If you are driving 80 km/h along a straight road and you look to the side for 1.7 s , how far do you travel during this inattentive period?
Express your answer using two significant figures.

Answers

Answer 1

Given the speed of the driver and the time elapsed, distance traveled during his inattentive period is 0.037 kilometers.

How far did the driver travel during the inattentive period?

Speed is simply referred to as distance traveled per unit time.

Mathematically, Speed = Distance ÷ time.

Given the data in the question;

Speed = 80 km/hTime = 1.7sDistance travelled = ?

First, convert 1.7 seconds to hours.

Time = 1.7s = (1.7 / (60×60)hr = 1.7/3600 hrs

Now, find the distance traveled during the inattentive period.

80 km/h = Distance ÷ 1.7/3600 hrs

Distance = 80 km/h × 1.7/3600 hrs

Distance = 136/3600 km

Distance = 0.037 km.

Given the speed of the driver and the time elapsed, distance traveled during his inattentive period is 0.037 kilometers.

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Related Questions

A ball is tossed upward. Neglecting air drag, the acceleration along its path is

Answers

The path would be Downwards

One group of nursing home patients drinks orange juice before watching an
informative film. A similar group drinks water before watching the same film.
After the film, both groups are tested for their memory of the film's contents.
Here, the independent variable is:

Answers

The Independent variable here in this sentence is the type of drink each group was subjected to intake.

Independent variable in a experiment is the one which stands independent and acts as a control which influences the experiment's outcome. Independent variables are employed in a experiment to test whether the dependent variable changes accordingly.

Here, in this experiment the dependent variable, i.e., the theory to be tested is the memory of the patients. The independent variable here is the type of juice given to the patients, which acts as a control of the experiment's outcome. Any variable which alters the final result of the experiment and which can be controlled through out the experiment is termed as Independent variable.

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A car was moving at 14 m/s. After 30 s, its speed increased to 20 m/s. What was the acceleration during this time?

Answers

Answer: 0.2 m/s²

Explanation:

Acceleration= (20-14)/30 = 0.2 m/s²

Question is down here
⬇️​

Answers

The property that the wire must have to be used in the game is high conductivity.

What is a wire?

A wire is a material that is used to carry electric current from one point to another. Thus, we know that when we connect wire to a source, there is a movement of electrons. We can thus see that the wire is able to conduct electricity.

Having said that, the game as shown in the image attached has to do with the formation of a circuit and the flow of electricity in that circuit that is formed.

We can see that the property that the wire must have to be used in the game is high conductivity.

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World should shift its energy dependence from nonrenewable sources to renewable sources. Write your opinion​

Answers

Answer:

•Renewable energy, often referred to as clean energy, comes from natural sources or processes that are constantly replenished. For example, sunlight or wind keep shining and blowing, even if their availability depends on time and weather.

• Renewable sources of energy are better than nonrenewable sources because they refill themselves over a short period of time.

I went driving the other day and I covered 100 miles in the first 2 hours. Then I had a picnic for an hourThen continue driving to my friends house and traveled 50 miles in 2 hoursWhat was my average speed?

Answers

100 miles in 2 hours = 50mph

0 miles in 1 hour = 0mph

50 miles in 2 hours = 25mph

50 + 25 = 75/2 = 37.5mph

A car travels at a speed of 55 km/h for 2.4 h. How far has the car traveled during this journey?

Answers

20.63km/h hope this helped

Naming lonic Compounds Practice
Write the name of the ionic compounds below:
1. KCI
2. MgO
3. K₂O
4. AlCl3
5. Cao
6. BaS
7 MẸCh
8. Al2S3
9. NaH
10. SrF₂
11. ZnS
12. Mgl₂
13. RbBr
14. Case
15. Al₂03
16. BaBrz
17. Na N
18. CsCl
19. Ca₂C
20. Mg3P₂

Answers

The name of the ionic compounds are

KCl - Potassium chlorideMgO - Magnesium oxideK₂O - Potassium oxideAlCl₃ - Aluminium chlorideCaO - Calcium oxideBaS - Barium sulfideAl₂S₃ - Aluminum sulfideNaH - Sodium hydrideSrF₂ - Strontium fluorideZnS - Zinc sulfideMgI₂ - Magnesium iodideRbBr - Rubidium bromideCaSe - Calcium selenideAl₂0₃ - Aluminum OxideBaBr₂ - Barium bromideNaN - Sodium azideCsCl - Cesium chlorideCa₂C - Calcium carbideMg₃P₂ - Magnesium phosphide

Ionic compounds that has only two types of elements are named by mentioning cation first followed by the anion name.

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1. A 10kg ball moving at a speed of 6 m/s strikes another 5kg ball at rest with a contact time of 0.5 seconds. The 10kg comes to a complete stop. (a) What average force was exerted on the 10kg ball? (b) What average force was exerted on the 5kg ball? (c) What is the final momentum of the 5kg ball? (d) What is the initial momentum of the 10kg ball?

Answers

a)average force was exerted on the 10kg ball= -120 N

b)average force was exerted on the 5kg ball= 60N

c)What is the final momentum of the 5kg ball= 2.5 kg m/s

d) the initial momentum of the 10kg ball=60 kg m/s

The average force can be calculated by formula F=m(vf–vi)/ Δt so, 10 (0-6)/0.5 = -120 N the minus is representing opposite direction

the average force on the 5 kg ball = 5(6-0)/0.5= 60 N

The final momentum of 5 kg ball is 5 X 0.5 =2.5 kg m/s and initial momentum of 10 kg ball = 10 X 6 = 60 Kg m/s.

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a half meter ruler is pivoted at its midpoint and balances whaen a weight of 20N is placed at the 10 cm mark and a weight W is placed at the 45 cm mark on the ruleer. Calculate the weight W

Answers

Answer:

W = 15 N

Explanation:

The moments from the center mus be equal for balance to occur

 center is at 25 cm  ( for a 1/2 meter stick)

moment one :   15 cm * 20 N

moment two :   20 cm * W         these must be equal

   15 * 20 = 20 * W        W = 15 N

an image produced by a concave mirror of a focal length 8.7cm, the object is 13.2cm tall and at a distance 19.3cm from the mirror, (a) find the location and height of the image, (b) find the height of the image produced by the mirror if the object is twice as far from the mirror.

Answers

Answer:

what is the best result of

The pressure at the top of Mount Everest is 0.31×105N/m2, and the temperature is −30∘C. The pressure at sea level is 1.0×105N/m2, and the temperature is 20∘C.

Determine how much more frequently you need to breathe on top of Mount Everest to inhale the same amount of oxygen as you do at sea level.

________ times more frequently.

Answers

It takes 2.7 times approximately to breath frequently on top of Mount Everest as you do at sea level.

What is Pressure ?

Pressure is force per unit area. Pressure in fluid is the product of fluid density, acceleration due to gravity and the height. Pressure is measured in  N/m²

Given that the pressure at the top of Mount Everest is 0.31×105N/m2, and the temperature is −30∘C.

From Ideal gas law, PV = nRT

Where

P = 31000 N/m²

T = - 30 + 273 = 243 K

n/V =  ?

R = 8.314 1/K/Mol

n/V = P / RT

n/V = 31000 / (8.314 × 243)

n/V = 31000 / 2020.302

n/V = 15.34

When the pressure at sea level is 1.0×105N/m2, and the temperature is 20∘C, then

T = 20 + 273 = 293 K

n/V = 100000 / (8.314 × 293)

n/V = 100000 / 2436.002

n/V = 41.05

To determine how much more frequently you need to breathe on top of Mount Everest to inhale the same amount of oxygen as you do at sea level, you divide the two values. That is,

41.05 / 15.34 = 2.67

Therefore, it takes 2.7 times approximately to breath frequently on top of Mount Everest as you do at sea level.

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Two football players hit each other head on in a football game. One football
player had an acceleration of 4 m/s² and a mass of 150 kg. The second football
player had an acceleration of 2 m/s² and a mass of 200 kg. Calculate the force
each player had when they hit. What will be the net force of both players and who
knocked who back.

Answers

Answer:

200N in Player 1's direction

Player 2 is knocked back

Explanation:

Let's start by working out the force exerted by each player by using the equation F=ma, Force = Mass x Acceleration.

Player 1 has an acceleration of 4m/s2 and a mass of 150kg, therefore their force will be: [tex]F = (150)(4)[/tex], [tex]F=600N[/tex]

Player 2 has an acceleration of 2m/s2 and a mass of 200kg, therefore their force will be: [tex]F = (200)(2)[/tex], [tex]F=400N[/tex]

They are both coming from different directions so we can find the net force by subtracting one from the other: [tex]600-400=200N[/tex]

Player 1 has the higher force so the net force will be 200N in the direction Player 1 is travelling in and Player 2 will be the one knocked back.

Hope this helped!

Which point charge in diagram A
produces the strongest field?

Answers

Point Z produces is the strongest electric field.

A notion that gives a "map" of electric field lines was put out by Michael Faraday. The lines of the electric field are also known as lines of force since the electric field is the electric force per unit charge. The distribution of electric field lines also reveals the size or strength of the field. Lines are more closely spaced, therefore the electric field is stronger.

The lines are more dispersed at greater separations when the electric field is weaker. In general, it is true that areas with tighter field lines have a greater electric field. The number of lines per unit area traveling perpendicularly through a surface is proportional to the strength of the electric field regardless of the number of charges present.

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story problem questions for newtons 3rd law

Answers

? Confused on what’s being asked?

A proton is moving in a circular orbit of radius 14 cm in a uniform magnetic field of
magnitude 0.35 T directed perpendicular to the velocity of the proton. Find the orbital speed
of the proton.

Answers

The orbital speed of the proton is 4.7 x 10⁶ m/s.

What is the orbital speed of the proton?

The orbital speed of the proton is calculated by applying the following equations as shown below.

Centripetal force of the proton = magnetic force of the proton

Fc = qVB

mv²/r = qvB

mv² = qvBr

v² = qvBr/m

v = qBr/m

where;

v is the speed of the protonB is the magnetic field strengthr is the radius of the circular pathm is mass of the proton

The orbital speed of the proton is calculated as follows;

v = (1.6 x 10⁻¹⁹ x 0.35 x 0.14) / (1.67 x 10⁻²⁷)

v = 4.7 x 10⁶ m/s

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Not sure if it’s right can someone make sure

Answers

Explanation:

stones a and b will hit the ground at the same time because they are of identical weight I think your right

What is the ratio of the nuclear densities of two nuclie having mass numbers in ratio 1:4

Answers

Answer:

[tex]1:1[/tex]

Explanation:

As the nuclear density is independent of mass number, so the ratio of nuclear densities of the two given nuclei is 1:1.

Given:-

Two nuclei have mass number in the ratio of 1:4.

To find:-

The ratio of nuclear densities of the two nuclei.

Answer:-

The nuclear densities for all atoms is same and it doesn't depends on mass number.

So the ratio of the nuclear densities will be 1:1 .

[tex]\implies \underline{\underline{\green{ \rm Ratio= 1:1}}}[/tex]

_________________________________________

Additional information:-

Here I would be providing you the proof for the same .

Firstly we know that the volume of nucleus is directly proportional to its mass number ( Why ? because as the mass number increases , nucleons increases so volume increases) . Therefore;

[tex]\implies V = \dfrac{4}{3}\pi r^3 \\[/tex]

And ,

[tex]\implies V \propto A \\[/tex]

So ,

[tex]\implies \dfrac{4}{3}\pi r^3 \propto A \\[/tex]

[tex]\implies r^3 \propto A \\[/tex]

[tex]\implies r\propto A^{\dfrac{1}{3}} \\[/tex]

[tex]\implies r = r_0 A^{\dfrac{1}{3}} \dots\rm{(1)} \\[/tex]

where,

[tex]r_0[/tex] is a constant and its value is 1.2 fm .[tex] A[/tex] = Mass number[tex] r[/tex] = radius of nucleus.

Now in order to find out the nuclear density , say [tex]\rho[/tex] ; we have ;

[tex]\implies \rho =\dfrac{Mass_{\rm nucleus}}{Volume_{\rm nucleus}} \\[/tex]

Now let's assume that there are m nucleons and their average mass number is A . So the average mass of m nucleons would be mA .

[tex]\implies \rho = \dfrac{mA}{ \dfrac{4}{3}\pi r^3}\\[/tex]

From equation (1) , we have ;

[tex]\implies \rho = \dfrac{mA}{\dfrac{4}{3}\pi (r_0 A^{\dfrac{1}{3}})^3} \\[/tex]

[tex]\implies \rho = \dfrac{mA}{\dfrac{4}{3}\pi r_0^3 A} \\[/tex]

[tex]\implies \underline{\underline{\green{\rho = \dfrac{3m}{4\pi r_0^3}}}} \\[/tex]

Now we can see that all the terms in the RHS of the equation are constants. This implies that nuclear density is also constant for all atoms independent of their mass numbers .

And we are done!

HELPPP

Is the density of an object always the same as its mass?


Yes


No

How would you calculate the density of an object?


mass/area


mass/volume

After completing the liquids portion of the lab, which liquid was the least dense?

oil


water


syrup
In a density column experiment (like the one explained in the lesson), the density of the liquids depends on the order in which they are added to the column.


True


False


Would a golf ball or a ping pong ball (of the same volume) have more density?

golf ball


ping pong ball

Answers

The correct responses are;

1) No the density of an object is not always the same as its mass

2) The density of an object is calculated from mass/volume

3) After completing the liquids portion of the lab, oil has the lest density

4) It is a false statement that the density of the liquids depends on the order in which they are added to the column.

5) The golf ball has a greater density than the ping pong ball

What is density?

The term density refers to the mass per unit volume of an object. This implies that if measure the mass of the object and divide this value by the volume of the object then we obtain the density of the object. The density is an empirical property hence it does not change.

Let us now answer the questions individually;

1) No the density of an object is not always the same as its mass

2) The density of an object is calculated from mass/volume

3) After completing the liquids portion of the lab, oil has the lest density

4) It is a false statement that the density of the liquids depends on the order in which they are added to the column.

5) The golf ball has a greater density than the ping pong ball

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Explain why sound connot travel in a vaccum

Answers

Sound cannot travel through a vacuum because there are no particles to carry the vibrations.

Explanation:

When travelling through air, the speed of sound is about 330 metres per second (m/s).

At the snail racing championship in England, the winner moved at an average velocity of 2.4 mm/s [fwd] for 140 s. Determine the winning snail’s displacement during this time interval.​

Answers

Answer: 1.2 mm

Explanation:

Time= 140 sec

Initial Velocity= 0

Average Velocity= 2.4 mm/s

Displacement=  0.5* (2.4/140) *140 = 1.2 mm

After 30s, a runner is sprinting at 3 m/s. But 10 s later, the runner is sprinting at 3.8 m/s. What is the runners acceleration during this time? Need to show step by step work please. Content is from physical science, 8th grade.

Need to show step by step - I need to check please

Answers

The acceleration of the runner 10 sec later will be 0.08 meters per second square (m/s²).

What is Average acceleration?

Average acceleration is defined as the average rate of change of velocity with time. Mathematically -

a = Δv/Δt

Given is a runner sprinting at 3 m/s and after 10 s the same runner is sprinting at 3.8 m/s. Therefore, we can write -

Initial velocity [u] = 3 m/s

Final velocity [v] = 3.8 m/s

time taken [Δt] = 10 s

The average acceleration of the sprinter can be calculated as follows -

a = Δv/Δt

Δv = v - u

Δv = 3.8 - 3 = 0.8 m/s

Δt = 10 s

Therefore -

a = 0.8/10 = 0.08 m/s².

Therefore, the acceleration of the runner 10 sec later will be 0.08 meters per second square (m/s²).

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2. (a) Joe walks a distance 28 meters in the +x direction, and then a distance 14 meters directed at an angle 25° above the +x axis. (i) Using a scale diagram, draw the displacement vectors for Joe's motion and label them A and B and also his total displacement from his initial position label it C. ​

Answers

Joe travels 28 kilometers in the direction of +x before travelling 14 metres at an angle of 25° above the +x axis.

Joe moves 6 m backwards throughout the entire excursion.

The parameters are as follows: Joe's initial location is x1 = 8 m, and his ultimate position is x2 = 2 m.

The alteration of an object's location is known as displacement. The shortest distance between the final location and the starting position is another way to express it.

Movement = x = x2 - x1

Make the forward motion negative.

Make the reverse direction positive.

Relocation = -2 - (-8)

Position = -2 + 8

displacement = 6 m in the reverse

Therefore, Joe moves 6 m backwards over the entire excursion.

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Two test charges are located in the – plane. If 1=−3.350 nC and is located at 1=0.00 m, 1=1.1600 m, and the second test charge has magnitude of 2=4.600 nC and is located at 2=1.100 m, 2=0.600 m, calculate the and components, x and y, of the electric field ⃗ in component form at the origin, (0,0). The Coulomb force constant is 1/(40)=8.99×109 N·m2/ C2.

Answers

E=15.23

Given:

Q=−3.350 nC

Position 1=0.00 m

Position 2=1.1600 m,

Second charge and location is 4.600 nC and is located at 2=1.100 m, 2=0.600

Formula :

E=kq/r^2

E=9×10^9×3.350×4.600/2^2=15.23 N/C

When charge is present in any form, an electric field is linked with each point in space. The value of E, also known as electric field strength, electric field intensity, or simply the electric field, expresses the magnitude and direction of the electric field. Knowing the value of the electric field at a place, without knowing what caused the field, is all that is required to predict what will happen to electric charges near that point.

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When will you say a body is in Uniform acceleration and Non Uniform acceleration?

Answers

Answer:

(i) A body has a uniform acceleration if its velocity changes by equal amounts in equal intervals of time. The motion of a freely falling body is an example of uniform acceleraion.

(ii) A body has a non-uniform acceleration if its velocity changes by unequal intervals of time. The motion of a car on a crowded city road is an example of non-uniform acceleration.

A speedboat moves west at 108 km/h for 22.0 min. It then moves at 60.0° south of west at 90.0 km/h for 10.0 min.

What is the average speed for the trip?


What is the magnitude of average velocity for the trip?


What is the direction of average velocity for the trip? Enter the angle in degrees where negative indicates north of west and positive indicates south of west.

Answers

Answer:

108 km/hr = 1.8 km/min

90 km/hr = 1.5 km/min

V1 = (-39.6, 0)

V2 = (-10.6, -10.6)         (10.6^2 + 10.6^2)^1/2 = 15    1.5 km/min * 10 min

V = (-50.2, -10.6)       final displacement vector

Total time = 22 + 10 = 32 min

S = (50.2^2 + 10.6^2)^1/2 = 51.3 km

Average speed = (1.8 * 22 + 1.5 * 10) / 32 = 1.71 km / min

Displacement = 51.3 km

Average velocity = 51.3 / 32 = 1.60 km/min

Direction = 10.6 / 50.2 = .211 or 11.9 deg S of W

A boat travels west at a speed of 20 m/s across a river that is flowing south at 9 m/s. What is the
resultant velocity of the boat?

Answers

Answer:

22m/s

Explanation:

This is a vector addition problem. First identify the two vectors: 20m/s in the -x direction and 9m/s in the -y direction. Since neither vector shares an axis component you can use the formula [tex]\sqrt{20^2+9^2\\[/tex] =21.93 (This is the Pythagorean Theorem where your non-hypotenuse sides are your velocity vectors). Round to 2 significant figures and you'll get 22m/s.

A train starts from rest and accelerates uniformly until it has traveled 7.1km and acquired a forward velocity of 36m/s. The train then moves at a constant velocity of 36m/s for 2.1min. The train then slows down uniformly at 0.072m/s^2, until it is brought to a halt. How long does the entire process take (in min )?

Answers

The time taken is 369.23 s

Length is measured in distance. For instance, the length of a road is its distance. The most popular units of measurement for distance in the metric system are millimeters, centimeters, meters, and kilometers.

First, the train travels 2.1 kilometers.

then travels for 400 seconds at a steady speed of 24 m/s

Distance = V x t = 24 x 400 = 9600m, or 9.6 km.

then it decreases in speed from 24 m/s to 0

hence it requires

Vf- V0=at

0-24=-0.065 xt => t=369.23 s

that indicates that it will endure for

Distance = V0t+1/2 gt2 = 24 x 369.23 - 1/2 (0.065 x 369.23) = 8849.52m, or 8.8km

2.1 + 9.6 + 8.8 = 20.5 kilometers.

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The displacement vector has scalar components of Ax = 80.0 m and Ay = 60.0 m. The displacement vector has a scalar component
of Bx = 60.0 m and a magnitude of B = 75.0 m. The displacement vector has a magnitude of C = 100.0 m and is directed at an angle of
36.9° above the +x axis. Determine which two of these vectors are equal.

Answers

Answer:

Vectors A and C are equal

Explanation:

If we have a certain vector [tex]\displaystyle \overrightarrow{A}[/tex], the components of the vector [tex]\displaystyle \mathrm {A_x\;and\;A_y}[/tex] are  given by:
[tex]A_x = Acos\theta\\A_y=Asin\theta[/tex]

where
[tex]\textrm {$\theta$ is the angle formed by the vector on the x-axis}[/tex]

For vector A, we have Ax = 80m and Ay=60m

For vector C we have
Cx = |C| cosθ = 100 cos(36.9) ≈ 100 x 0.8 = 80m
Cy = |C| sinθ = C sin(36.9) ≈ 100 x 0.6 = 60m

So A and C are equal

We can eliminate B since Bx = 60 and Ax=80 so they cannot be equal

Expert help me

When firing a rifle, the bullet is shot at a high speed as the rifle recoils in the opposite directo How do you compare the force which they impart on each other? • How do you then explain the big difference in the velocities of two? Would you care to fire a rifle whose bullet is ten times as massive as the rifle? Why or why not?​

Answers

Answer:

See below

Explanation:

The forces are equal but opposite directions

the velocities of the two differ because of the different masses of the two objects....the bullet is light compared to the rifle/shooter.

I would not want to fire a heavy projectile...the 'kick' or 'recoil' of the rifle would be quite large.....and painful!

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