Answer:
-1m/s
Explanation:
We can calculate the speed of block A after collision
According to collision theory:
MaVa+MbVb = MaVa+MbVb (after collision)
Substitute the given values
5(3)+10(0) = 5Va+10(2)
15+0 = 5Va + 20
5Va = 15-20
5Va = -5
Va = -5/5
Va = -1m/s
Hence the velocity of ball A after collision is -1m/s
Note that the velocity of block B is zero before collision since it is stationary
The rocket engine works on the basic principle proposed by Newton which is Newton’s Third Law.
===> Distance fallen from rest in free fall =
(1/2) (acceleration) (time²)
(122.5 m) = (1/2) (9.8 m/s²) (time²)
Divide each side by (4.9 m/s²): (122.5 m / 4.9 m/s²) = time²
(122.5/4.9) s² = time²
Take the square root of each side: 5.0 seconds
===> (Accelerating at 9.8 m/s², he will be dropping at
(9.8 m/s²) x (5.0 s) = 49 m/s
when he goes 'splat'. We'll need this number for the last part.)
===> With no air resistance, the horizontal component of velocity
doesn't change.
Horizontal distance = (10 m/s) x (5.0 s) = 50 meters .
===> Impact velocity = (10 m/s horizontally) + (49 m/s vertically)
= √(10² + 49²) = 50.01 m/s arctan(10/49)
= 50.01 m/s at 11.5° from straight down,
away from the base of the cliff.
If the circuit is open, then there will be no light or sound. No current will flow and no air will move.
To fill up all of the valence electrons we must have 8 electrons total to have an ionic bond. In this case Lithium is in group 1 and so only has 1 valence electron and so it can bond with elements from group 17, since 7 + 1 = 8. However, this unknown element did bond with beryllium, and beryllium is in group 2 so it must bond with an atom that has 6 valence electrons. The elements that have 6 valence electrons are in group 16 not group 2. So the result does not support her conclusion.
