G(2)=2
For this, you can plug in 2 everywhere you see an n. So the equation will read:
g(2)=g(2-1)+2 -> g(2)=g(1)+2. Since we are given g(1)=0, we can plug in 0 where we see g(1). The equation is now. g(2)=0+2. So, g(2)=2.
Answer:
The required mass to prepare 2.5 L of 1.0 M NaOH solution is 100 g
Explanation:
We do this by preparing the equation:
Mass = concentration (mol/L) x volume (L) x Molar mass
Mass = 1.0 M x 2.5 L x 40 g/mol
Mass = 100 g
Answer:
The result is 3.859 in which we use four significant figures.
Explanation:
We start by solving the mathematical operation :
The result for the operation is 3.859438 but the numbers in the operation are given with four significant figures and that is why we are going to use four significant figures to express the result
To express 3.859438 with four significant figures we use the first four digits that appear from left to right starting by the first digit that is different to zero
In this case : 3.859 will be the result with four significant figures.
We also use a rule that says : To decide if the last significant figure remains the same we look for the value of the digit at its right.
If that number is greater than or equal to 5 ⇒ we sum one to the last significant figure
For example 3.859738 = 3.860 with four significant figures because the ''7'' is greater that 5
If that number is less than 5 ⇒ the last significant figure remains the same
In our case : 3.859438 = 3.859 because ''4'' is less than ''5''
Answer: The distance is slightly less than 3.5 m
Explanation: assuming wall and target are the same thing, and the bullet has constant velocity, the bullet will travel 7 m in half a second, so half that distance is 3.5 m.
In reality, the bullet is decelerating (at an unknown rate) so the distance is slightly less than 3.5 m.
There is also a vertical velocity component, which means it hits the target/wall at an angle. The trajectory is such that it hits the wall above the shooter because the ricochet hits at ~the level at which it left the firearm.
If the wall was absent, the bullet would have described a parabola which brough it back to the initial level after 7m. This could be calculated, but it means that the actual distance between the shooter and the wall is slightly less than 3.5 m
In addition, the collision with the wall is not 100% elastic, so the velocity aftercthe ricochetvis further reduced.
A calculation would be complex because these confounding factors are not completely independent of each other, but all reduce the average velocity and therefore the distance.
Therefore it is only possible to say that the distance was somewhat less than 3.5 m
<span>Hazards from huge explosive eruptions comprise extensive ash-falls with small glass particles in it, pyroclastic flows (combinations of hot gases and pumice chunks), and immense lahars which are volcanic mud flows which can even include debris. These put people within the area in danger. Nearby properties get damaged as well within a circumference of tens to hundreds of miles away from the volcano. These eruptions can extensively cause a change to the global climate too. Hazards from quiet eruptions include lava flows which contain exploding fires and creating gas clouds which are rich in chlorine where lava pours into the sea. This can cause damage to wildlife and sea creatures.</span>
