Answer:
AB = 29
Explanation:
For a better understanding, we must work this problem in a graphic way. In the attached image we can see the solution.
First, we draw a vertical dotted lines from the point J & K to the line ML, then we can see two new portions with the same length. Then with this simple analysis:
2x = 39 - 19
x = 10
Then we know that x = 10, another important data to find the answer is that the AB line is located in the midpoints of the legs. We also can see the right triangle MJ and the dotted line.
Now for every single right triangle, no matter its size and relationship between the vertical and the horizontal lengths, if some point is located in the hypotenuse (leg) at the middle of its length. This will be proportional to the vertical and the horizontal cathetus, therefore we will have the middle point on those two lines.
So, the AB line will be the sum of JK plus two times 5
AB = 19 + 5 + 5 = 29
The answer is b) false. Stars are mainly classified by their spectra (the elements they absorb) and their temperature.
Answer:
D
Explanation:
F = G m1 m2 / r^2 now double r
F = G m1m1/ (2r)^2
F = 1/4 G m1m2/r^2 <===== this is 1/4 of the original
Since the two trains are passing in opposite
directions, so this means that their relative velocities will be the sum of the
two trains that is:
<span>
relative velocity = (13 + 28) = 41m/s</span>
<span>
a. The passengers aboard on train B will see that train A is
moving at 41m/sec due east</span>
<span>b. The passengers aboard on train A will see that
train B is moving at 41m/sec due west</span>
Find the velocity of the object after one second.
v = vo + at
v = (0 m/s) + (9.8 m/s^2)(1 s)
v = 9.8 m/s
Now, using that, you can find the displacement in that one second between 1 and 2.
d = vot + (1/2)at^2
d = (9.8 m/s)(1 s) + (1/2)(9.8 m/s^2)(1 s)^2
d = 14.7 m