1). I started up my car. Gasoline was spritzed into the cylinders, mixed with air, and then exploded with an electrical spark. As the gasoline vapor instantly burned in the air, several new things were formed that weren't there before, like carbon dioxide, carbon monoxide, water, and oxides of nitrogen.
2). I left my dinner on the stove a little too long, and it got a layer of crunchy crackly sooty carbon on the bottom. That part of it didn't taste too good. This isn't exactly something that happens every day, but more often than I'd like it too.
3). All day, every day, and all night, every night, about 10 or 20 times every minute, I pull air into my lungs. I keep it there for a while, then I blow it out and pull in some fresh stuff. The air I blow out has less oxygen and more carbon dioxide in it than it had when I pulled it in. That's because of the hundreds of chemical reactions going on inside my body, to keep me alive and functioning. I hope these keep going on for many many more days in the future.
Answer : The volume of a sample of 4.00 mol of copper is ![28.5cm^3](https://tex.z-dn.net/?f=28.5cm%5E3)
Explanation :
First we have to calculate the mass of copper.
![\text{ Mass of copper}=\text{ Moles of copper}\times \text{ Molar mass of copper}](https://tex.z-dn.net/?f=%5Ctext%7B%20Mass%20of%20copper%7D%3D%5Ctext%7B%20Moles%20of%20copper%7D%5Ctimes%20%5Ctext%7B%20Molar%20mass%20of%20copper%7D)
![\text{ Mass of copper}=(4.00moles)\times (63.5g/mole)=254g](https://tex.z-dn.net/?f=%5Ctext%7B%20Mass%20of%20copper%7D%3D%284.00moles%29%5Ctimes%20%2863.5g%2Fmole%29%3D254g)
Now we have to calculate the volume of copper.
Formula used :
![Density=\frac{Mass}{Volume}](https://tex.z-dn.net/?f=Density%3D%5Cfrac%7BMass%7D%7BVolume%7D)
Now put all the given values in this formula, we get:
![8.92\times 10^3kg/m^3=\frac{254g}{Volume}](https://tex.z-dn.net/?f=8.92%5Ctimes%2010%5E3kg%2Fm%5E3%3D%5Cfrac%7B254g%7D%7BVolume%7D)
![Volume=\frac{254g}{8.92\times 10^3kg/m^3}=2.85\times 10^{-2}L=2.85\times 10^{-2}\times 10^3cm^3=28.5cm^3](https://tex.z-dn.net/?f=Volume%3D%5Cfrac%7B254g%7D%7B8.92%5Ctimes%2010%5E3kg%2Fm%5E3%7D%3D2.85%5Ctimes%2010%5E%7B-2%7DL%3D2.85%5Ctimes%2010%5E%7B-2%7D%5Ctimes%2010%5E3cm%5E3%3D28.5cm%5E3)
Conversion used :
![1kg/m^3=1g/L\\\\1L=10^3cm^3](https://tex.z-dn.net/?f=1kg%2Fm%5E3%3D1g%2FL%5C%5C%5C%5C1L%3D10%5E3cm%5E3)
Therefore, the volume of a sample of 4.00 mol of copper is ![28.5cm^3](https://tex.z-dn.net/?f=28.5cm%5E3)
Answer:
For the car to move with constant velocity the additional force required is ![F__{dg} }= -3674 \ N](https://tex.z-dn.net/?f=F__%7Bdg%7D%20%7D%3D%20%20-3674%20%5C%20%20N)
Explanation:
From the question we are told that
The net force of the car is ![F_{net} = 3674 \ N](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%203674%20%5C%20%20N)
Generally the total force acting on the car is the net force plus the force due to gravity acting in direction of the car (Let denote it as
)
So the total force acting on the car is mathematically represented as
![F_{net} + F__{dg}} = F](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%2B%20F__%7Bdg%7D%7D%20%3D%20F)
Here this F representing the total force can be mathematically represented as
![F = ma](https://tex.z-dn.net/?f=F%20%3D%20%20ma)
Now for constant velocity to be attained, the acceleration of the car will be zero
So at constant velocity
![F = m * 0](https://tex.z-dn.net/?f=F%20%3D%20%20m%20%2A%200)
=> ![F = 0 \ N](https://tex.z-dn.net/?f=F%20%3D%20%20%200%20%5C%20%20N)
So
![F_{net} + F__{dg}} = 0](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%2B%20F__%7Bdg%7D%7D%20%3D%200)
=> ![F__{dg} }= -F_{net}](https://tex.z-dn.net/?f=F__%7Bdg%7D%20%7D%3D%20%20-F_%7Bnet%7D)
=> ![F__{dg} }= -3674 \ N](https://tex.z-dn.net/?f=F__%7Bdg%7D%20%7D%3D%20%20-3674%20%5C%20%20N)
Answer:
As a pendulum moves toward the equilibrium position, velocity increases and acceleration decreases. As the pendulum moves away from the equilibrium position, velocity decreases and acceleration increases.
Explanation:
Using the law of conservation of energy, we know that Em1=Em2.
Em1 (at the highest point) = Eg + Ek, where Ek is 0
Em2 (at the equilibrium point) = Eg +Ek, where Eg is 0
This makes sense. At the highest point, the pendulum is at its maximum height. At this point, however, it stops moving, so its velocity is 0. At the equilibrium point, the pendulum is at its lowest height (i.e. h=0). At this point, however, its moving at its maximum velocity. This velocity is constant, which means that acceleration is 0.