Change the following as indicated in the brackets.
8m (km,cm)​

Answers

Answer 1

metres to kilometres = 1/1000

8 m ⇒ 0.008 km

metres to centimetres = × 100

8 m ⇒ 800 cm

Answer 2

Answer:

1000m=1km

so, 8m=8/1000 = 0.008km

1m=100cm

so, 8m=8×100=800


Related Questions

The rhinestones in costume jewelry are glass with index of refraction 1.50. To make them more reflective, they are often coated with a layer of silicon monoxide of index of refraction 2.00. What is the minimum coating thickness needed to ensure that light of wavelength 576 nm and of perpendicular incidence will be reflected from the two surfaces of the coating with fully constructive interference

Answers

Answer:

[tex]T=62.9*10^{-9}[/tex]

Explanation:

From the question we are told that:

Index of refraction of Rinestones [tex]\gamma_1 =1.5[/tex]

Index of refraction of silicon [tex]\gamma_2 =2.0[/tex]

Wavelength  [tex]\lambda=576nm=576*10^{-9}[/tex]

Let each layer have thickness T

Therefore

Total Thickness =2T

Generally the equation for  Constructive interference is mathematically given by

[tex]2T=(m+0.5)\frac{l\lambda}{\gamma_2}[/tex]

Where

 [tex]M=0[/tex]

 [tex]2T=(0+0.5)\frac{576*10^{-9}}{2*2.0}[/tex]

 [tex]T=62.9*10^{-9}[/tex]

The speed of light in a solid is 1.24 x 108 m/s.
Calculate the index of refraction

Answers

Answer:

125

Explanation:

According to the question, the refractive index of the solid is approximately 2.42.

What is meant by the Refractive index?

The refractive index may be defined as the measure of the bending of a light ray when passing from one medium to another. It can also be defined as the ratio of the velocity of a light ray in an empty space to the velocity of light in a substance, n = c/v.

The Refractive index for a medium may be defined as the ratio of the speed of light in a vacuum to the speed of light in that particular medium. It is a pure ratio and hence it does not have a unit.

According to the question,

The speed of light in a solid = 1.24 × 10⁸ m/s.

The refractive index of the solid = ?

The formula for the Refractive index of a medium = (Speed of light in vacuum) / (Speed of light in the medium).

The Speed of light in vacuum = 3 × 10⁸ m/s

[(3 × 10⁸)m/s] ÷ [(1.24×10⁸)m/s]

= (3×10⁸)/(1.24×10⁸)

= 3/1.24

≈ 2.42

Therefore, the refractive index of the solid is approximately 2.42.

To learn more about the refractive index, refer to the link:

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A truck travelling at 30m/s decelerates at 1.5m/s². How far does it travel during the 10th second after the brakes are applied?​

Answers

Answer

225 meters.

Explanation:

x=x0+30t-(1/2)(1.5)t^2

x=0+30(10)-(1/2)(1.5)(10)^2

x=300-75

x=225

The force of friction acting on a sliding crate is 223 N.
How much force must be applied to main- tain a constant velocity?

Answers

Answer:

Friction Opposes Motion of an Object.

Now

To get the Net force that Moves an Object and causes acceleration....You subtract the Frictional force

Net force = Pushing Force - Frictional Force

Recall

Net Force; F=Ma

Ma = P - Fr

Now the question asked for How Much force Must be applied to Maintain a Constant velocity.

In a Constant Velocity Motion... Acceleration do not change... Its Zero

So Putting this into the formula above

M(0) = P - Fr

0=P - Fr

Fr = P.

This means

That The force needed to keep this object Moving at Constant Velocity Must be equal to its Frictional Force

Since Frictional Force; Fr =223N

The Applied Force(Pushing Force) Must be equal to 223N too.

Explain how you would apply any five pricing techniques to attract customers​

Answers

Answer:

Explanation:

1. DECOY PRICING

This occurs when customers make a purchase they must often choose between products with different prices and attributes.

This method of pricing is meant to influence the consumer's purchasing decision and maximise the sales of one particular product. The seller will offer at least three products; two of the products will have a similar or equal price. The two products with the similar prices should be the most expensive ones, and one of the two should be less attractive than the other.

2. LOSS LEADER

This is when a product is sold at a low price (often without profit) in order to stimulate other profitable sales or to attract new customers.

The main is that it will help the business to expand their market share as a whole. It's common practice when first entering a market as it introduces new customers to a service or product in the hope of building a customer base and securing future

3.ODD PRICING

This is a method of psychological pricing a product. Prices ending in 9, 95, 97, 99 are sometimes called “charm prices” and in this type of pricing, the seller fixes a price where the last digits are odd numbers. This is intended to give the buyer no room for manœuvering or for bargaining as the price appears to be less - a product priced at £9.99 will seems much cheaper than one priced at £10.00

4. PRICE DISCRIMINATION

The purpose of price discrimination is to capture the market's consumer surplus and generate the most revenue possible for a product. Identical goods or services are sold at different prices from the same provider to different segments of the market. Industries that commonly use price discrimination include the travel industry, pharmaceuticals and textbook publishers.

5. PRODUCT BUNDLE PRICING

Using this method, sellers will combine several products in the same package. It also serves to move old stock. Blu-ray and videogames are often sold using the bundle approach once they reach the end of their product life cycle. This technique is used at auctions where one attractive item may be included in a lot with a box of less interesting things. Buyers must bid for the entire lot. It’s a good way of moving slow selling products, and in a way is another form of promotional pricing.

name the device of measurement and write its used or its function?​

Answers

Answer:

There is a lot of instruments used for measurement, may I ask which one are you referring to?

Explain: What happens to the velocity of a stream as the size of the sediment increases?

Answers

Answer:

Also, as stream depth increases, the hydraulic radius increases thereby making the stream more free flowing. Both of these factors lead to an increase in stream velocity. The increased velocity and the increased cross-sectional area mean that discharge increases.

You have 1 hour to answer this question or you will be logged out.
How much force (in Newtons) does a baseball pitcher have to exert on a 250g baseball to make it accelerate to 50 m/s the
Instant that it leaves his hand?

Answers

Answer:

Force = 12.5 Newton

Explanation:

Given the following data;

Mass = 250 g to kilograms = 250/1000 = 0.25 kg

Acceleration = 50 m/s²

To find the force;

Newton's Second Law of Motion states that the acceleration of a physical object is directly proportional to the net force acting on the physical object and inversely proportional to its mass.

Mathematically, it is given by the formula;

[tex] Acceleration = \frac {Net \; force}{mass} [/tex]

Making force the subject of formula, we have;

[tex] Force = mass * acceleration [/tex]

Substituting into the formula, we have

[tex] Force = 0.25 * 50 [/tex]

Force = 12.5 Newton

In Part l, the independent variable was

Answers

Answer:

The independent variable is the variable the experimenter manipulates or changes, and is assumed to have a direct effect on the dependent variable. ... The dependent variable is the variable being tested and measured in an experiment, and is 'dependent' on the independent variable.

A car accelerates for 10 seconds. During this time, the angular
velocity of the wheels of the car increases from 10 rad/s to 25 rad/s.
What is the angular acceleration of the wheels during this time?
e

Answers

Answer:

the angular acceleration of the car is 1.5 rad/s²

Explanation:

Given;

initial angular velocity, [tex]\omega_i[/tex] = 10 rad/s

final angular velocity, [tex]\omega_f[/tex] = 25 rad/s

time of motion, t = 10 s

The angular acceleration of the car is calculated as follows;

[tex]a_r = \frac{\omega_f - \omega_i }{t} \\\\a_r = \frac{25-10}{10} = 1.5 \ rad/s^2[/tex]

Therefore, the angular acceleration of the car is 1.5 rad/s²

What happens when a negatively charged object A is brought near a neutral object B?
A.
Object B gets a negative charge.
Ο
O o
B.Object B gets a positive charge.
O C.
Object B stays neutral but becomes polarized.
D.
Object A gets a positive charge.
O
E.
Object A loses all its charge.
Reset
Next

Answers

Answer:

A.

Explanation:Object b will get a negative charge .

Rewrite the false statements correctly
1.If an object sinks in one liquid and floats on another liquid,it implies that the density of second liquid is less than the first liquid.
2.The immersed volume of body in a liquid depends on density of the liquid.
3.Relative density of a body is usually expressed in kgm^-3​

Answers

Explanation:

1. if an object sjnks in one liquid and floats on another liquid it implies that the density of second liquid is greater than the density of first liquid

What type of electromagnetic waves do heat lamps give off?

A. infrared

B. ultraviolet

C. microwaves

D. radio waves

Answers

Microwaves c…………….nnsjksisisysgxgd

help me please
only if you really know​

Answers

Assume R is measured in meters (m) and M in kilograms (kg). Then

R ² / (GM) = [m]² / ([N•m²/kg²] [kg]) = m•kg / N = m•kg / (kg•m/s²) = s²

so t ² is indeed proportional to R ²/(GM).

Which type of wave causes particles of matter to vibrate in a direction
perpendicular to the direction of its motion?
O A. Sound
B. Transverse
C. Longitudinal
D. Compression

Answers

It’s transverse so the answer is B

Answer:

C.) Longitudinal

A ship anchored at sea is rocked by waves that have crests Lim apart the waves travel at 70m/S, at what frequency do the waves reach the ship?

Answers

Question: A ship anchored at sea is rocked by waves that have crests 100 m apart the waves travel at 70m/S, at what frequency do the waves reach the ship?

Answer:

0.7 Hz

Explanation:

Applying,

v = λf............... Equation 1

Where v = velocity of the wave, f = frequency fo the wave, λ = wavelength of the wave

make f the subject of the equation

f = v/λ................. Equation 2

From the question,

Given: v = 70 m/s, λ = 100 m ( distance between successive crest)

Substitute these values into equation 2

f = 70/100

f = 0.7 Hz

Hence the frequency at which the wave reach the ship is 0.7 Hz

how do you do question 2?

Answers

Answer:

Look at explanation

Explanation:

a) Weight is another term for how much gravity is on an object and can be calculated by using the local gravity*mass so in order to find weight divide 1200 by g

m= 1200/9.8= 122.45kg

b) We know Fnet=ma and Fnet=Fapp+Fresistivity so Fapp+Fresistivity=ma

Plug in values

500+Fresistivity=122.45*2

solve for Fresistivty= 244.9-500=-255.1N (the reason it is negative is because it is in the opposite direction

c)Power= ΔE/Δt and we also know ΔE=ΔWork so Power= ΔWork/Δt. If a person pulls harder, they have a greater force and since mass is constant, acceleration is greater and since the amount of time needed to cover A to B is reduced since x-x0=v0t+1/2at²(v0=0, when you solve for t it will be lower because acceleration increases). If t decreases than Power increases by inverse proportionality. Work =Fd if the amount of Force increases by distance travelled remains constant than work also increases so power will also increase.

Car B is being pushed by a force of 22000 N. If it has a mass of 1375 kg.,
what is the acceleration of this car?

Answers

Answer:

a = 16 m/s²

General Formulas and Concepts:

Dynamics

Newton's Law of Motions

Newton's 1st Law of Motion: An object at rest remains at rest and an object in motion stays in motion Newton's 2nd Law of Motion: F = ma (Force is equal to [constant] mass times acceleration) Newton's 3rd Law of Motion: For every action, there is an equal and opposite reaction

Explanation:

Step 1: Define

Identify variables

[Given] F = 22000 N

[Given] m = 1375 kg

[Solve] a

Step 2: Find Acceleration

Substitute in variables [Newton's 2nd Law of Motion]:                                 22000 N = (1375 kg)aIsolate a:                                                                                                            16 m/s² = aRewrite:                                                                                                             a = 16 m/s²

Una muestra de agua (líquida) de 1220 kg se encuentra a 0° C y baja se temperatura hasta -29°C mientras se congela en el proceso. ¿Cuánta energía es liberada al ambiente (Mega Joules) ?

Answers

Answer:

-148,6 MJ

Explanation:

Dado que la liberación de calor al medio ambiente es;

H = mcθ

Dónde;

m = masa de agua

c = capacidad calorífica específica del agua

θ = aumento de temperatura

H = 1220 kg × 4200 × [-29-0]

H = -148,6 MJ

Your friend has informed you that his/her uncle has decided not to look after him anymore.Write to your friends uncle giving him at least two reasons why he should change his mind

Answers

Answer:

Cause you are their family and they need you. Do you wanna be someone who abandoned them or be the one they look to and dont dislike into adulthood. Also you signed onto this job no one said it was going to be easy but you making your life easier but harder for someone who is just a kid who still needs you.

why are circuit breakers used in parts of national grid ?

Answers

Answer:

To protect control circuits or small devices with insufficient cutting power

Explanation:

what is the speed of a wave with a wavelength of 3.0 m and a period of 0.40 s?

Answers

The formula you need for this is
v = f λ
(velocity = frequency • wavelength)
A built-in reminder for this relationship is the units:
meters / second = meters • 1/seconds (aka hz, frequency)
therefore v = 3m / 0.4 seconds = 7.5 m/s

In the picture below, a car hits a wall. Using what you know about Newton’s Third Law, which is true?

The force of the wall on the car and the car on the wall are equal

The force of the wall on the car is greatest

The force of the car on the wall is greatest

There is not enough information to tell

Answers

Answer:

A...................................

The force of the wall on the car and the car on the wall are equal is true about Newton’s Third Law. Option A is the correct answer.

According to Newton's Third Law of Motion, for every action, there is an equal and opposite reaction. This means that if the car hits the wall, there will be a force exerted by the car on the wall, and an equal and opposite force exerted by the wall on the car. Option A is the correct answer.

The forces involved in the interaction between the car and the wall are equal in magnitude but opposite in direction, as dictated by Newton's Third Law. Newton's Third Law of Motion states that for every action, there is an equal and opposite reaction. This means that when an object exerts a force on another object, the second object exerts a force of equal magnitude but in the opposite direction on the first object.

Learn more about Newton here:

https://brainly.com/question/28171613

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The complete question is, "In the picture below, a car hits a wall. Using what you know about Newton’s Third Law, which is true?

a. The force of the wall on the car and the car on the wall are equal

b. The force of the wall on the car is greatest

c. The force of the car on the wall is greatest

d. There is not enough information to tell"

Lamp is placed in the lamp holder. The switch is closed. The lamp glows brightly for a short time and then the lamp does not work. Explain these observations

Answers

Solution :

It is given that the lamp glows brightly for a shorter period of time when the switch is closed on it the switch is put on. But after the some time the lamp goes off and it stops working.

This is because as soon as we on the switch, the current start flowing to the lamp which makes the filament of the lamp to glow, but due to some issue, the current stop flowing even when the switch is on and this stops the lamp from glowing and hence the lamp does not work.

At the local grocery store, you push a 14.5-kg shopping cart. You stop for a moment to add a bag of dog food to your cart. With a force of 12.0 N you now accelerate the cart from rest through a distance of 2.29 m in 3.00 s. What was the mass of the dog food?

Answers

Answer:

The mass of the dog food added is 9.03 kg

Explanation:

Given;

mass of the shopping cart, m₁ = 14.5 kg

let the mass of the bag added = m₂

the force applied, F = 12 N

initial velocity of the cart-bag system, u = 0

distance traveled by the system, d = 2.29 m

time of motion of the system, t = 3.0 s

The acceleration of the system is calculated as;

[tex]d = ut + \frac{1}{2} at^2\\\\2.29 = 0 + (\frac{1}{2} \times 3^2)a\\\\2.29 = 4.5 a\\\\a = \frac{2.29}{4.5} \\\\a = 0.51 \ m/s^2[/tex]

The total mass of the system (M) is calculated as follows;

F = Ma

M = F/a

M = (12)/(0.51)

M = 23.53 kg

The mass of the dog food added is calculated as;

m₂ = M - m₁

m₂ = 23.53 kg - 14.5 kg

m₂ = 9.03 kg

In this graph, calculate the speed of
segment A in m/s?

Answers

Answer:

The answer is Speed=2m/s

Explanation:

S=D/T

S=10m/5s

S=2m/s

Which energy store is increased when an object is heated?

Answers

Answer:

Kinetic Energy

Explanation:

which forces are capable of affecting particles or objects from large distance

Answers

Answer:

only long-range force that affects all particles is the gravitational force.

Explanation:

In nature there are four fundamental forces: nuclear, weak, gravitational and electrical.

The last two are long-range, that is, the forces are zero for infinite distances, the current gravitational on all the particles and the electric one acts on the charged particles, without the chosen charge it is zero, the forces is also zero.

Consequently the only long-range force that affects all particles is the gravitational force.

A car is travelling at 60m/s. and brakes to a speed of 14m/s, in 2.7 seconds. What is the deceleration?​

Answers

Answer:

by using v = u + at equation we can find "a"

14 = 60 - 2.7a

2.7a = 60 - 14

2.7a = 46

decceleration = 17.03

Use the universal law of gravitation to solve the following problem.


Hint: mass of the Earth is = 5.97 x 1024 kg


A scientific satellite of mass 1300 kg orbits Earth 200 km above its surface. If Earth has a radius of 6378 km, what is the force of gravity acting on the scientific satellite?


a. Write out the formula for this problem.


b. Plug in the values from this problem into the formula.


c. Solve the problem, writing out each step.


d. Correct answer

Answers

Answer:

a.

[tex]F=G\cdot\dfrac{M \cdot m}{r^{2}}[/tex]

b.

[tex]F=6.67430 \times 10^{-11} \dfrac{N \cdot m^2}{kg^2} \times \dfrac{5.97 \times 10^{24} \ kg \times 1,300 \ kg}{(6,578 \ m)^{2}}[/tex]

c.

[tex]F=6.67430 \times 10^{-11} \dfrac{N \cdot m^2}{kg^2} \times \dfrac{5.97 \times 10^{24} \ kg \times 1,300 \ kg}{(6,578 \ m)^{2}} \approx 1.144 \times 10^{13} \ N[/tex]

d. The force of gravity acting on the satellite is approximately 1.144 × 10¹³ N

Explanation:

a. The formula for finding the force of gravity, F, acting object on an object is given as follows;

[tex]F=G\cdot\dfrac{M \cdot m}{r^{2}}[/tex]

Where;

F = The force acting between the Earth and the object

G = The gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

M = The mass of the Earth = 5.97 × 10²⁴ kg

m = The mass of the object

r = The distance between the center of the Earth and the object

b. Finding the gravitational force, 'F', between the Earth and the scientific satellite, we have;

The given mass of the satellite, m = 1,300 kg

The distance between the center of the Earth and the center of the satellite,  r = The length of the radius of the Earth + The height of orbit of the satellite

The given height of orbit of the satellite, h = 200 km

∴ r = R + h = 6,378 km + 200 km = 6,578 m

Therefore, by plugging in the values, we get;

[tex]F=6.67430 \times 10^{-11} \dfrac{N \cdot m^2}{kg^2} \times \dfrac{5.97 \times 10^{24} \ kg \times 1,300 \ kg}{(6,578 \ m)^{2}}[/tex]

c. Solving the above equation gives;

[tex]F=6.67430 \times 10^{-11} \dfrac{N \cdot m^2}{kg^2} \times \dfrac{5.97 \times 10^{24} \ kg \times 1,300 \ kg}{(6,578 \ m)^{2}} \approx 1.144 \times 10^{13} \ N[/tex]

d. The force of gravity acting on the satellite, F ≈ 1.144 × 10¹³ Newton

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