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2 years ago

Impact of Synthetic Materials Discussion

Physics

1 answer:

Wittaler [7]2 years ago

7 0

With the below stated fact I think that the pros of using synthetic materials outweighs the cons.

Before we answer the question

If the pros of using synthetic materials outweigh the cons or otherwise, let us first establish the what a synthetic material is and its pros and cons

Synthetic material

Synthetic material is a material in which a substances are changed chemically to produce a material with different characteristics.

Types of Synthetic material

Plastic bag
Synthetic fiber/cloth
Artificial sweetener
Synthetic fuel
Aspirin

Pros of Synthetic materials

They are used in construction because of there strong bonding
They can be molded
They can be colored into any shade
They are cheap

Cons of Synthetic materials

Sometimes none degradable
They do not absorb water.
They are highly flammable
Melts easily

In conclusion

Although Synthetic materials has so very pressing disadvantages to human kind it is unrefusable the importance of Synthetic materials to the day to day live of all humans

Therefore, I think that the pros of using synthetic materials outweighs the cons

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Water has a mass per mole of 18.0 g/mol, and each water molecule (H20) has 10 electrons. (a) How many electrons are there in one

Allushta [10]

Answer:

total number of electron in 1 litter is 3.34 × $10^{26}$ electron

Explanation:

given data

mass per mole = 18 g/mol

no of electron = 10

to find out

how many electron in 1 liter of water

solution

we know molecules per gram mole is 6.02 × $10^{23}$ molecules

no of moles is 1

total number of electron in water is = no of electron ×molecules per gram mole × no of moles

total number of electron in water is = 10 × 6.02 × $10^{23}$ × 1

total number of electron in water is = 6.02× $10^{24}$ electron

and

we know

mass = density × volume ..........1

here we know density of water is 1000 kg/m

and volume = 1 litter = 1 × $10^{-3}$ m³

mass of 1 litter = 1000 × 1 × $10^{-3}$

mass = 1000 g

total number of electron in 1 litter = mass of 1 litter × $\frac{molecules per gram mole}{mass per mole}$

total number of electron in 1 litter = 1000 × $\frac{6.02*10{24}}{18}$

total number of electron in 1 litter is 3.34 × $10^{26}$ electron

8 0

3 years ago

The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.320 with the floor. If

mart [117]

Answer: 29.50 m

Explanation: In order to calculate the higher accelation to stop a train without moving the crates inside the wagon which is traveling at constat speed we have to use the second Newton law so that:

f=μ*N the friction force is equal to coefficient of static friction multiply the normal force (m*g).

f=m.a=μ*N= m*a= μ*m*g= m*a

then

a=μ*g=0.32*9.8m/s^2= 3.14 m/s^2

With this value we can determine the short distance to stop the train

as follows:

x= vo*t- (a/2)* t^2

Vf=0= vo-a*t then t=vo/a

Finally; x=vo*vo/a-a/2*(vo/a)^2=vo^2/2a= (49*1000/3600)^2/(2*3.14)=29.50 m

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2 years ago

Important result of french revolution

qaws [65]

The french revolution led to many deaths and impacted history because of the amount of life lost

8 0

3 years ago

A small bag of sand is released from an ascending hot-air balloon whose upward constant velocity is v0 = 2.45 m/s. knowing that

lakkis [162]

When you draw an illustration for this problem, you would come up with the same drawing as shown in the picture. As the hot-air balloon travels upwards, there is a slight time when the bag of sand rises up until it reaches the maximum height. Then, it goes back down to the ground. The total time would be t₁ + t₂. The solution is as follows:

H = v₀²/2g = (2.45)²/2(9.81) = 0.306 m
t₁ = H/v₀ = 0.306 m/2.45 m/s = 0.125 s

t₂ = √2(H + 98.8)/g = √2(0.306+ 98.8)/9.81
t₂ = 4.495 s

Total time = 0.125 s + 4.495 s = 4.62 seconds

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3 years ago

Pedro e Maria saíram para passear de carro. Eles partiram de São Paulo às 10 h em direção à Braúna, localizada a 500 km da capit

Salsk061 [2.6K]

Answer:

This can be translated to:

"P edro and Maria went for a drive. They left São Paulo at 10 am towards Braúna, located 500 km from the capital. As the journey was long, they made two 15-minute stops for gas and also spent 45 minutes for lunch. When arriving at the final destination, Maria looked at the clock and saw that it was 6 pm. What is the average speed of the trip?"

Ok, the first thing we know is, that for the average speed we can write:

Speed = Distance/time.

First, we know that Distance = 500km.

And for the time we have two possiblities:

The total average speed will decrease because they were stopped a total of:

15min + 15min + 45min = 75min = 1.25 hours

Then if the travel starts at 10am, and ends at 6 pm, the total time that has passed is:

6pm = 18hs

6pm - 10am = 18 - 10 = 8hs

Then the average speed will be:

Speed = 500km/8h = 62.5 km/h.

Now, if we considerate the average speed only when they are moving, the total time that they are moving is:

Total time in travel - time that they where stopped

8h - 1.25h = 6.75h

Then the average speed would be:

Speed = 500km/6.75h = 74.1 km/h

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3 years ago