Answer:
The correct answer is 12.8 J.
Step-by-step explanation:
To solve this problem, we must remember how to calculate the kinetic energy of an object.
The kinetic energy is represented by the formula K = 1/2 * m * v^2, where m represents the mass of the object and v represents the speed or velocity of the object. If we plug in the given values into the formula, we get:
K = 1/2 * m * v^2
K = (1/2) * (0.40) * (8.0)^2
Our first step is to square the velocity. After doing this step, we get the following:
K = (1/2) * (0.40) * (64)
Finally, we can perform the multiplication, to get:
K = 12.8
The unit for kinetic energy is joules, so the correct answer is 12.8 J.
Hope this helps!
(5-5) x
It is different because when you distribute the x it will make it 5x-5x while all the other equations are 5-5x
This question uses trig
sin(55) = height/195
Therefore, height = 195sin(55) meters (just plug in calculator)
Let's call the stamps A, B, and C. They can each be used only once. I assume all 3 must be used in each possible arrangement.
There are two ways to solve this. We can list each possible arrangement of stamps, or we can plug in the numbers to a formula.
Let's find all possible arrangements first. We can easily start spouting out possible arrangements of the 3 stamps, but to make sure we find them all, let's go in alphabetical order. First, let's look at the arrangements that start with A:
ABC
ACB
There are no other ways to arrange 3 stamps with the first stamp being A. Let's look at the ways to arrange them starting with B:
BAC
BCA
Try finding the arrangements that start with C:
C_ _
C_ _
Or we can try a little formula; y×(y-1)×(y-2)×(y-3)...until the (y-x) = 1 where y=the number of items.
In this case there are 3 stamps, so y=3, and the formula looks like this: 3×(3-1)×(3-2).
Confused? Let me explain why it works.
There are 3 possibilities for the first stamp: A, B, or C.
There are 2 possibilities for the second space: The two stamps that are not in the first space.
There is 1 possibility for the third space: the stamp not used in the first or second space.
So the number of possibilities, in this case, is 3×2×1.
We can see that the number of ways that 3 stamps can be attached is the same regardless of method used.
Could you take another photo with you zoomed a lil bit more so I can answer it?
