A spring has a spring constant of 65.5 N/m and it is
stretched with a force of 15.3 N. How far will it stretch?

Answers

Answer 1
0.234m !!!!!! hope this helps :)))

Related Questions

[O.05]Which of these Earth spheres interact with oceans during beach erosion?

geosphere
hydrosphere
cryosphere
atmosphere

Answers

Answer:

hydrosphere of these Earth spheres interact with oceans during beach erosion.

Answer:

Geosphere

Explanation:

The geosphere is the solid parts of the earth including rocks, minerals, and sand on the beach. The shoreline is where the geosphere meets the ocean. This is how they interact during erosion.

It is not the hydrosphere because the ocean is a part of the hydrosphere.

Why does a person feel cooler under a rotating fan?

Answers

What a fan does is create a wind chill effect. ... By blowing air around, the fan makes it easier for the air to evaporate sweat from your skin, which is how you eliminate body heat. The more evaporation, the cooler you feel.

The surface of the sun has a temperature of about 5,800 Kelvin. What is the average kinetic energy of particles on the surface of the sun? Please show your work

Answers

Answer:

273.15

Explanation:

So that's three over two times 1.38 times ten to the minus twenty-three joules per Kelvin, times 5500 degrees Celsius, the surface of the sun converted into Kelvin by adding 273.15. This works out to 1.20 times ten to the minus nineteen joules. So that's the average kinetic energy of hydrogen atoms.

You wish to make a simple amusement park ride in which a steel-wheeled roller-coaster car travels down one long slope, where rolling friction is negligible, and later slows to a stop through kinetic friction between the roller coaster's locked wheels sliding along a horizontal plastic (polystyrene) track. Assume the roller-coaster car (filled with passengers) has a mass of 743.0 kg and starts 83.4 m above the ground. (a) Calculate how fast the car is going when it reaches the bottom of the hill. m/s (b) How much does the thermal energy of the system change during the stopping motion of the car

Answers

Answer:

(a) The car is going approximately 40.43 m/s at the bottom of the hill

(b) The thermal energy will increase by 607,268.76 J

Explanation:

In the question, we have;

The height of the roller coaster above ground = 83.4 m

The mass of the roller coaster, m = 743.0 kg

(a) By the conservation of energy principle, we have;

The potential energy at the top of the hill, P.E., is equal to the kinetic energy at the bottom of the hill, K.E.

∴ P.E. = K.E.

P.E. = m·g·h

Where;

m = The mass of the roller coaster = 743.0 kg

g = The acceleration due to gravity = 9.8 m/s²

h = The height of the roller coaster = 83.4 m

Therefore, we have;

P.E. = 743.0 kg × 9.8 m/s² × 83.4 m = 607,268.76 J

P.E. = 607,268.76 J

K.E. = 1/2·m·v²

∴ K.E. = 1/2 × 743.0 kg × v²

P.E. = K.E.

∴ P.E. = K.E. = 607,268.76 J

1/2 × 743.0 kg × v² = 607,268.76 J

v² = 607,268.76 J/(1/2 × 743.0 kg) = 1,634.64 m²/s²

v = √(1,634.64 m²/s²) ≈ 40.43 m/s

(b) Given that the material wheel moves along polystyrene track, the sound released will be minimal and almost all the kinetic energy will be converted to heat energy when the train stops, therefore, the thermal energy will increase by K.E. = 607,268.76 J

The thermal energy change of the system is 624,492 J.

We know that in the roller coaster, there is an energy transformation from gravitational potential energy to kinetic energy. As such we can write;

mgh = 1/2mv^2

Where we cancel out the mass from both sides, we are left with;

gh=0.5v^2

v= √gh/0.5

v = √10 × 83.4 m/0.5

v = 41 ms-1

Now the kinetic energy is converted also into heat energy hence;

Thermal energy change of the system = 1/2 mv^2 = 0.5 × 743.0 kg × ( 41 ms-1)^2 = 624,492 J

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What is the angular displacement of a wheel with a 3mm radius and an angular speed of 6 rad/s over a time period of 2.5 seconds?

Answers

Answer:

The answer should be uranus

Explanation:

During a blood transfusion, gravity is used to provide the pressure to overcome the blood pressure and force the flow through a small needle into a vein. Consider a case where the needle is 3.0 cm long and has an internal diameter of 0.75 mm. If the required rate of flow is 0.03 cm3/s and the blood pressure in the vein is 11.00 kPa higher than atmospheric pressure (Pa), how high should the bottle be placed above the needle

Answers

Density of blood = 1.06x10³

Viscosity = 4 mpas

Answer:

It should be 1.10 higher

Explanation:

L = 0.93

D = 0.75

R = 0.75/2 = 0.375

Q = 0.03x10^-3

Blood pressure = 11x10³

Pn = 4x10^-3

n = 4 x 10^-3

Density of blood = 1.06x10³

Pn - Pv = 8*Q*n*L/pi*r⁴

Pn - pv = 463.57pa

Pn - pv = 463.57pa

Make pn subject

Pn = Pv + 463.57pa ----1

Vn = Q/An

= 0.0679m/s

To get height above needle

Pn + 1/2pv²n = Pa + pgh ----2

We equate 1 and 2 together

We get

Pgh = 11466

To get h we divide through by pg

h = 11466/pg

h = 11466/(1.06x10³)x9.81

h = 1.1026

Approximately

Height = 1.1 meters

So it should be 1.10 meters higher

the full calculations are in the attachment.

thank you!

Why are dominant alleles always shown as capital letters?​

Answers

Answer:

When writing a genotype, the dominant allele is usually represented by a capital letter, while the recessive allele has a lowercase letter.

Explanation:

Please mark me brainlest.

Answer:

Because in a punnett square, dominant alleles need to be differentiated from the recessive alleles, to do this, we write the dominant alleles as capital letters.

Explanation:

14
How many
electrons
are in the
atom
pictured?
9 p*
10 nº

Answers

Answer:

9 electrons

Explanation:

The structure has 9 protons and hence the number of electrons equals the number of protons that's why it is said to be electrical neutral

A wave with a frequency of 5Hz travels a distance of 40mm in 2 seconds.What is the speed of the wave​

Answers

Answer:

20mm per second

Explanation:

A long, straight, current-carrying wire runs from north to south.
a. A compass needle placed above the wire points with its north pole toward the east. In What direction is the current flowing
b. If a compass is put underneath the wire, in which direction will the compass needle point?

Answers

Answer:

a. The current is flowing from South to North. So, the current flows in the North direction

b. West

Explanation:

a. A compass needle placed above the wire points with its north pole toward the east. In What direction is the current flowing?

Using Maxwell's corkscrew rule, with the thumb of the right hand pointing in the direction of the current and the closed fingers pointing in the direction of the magnetic field.

Since the compass is placed above the wire and points east, the direction of the magnetic field at that point is east.

Since the magnetic field is tangential to the circular path around the wire, to produce an eastward magnetic field above the wire, the current must go from South to North. So, the current flows in the North direction.

b. If a compass is put underneath the wire, in which direction will the compass needle point?

If the compass is put underneath the wire, using Maxwell's corkscrew rule, since the current points northward, and the magnetic field is tangential to the circular path around the wire, the magnetic field below the wire points west.

So, the direction of the compass needle when the compass is put beneath the wire is west.

The earth rotates through one complete revolution every 24 hours. Since the axis of rotation is perpendicular to the equator, you can think of a person standing on the equator as standing on the edge of a disc that is rotating through one complete revolution every 24 hours. Find the angular and linear velocity of a person standing on the equator. The radius of earth is approximately 4000 miles.

Answers

Answer:

ω = 7.27 x 10⁻⁵ rad/s

v = 467.99 m/s

Explanation:

First, we will find the angular velocity of the person:

[tex]Angular\ Velocity = \omega = \frac{Angular\ Distance}{Time}[/tex]

Angular distance covered = 1 rotation = 2π radians

Time = (24 h)(3600 s/ 1 h) = 86400 s

Therefore,

[tex]\omega = \frac{2\pi\ rad}{86400\ s}[/tex]

ω = 7.27 x 10⁻⁵ rad/s

Now, for the linear velocity:

[tex]v = r\omega[/tex]

where,

v = linear velocity = ?

r = radius of earth = (4000 miles)(1609.34 m/1 mile) = 6437360 m

Therefore,

[tex]v = (6437360\ m)(7.27\ x\ 10^{-5}\ rad/s)[/tex]

v = 467.99 m/s

6xy from -12xy
please give me a answer this question​

Answers

6 floors down from 12 floors underground = 18 floors underground.

6 degrees colder than 12 degrees below zero  =  18 degrees below zero

6 brown cows taken away from -12 brown cows = -18 brown cows

6 cars sold from a dealer that 12 cars were stolen from = 18 cars gone

6xy taken away from -12xy  =  -18xy

Which of these statements is true about the effect of a force exerted upon an object?
A. A large force always produces a large change in the object’s momentum.
B. A small force always produces a large change in the object’s momentum.
C. A small force applied over a long time interval can produce a large change in the object’s momentum.
D. A large force produces a large change in the object’s momentum only if the force is applied over a very short time interval.

Answers

Answer:

D. A large force produces a large change in the object’s momentum only if the force is applied over a very short time interval.

Explanation:

Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.

Mathematically, momentum is given by the formula;

[tex] Momentum = mass * velocity [/tex]

Also, the impulse of an object is given by the formula;

[tex] Impulse = force * time [/tex]

In accordance with the impulse-momentum theorem, the statement which is true about the effect of a force exerted upon an object is that a large force produces a large change in the object’s momentum only if the force is applied over a very short time interval.

A custodian pulls a vacuum 13.5 m
with a 33.9 N force at a 55.0°
angle, against a 14.2 N friction
force. Find the total work done on
the vacuum.
(Unit = J)

Answers

Answer:

Horizontal component of pull = (cos 55 x 33.9) = 19.4N.

Net horizontal force = (19.4 - 14.2) = 5.2N.

Work = (fd) = (5.2 x 13.5) = 70.2 Joules.

Rounded to 1 decimal place throughout.

Explanation:

The total work done on the vacuum is 70.2 J.

What is work done?

Work done is equal to product of force applied and distance moved.

Work = Force x Distance

Given is a custodian pulls a vacuum 13.5 m with a 33.9 N force at a 55.0° angle, against a 14.2 N friction force.

Horizontal component of pull = (cos 55 x 33.9) = 19.4N.

Net horizontal force = (19.4 - 14.2) = 5.2N.

Work done by vacuum will be

Work =5.2 x 13.5

Work =70.2 J

Thus, the total work done on the vacuum is 70.2 J.

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How effective are child care programs for children compared to at-home care ?

Answers

Answer:

in my opinion i would say that at home cares of more effective because half of the time your child is safe and sound but in child care programs sometimes you have no idea what is going on when you are not around

Explanation:

What is the terminal velocity of blood? A. 8.9 feet/seconds B. 9.8 feet/seconds C. 25.1 feet/second D. 52.1 feet/seconds

Answers

Answer:

the answer is c) 25.1 feet/second

Explanation:

because the blood droplets can not increase speed past terminal velocity 100.

The terminal velocity of blood is 25.1 ft/s in air. So, option C is correct.

What is meant by terminal velocity ?

Terminal velocity is defined as the maximum velocity to which a freely falling drop can accelerate.

Here,

When a drop of blood is observed, it can be found that due to the effect of gravity, the velocity of the freely falling blood drop increases. So, it moves with increasing velocity, which is due to the influence of acceleration due to gravity. During the free fall of blood drop, there are resistive forces in air that opposes the motion or tries to reduce the velocity of the blood drop. The friction in air affects the freely falling blood drop and the blood drop tends to reduce its velocity and reaches a maximum of velocity after which it cannot accelerate, which is due to the frictional force in air. This maximum velocity is called terminal velocity. The terminal velocity of blood is 25.1 feet/second.

Hence,

The terminal velocity of blood is 25.1 ft/s in air.

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name and explain two sleeping disorders

Answers

insomnia and night terrors

a.
2. The following examples are good conductors except:
a. Silver
b. Metal
Plastic
d. Copper ​

Answers

Copper!! Because batteries need copper to operate!!

Answer:

plastic

Explanation:

The local church is hosting a carnival which includes a bumper car ride. Bumper car A and its driver have a mass of 300 kg; bumper car B and its driver have a mass of 200 kg. Bumper car A has a velocity to the right of 2 m/s and bumper car B is at rest. At t = 0 s, bumper car A and B are separated by 10 m. Bumper car A accelerates at 1 m/s2 to a velocity of 4 m/s and continues at this constant speed until colliding with bumper car B.



Calculate the time required for bumper car A to travel the 10 m to collide with bumper car B.
Calculate the speed of bumper car A following the collision with bumper car B, which now has a velocity to the right of 3 m/s.
Is the direction of motion for bumper car A following the collision with bumper car B to the right, to the left, or is bumper car A at rest?
Is the collision elastic? Justify your answer.

Answers

Answer:

a. 20 s

b. 0 m/s  

c. right

d.no its inelastic because when the car b was at rest and a was coming in at it, since b had no force what so ever car a swept it away with it moving to the right

Explanation:

im not sure though

By applying conservation of linear momentum, the answers are:

1. Time = 2 s

2. 3 m/s

3. same direction

4. Inelastic collision

COLLISION

There are for types of collision. They are;

Elastic CollisionPerfectly elastic collisionInelastic collisionPerfectly Inelastic collision

Given that a local church is hosting a carnival which includes a bumper car ride. Bumper car A and its driver have a mass of 300 kg; bumper car B and its driver have a mass of 200 kg. Bumper car A has a velocity to the right of 2 m/s and bumper car B is at rest. At t = 0 s, bumper car A and B are separated by 10 m. Bumper car A accelerates at 1 m/s2 to a velocity of 4 m/s and continues at this constant speed until colliding with bumper car B.

1. The time required for bumper car A to travel the 10 m to collide with bumper car B can be calculated by using first equation of linear motion.

V = U + at

Where

V = 4 m/s

U = 2 m/s

a = 1 m/[tex]s^{2}[/tex]

Substitute all the parameters into the formula

4 = 2 + t

t = 4 - 2

t = 2s

2. To calculate the speed of bumper car A following the collision with bumper car B, which now has a velocity to the right of 3 m/s, we will apply conservation of linear momentum

[tex]m_{1}u_{1}[/tex] = [tex]m_{1}v_{1}[/tex] + [tex]m_{2}v_{2}[/tex]

300 x 4 = 300V + 200 x 3

1200 = 300V + 300

300V = 1200 - 300

300V = 900

V = 900/300

V = 3 m/s

3. Since the final velocity of car A is positive, the direction of motion for bumper car A follows the collision with bumper car B to the right.

4. Since the both move at the same velocity, the collision inelastic.

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A ball of mass m=10g, carrying a charge q =-20μe is suspended from a string of length L= 0.8m above a horizontal uniformly charged infinite plane sheet of charge density σ = 4μe/m^2. The ball is displaced from the vertical by an angle and allowed to swing from rest.

Required:
a. Obtain the equations of motion of the charged ball based on Newtonian laws of motion.
b. Assume the displaced angle θ is small and simplify the results obtained in part (a) to obtain the frequency of oscillations of the charged ball.

Answers

Answer:

a)       [tex]- ( g - \frac{q}{m} \frac{\sigma }{ 2 \epsilon_o} ) \frac{sin \theta}{R }[/tex] =  [tex]\frac{d^2 \theta}{d t^2}[/tex]

b)     f = 2π  [tex]\sqrt{ \frac{R}{ g - \frac{q}{m} \frac{\sigma }{2 \epsilon_o} } }[/tex]  

Explanation:

a) To have the equations of motion, let's use Newton's second law.

Let's set a reference system where the x-axis is parallel to the path and the y-axis is in the direction of tension of the rope.

For this reference system the tension is in the direction of the y axis, we must decompose the weight and the electrical force.

Let's use trigonometry for the weight that is in the vertical direction down

             sin θ = Wₓ / W

             cos θ = W_y / w

             Wₓ = W sin θ

             W_y = W cos θ

we repeat for the electric force that is vertical upwards

              F_{ex} = F_e sin θ

              F_{ey} = F_e cos θ

the electric force is

               F_e = q E

where the field created by an infinite plate is

               E = [tex]\frac{ \sigma}{2 \epsilon_o}[/tex]  

let's write Newton's second law

Y  axis  

           T - W_y = 0

            T = W cos θ

X axis

            F_{ex} - Wₓ = m a                   (1)

           

we use that the acceleration is related to the position

            a = dv / dt

            v = dx / dt

where x is the displacement in the arc of the curve

substituting

            a = d² x /dt²

we substitute in 1

           q E sin θ - mg sin θ = m [tex]\frac{d^2 x}{dt^2}[/tex]

we have angular (tea) and linear (x) variables, if we remember that angles must be measured in radians

           θ = x / R

           x = R θ

we substitute

           sin θ (q E - mg) = m \frac{d^2 R \ theta}{dt^2}  

           

          [tex]- ( g - \frac{q}{m} \frac{\sigma }{ 2 \epsilon_o} ) \frac{sin \theta}{R }[/tex] =  [tex]\frac{d^2 \theta}{d t^2}[/tex]

this is the equation of motion of the system

b) for small oscillations

         sin θ = θ

therefore the solution is simple harmonic

      θ = θ₀ cos (wt + Ф)

if derived twice, we substitute

- ( g - \frac{q}{m} \frac{\sigma }{ 2 \epsilon_o}  ) \frac{\theta}{R } θ₀ cos (wt + Ф) = -w² θ₀ cos (wt + Ф)

     

       w² =  [tex]\frac{g}{R}[/tex] - [tex]\frac{q}{m} \frac{ \sigma }{2 \epsilon_o} \frac{1}{R}[/tex]  

angular velocity is related to frequency

       w = 2π f

        f = 2π / w

        f = 2π/w

     

        f = 2π  [tex]\sqrt{ \frac{R}{ g - \frac{q}{m} \frac{\sigma }{2 \epsilon_o} } }[/tex]  

a sports car has a mass of 1300kg.Starting from the rest the car generates a force of 4400 N.The frictional force opposing this motion is 280 N.What is the car’s acceleration

Answers

since the car moves, the force needed to move is greater than the frictional forces opposing it

a = 3.17m/s²

The acceleration of the car with a mass of 1300 kg which starts from rest generating a force of 4400 N and having a frictional force of 280 N is 3.17 m/s²

We'll begin by calculating the net force acting on the car. This can be obtained as follow:

Force (F) = 4400 N

Frictional force (Fբ) = 280 N

Net force (Fₙ) =?

Fₙ = F – Fբ

Fₙ = 4400 – 280

Fₙ = 4120 N

Thus, the net force acting on the car is 4120 N

Finally, we shall determine the acceleration of the car. This can be obtained as follow:

Mass (m) = 1300 Kg

Net force (F) = 4120 N

Acceleration (a) =?

Force = Mass × Acceleration

4120 = 1300 × a

Divide both side by 1300

[tex]a = \frac{4120}{1300}\\\\[/tex]

a = 3.17 m/s²

Therefore, the acceleration of the car is 3.17 m/s².

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g Design an experiment you can use to determine the mass of the metal cylinder. When you explain your experiment, be sure to mention: What is the underlying model (equation) that you can use to determine the mass from your measurements

Answers

Answer:

m = [tex]\frac{k}{g}[/tex] x,

graph of x vs m

Explanation:

For this exercise, the simplest way to determine the mass of the cylinder is to take a spring and hang the mass, measure how much the spring has stretched and calculate the mass, using the translational equilibrium equation

              F_e -W = 0

              k x = m g

              m = [tex]\frac{k}{g}[/tex] x

We are assuming that you know the constant k of the spring, if it is not known you must carry out a previous step, calibrate the spring, for this a series of known masses are taken and hung by measuring the elongation (x) from the equilibrium position, with these data a graph of x vs m is made to serve as a spring calibration.

  In the latter case, the elongation measured with the cylinder is found on the graph and the corresponding ordinate is the mass

A storage tank has the shape of an inverted circular cone with height 15 m and base radius of 5 m. It is filled with water to a height of 11 m. Find the work required to empty the tank by pumping all of the water to the top of the tank. (The density of water is 1000 kg/m3. Assume g = 9.8 m/s2 is the acceleration due to gravity. Round your answer to the nearest integer.) g

Answers

Answer:

THE CORRECT ANSWER FOR THIS KS

Explanation:

STOP USING THIS APP FOR ANSWERS U KNOW NOTHING

A car can go from 0 m/s to 27 m/s in 4.5 seconds. If a net force of 6600 N acted on the car, what is its mass? (Hint: You must first calculate the acceleration then plug into Newton’s second law.)

Answers

Answer:

1100 kg

Explanation:

acceleration:

[tex]a=(27-0)/4.5[/tex]

[tex]a=6ms^{-2}[/tex]

Newton's second law:

[tex]F=ma[/tex]

[tex]6600=6m\\m=1100 kg[/tex]

A woman stands still holding a 350 Newton bag (about 80 pounds) 2 meters off the ground. How much work does the woman do?

Answers

Answer:

700 Joules

Explanation:

What we know:

Force = 350 Newton

Distance = 2 meters

Work = ?

The formula for work is:

Work = force x distance

Plugging values into equation:

Work = 350 Newtons x 2 meters

Work = 700 Joules

I had the same question on the homework assignment with the answers provided:

a) It depends on how long she holds the bag

b) 350 Joules

c) none

d) 350 Newtons

The final answer would be c) none. The work done is 700 Joules. Hope this helps!

A scientist plans to release a weather balloon from ground level, to be used for high-altitude atmospheric measurements. The balloon is spherical, with a radius of 2.10 m, and filled with hydrogen. The total mass of the balloon (including the hydrogen within it) and the instruments it carries is 24.0 kg. The density of air at ground level is 1.29 kg/m3. (a) What is the magnitude of the buoyant force (in N) acting on the balloon, just after it is released from ground level

Answers

Answer:

Fb = 490.4 N

Explanation:

According to Archimedes' principle, any object submerged in a fluid, receives a push upward (which we call buoyant force) equal to the weight of the volume of the fluid removed by the object.We can express this force (Fb), in terms of the density and the volume of the fluid, as follows:

       [tex]F_{b} = \rho * V * g (1)[/tex]

The volume removed from the fluid by the balloon is just the volume of  the balloon, assuming that it is a perfect sphere, as follows:

       [tex]V = \frac{4}{3}*\pi * R^{3} = \frac{4}{3} *\pi *(2.1m)^{3} = 38.8 m3 (2)[/tex]

Replacing by the givens and (2) in (1), we get:

       [tex]F_{b} = \rho * V * g = 1.29 kg/m3* 38.8m3*9.8m/s2 = 490.4 N (3)[/tex]

A 0.242 g sample of potassium is heated in oxygen. The result is 0.292 g of a crystalline compound. What is the formula of this compound?

A.
KO3

B.
KO2

C.
KO

D.
K2O

Answers

Answer:

Hello there Dude answer is B :D hope it helped mark me brainliest.

The formula of the compound formed has been  [tex]\rm \bold {K_2O}[/tex]. Thus, option D is correct.

The sample of potassium has mass of 0.242 g. Since, the substance has been heated in the presence of oxygen, the gain in the weight has been corresponds with the mass of oxygen.

The given sample has:

Mass of potassium, [tex]m_K=0.242\;\text g[/tex]

Mass of heated sample, [tex]m_S=0.292\;\text g[/tex]

The mass of oxygen ([tex]m_O[/tex])  in the sample has been given as:

[tex]m_O=m_S-m_K[/tex]

Substituting the values:

[tex]m_O=0.292\;-\;0.242\;\text g\\m_O=0.05\;\text g[/tex]

The mass of oxygen in the sample has been 0.05 g.

The moles (M) of compounds in the sample has been given as:

[tex]M=\dfrac{m}{mwt}[/tex]

Where, m has been the mass of the compound, and

mwt has been the molecular weight of the compound.

The moles of potassium ([tex]M_K[/tex]) has been given as:

[tex]M_K=\dfrac{0.242}{39.098}\\M_K=0.006\;\text mol[/tex]

The moles of oxygen ([tex]M_O[/tex]) has been given as:

[tex]M_O=\dfrac{0.05}{16}\\M_O=0.003\;\text mol[/tex]

The molecular compound has been formed with Potassium and oxygen in the ratio of their moles as:

[tex]\rm \dfrac{K}{O}=\dfrac{0.006}{0.003}\\ \dfrac{K}{O}= \dfrac{2}1}[/tex]

Thus, the molecular formula of the compound has been [tex]\rm \bold {K_2O}[/tex]. Thus, option D is correct.

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You are driving your car on a very cold late Fall day. You clear a turn and see a couple of pedestrians standing at the cross walk. They are eager to cross the road and to get into the warmth of their apartment as soon as possible. You have two options: continue driving your car as you were without lowering your speed and drive right by the pedestrians OR slow down, stop right at the crosswalk, and yield to the pedestrians. Although by Virginia law the choice is clear, what about Physics laws? Which scenario (passing by or slowing down and stopping at the crosswalk to yield) will minimize the time the pedestrians are out in the cold freezing before they can cross the road?
Make the following assumptions in your argument. Before you noticed the pedestrians, you are moving with a constant velocity v=22 miles/hour. The distance at which you noticed the pedestrians is D=23 meters. Write down a symbolic expression for the amount of time, tpass , the pedestrians will have to wait till they cross the road if you simply drive by without slowing down or speeding up.
Write down a symbolic expression for the amount of time, tstop, the pedestrians will have to wait till they cross the road if you slow down, come to a complete stop at the crosswalk and yield to the pedestrians.

Answers

Answer:

t_pass = 2.34 m

t_stop = 4.68 s

Thus, for the car passing at constant speed the pedestrian will have to wait less.

Explanation:

If the car is moving with constant speed, then the time taken by it will be given as:

[tex]t_{pass} = \frac{D}{v}[/tex]

where,

t_pass = time taken = ?

D = Distance covered = 23 m

v = constant speed = (22 mi/h)(1609.34 m/1 mi)(1 h/3600 s) = 9.84 m/s

Therefore,

[tex]t_{pass} = \frac{23\ m}{9.84\ m/s} \\[/tex]

t_pass = 2.34 m

Now, for the time to stop the car, we will use third equation of motion to get the acceleration first:

[tex]2as = v_{f}^{2} - v_{i}^2\\a = \frac{v_{f}^{2} - v_{i}^2}{2D}\\\\a = \frac{(0\ m/s)^{2}-(9.84\ m/s)^2}{(2)(23\ m)}\\\\a = -2.1\ m/s^2[/tex]

Now, for the passing time we use first equation of motion:

[tex]v_{f} = v_{i} + at_{stop}\\t_{stop} = \frac{v_{f}-v_{i}}{a}\\\\t_{stop} = \frac{0\ m/s - 9.84\ m/s}{-2.1\ m/s^2}[/tex]

t_stop = 4.68 s

Constant velocity is the velocity which covers the same distance for each interval of the time.

The time required to pass is 2.34 seconds and the time to stop is 4.68 seconds.

What is constant velocity?

Constant velocity is the velocity which covers the same distance for each interval of the time.

It can be given as,

[tex]v=\dfrac{x}{t}[/tex]

As the distance covered by the car is 23 meters and the constant velocity of the car is 22 miles per second.

Convert the unit of velocity in m/s the value obtained will be 9.84 m/s.

Thus amount of time, [tex]t_{pass}[/tex] is,

[tex]9.84=\dfrac{23}{t_{pass} } \\t_{pass} =\dfrac{23}{9.84} \\t_{pass} =2.34[/tex]

As the distance covered by the car is 23 meters and the constant velocity of the car is 22 miles per second.

Convert the unit of velocity in m/s the value obtained will be 9.84 m/s.

Thus amount of time, [tex]t_{pass}[/tex] is,

[tex]9.84=\dfrac{23}{t_{pass} } \\t_{pass} =\dfrac{23}{9.84} \\t_{pass} =2.34[/tex]

According to the third equation of the motion acceleration can be given as,

[tex]v^2-u^2=2ax\\a=\dfrac{v^2-u^2}{2x}\\a=\dfrac{0^2-9.84^2}{2\times 23}\\a=-2.1 \rm \; m/s^2[/tex]

Now, use the first equation of motion, to get the required time,

[tex]v=u+at\\0=9.84+(-2.1)t\\t=4.68\rm \; s[/tex]

Therefore, the time required to pass is 2.34 seconds and the time to stop is 4.68 seconds.

For more details about equation of motion, refer to the link:

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