particular points in a unit of time is the speed.

Therefore the correct option is B.

Read more about waves, here

https://brainly.com/question/3639648

#SPJ5

Type your answer in the box.
An organ is a group of two or more
function.
that work together to perform a common function

Answers

An organ is a group of two or more tissue. They function to maintain homeostasis in the body (keep the person alive) all body systems work together to perform functions (switching between throat and windpipe etc)

Suppose you spray your sister with water from a garden hose. The water is supplied to the hose at a rate of 0.625×10−3 m3/s and the diameter of the nozzle you hold is 5.19×10−3 m. At what speed v does the water exit the nozzle?

Answers

Answer:

0.153 m/s

Explanation:

The flowrate Q = 0.625 x 10-3 m^3-/s

The diameter of the nozzle d = 5.19 x 10^-3 m

the velocity V = ?

The cross-sectional area of the flow A = [tex]\pi d^{2}/4[/tex]

==> (3.142 x 5.19 x 10^-3)/4 = 4.077 x 10^-3 m^2

From the continuity equation,

Q = AV

V = Q/A = (0.625 x 10-3)/(4.077 x 10^-3) = 0.153 m/s

Define fluid flow. What are the types of fluid flow?​

Answers

Answer:

The different types of fluid flow are: Steady and Unsteady Flow. Uniform and Non-Uniform Flow. ... Compressible and Incompressible Flow. Rotational and Irrotational Flow.

A photoelectric-effect experiment finds a stopping potentialof 1.93V when light of 200nm is used to illuminate thecathode.
a) From what metal is the cathode made from?
b) What is the stopping potential if the intensity of thelight is doubled?

Answers

Answer:

a) Tantalum

b) 1.93 V

Explanation:

The energy of the incident photon= hc/λ

h= Plank's constant=6.63×10^-34 Is

c= speed of light = 3×10^8 ms-1

λ= wavelength of incident photon

E= 6.63×10^-34 × 3×10^8/ 200×10^-9

E= 0.099×10^-17

E= 9.9×10^-19 J

The kinetic energy of the electron = eV

Where;

e= electronic charge = 1.6×10^-19 C

V= 1.93 V

KE= 1.6×10^-19 C × 1.93 V

KE= 3.1 ×10^-19 J

From Einstein's photoelectric equation;

KE= E -Wo

Wo= E -KE

Wo=9.9×10^-19 J - 3.1 ×10^-19 J

Wo= 6.8×10^-19 J

Wo= 6.8×10^-19 J/1.6×10^-19

Wo= 4.25 ev

The metal is Tantalum

b) the stopping potential remains 1.93 V because intensity of incident photon has no effect on the stopping potential.

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