1.A) 4.9 m
AL2006 Ace
The instant it was dropped, the ball had zero speed.
After falling for 1 second, its speed was 9.8 m/s straight down (gravity).
Its AVERAGE speed for that 1 second was (1/2) (0 + 9.8) = 4.9 m/s.
Falling for 1 second at an average speed of 4.9 m/s, is covered 4.9 meters.
ANYTHING you drop does that, if air resistance doesn't hold it back.
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2 idk sorry
Answer:

Explanation:
The temperature in stratosphere is generally about 270 K
molecular weight of an ozone molecule = 48 gm/mole
now formula for most probable velocity

plugging the values we get


Answer: mass x height x gravitational field strength (g)
note: gravitational field strength (g) = 10 N/Kg
55 x 15 x 10 = 8250
gpe = 8250j
Explanation:
Refer to the diagram shown below.
m = the mass of the object
x = the distance of the object from the equilibrium position at time t.
v = the velocity of the object at time t
a = the acceleration of the object at time t
A = the amplitude ( the maximum distance) of the mass from the equilibrium
position
The oscillatory motion of the object (without damping) is given by
x(t) = A sin(ωt)
where
ω = the circular frequency of the motion
T = the period of the motion so that ω = (2π)/T
The velocity and acceleration are respectively
v(t) = ωA cos(ωt)
a(t) = -ω²A sin(ωt)
In the equilibrium position,
x is zero;
v is maximum;
a is zero.
At the farthest distance (A) from the equilibrium position,
x is maximum;
v is zero;
a is zero.
In the graphs shown, it is assumed (for illustrative purposes) that
A = 1 and T = 1.