Explanation:
(a)
The initial vertical velocity is 13 m/s. At the maximum height, the vertical velocity is 0 m/s.
v = at + v₀
0 = (-9.8) t + 13
t ≈ 1.33 s
(b)
Immediately prior to the explosion, the ball is at the maximum height. Here, the vertical velocity is 0 m/s, and the horizontal velocity is constant at 25 m/s.
v = √(vx² + vy²)
v = √(25² + 0²)
v = 25 m/s
(c)
Momentum is conserved before and after the explosion.
In the x direction:
m vx = ma vax + mb vbx
m (25) = (⅓ m) (0) + (⅔ m) (vbx)
25m = (⅔ m) (vbx)
25 = ⅔ vbx
vbx = 37.5 m/s
And in the y direction:
m vy = ma vay + mb vby
m (0) = (⅓ m) (0) + (⅔ m) (vby)
0 = (⅔ m) (vby)
vby = 0 m/s
Since the vertical velocity hasn't changed, and since Fragment B lands at the same height it was launched from, it will have a vertical velocity equal in magnitude and opposite in direction as its initial velocity.
vy = -13 m/s
And the horizontal velocity will stay constant.
vx = 37.5 m/s
The velocity vector is (37.5 i - 13 j) m/s. The magnitude is:
v = √(vx² + vy²)
v = √(37.5² + (-13)²)
v ≈ 39.7 m/s
Answer:
The waves are refracted as they travel through the Earth due to a change in density of the medium. This causes the waves to travel in curved paths. When the waves cross the boundary between two different layers, there is a sudden change in direction due to refraction
Answer:
D) ¼ as large
Explanation:
The magnitude of the electric force between two charges is
where
k is the Coulomb's constant
q1 and q2 are the magnitudes of the two charges
r is the separation between them
From the formula we see that the magnitude of the force is inversely proportional to the square of the distance.
In this problem, the distance is doubled:
So the new force will be
So the force will decrease to 1/4 of its original value.
Hmmmm, you either tell your parent, try to fix it, or blame it on the dog with a baseball bat that always comes on your yard and tries to eat you.
o(*￣▽￣*)ブ
