Water is a liquid and it evaporates into the sky as a gas and than it can fall as hail or a pond can freeze over making ice which is a solid
Answer:
I think you just have to input the horizontal velocity component.
Explanation:
Assuming no air resistance
The horizontal velocity is constant at
vx = 17.5cos71 = 5.6974... = 5.70 i m/s
The vertical velocity varies with gravity
vy = 17.5sin71 + -9.81(22) = -199.273... = -199 j m/s
v = 5.70i - 199j m/s
arguably one should round to only two significant digits
Completing the square gives the answer right away.
which indicates a maximum height of 87 when 
.
assuming the reference line to measure the height for gravitational potential energy lying at the equilibrium position
m = mass attached to the spring = 10.00 kg
k = spring constant of the spring = 250 N/m
h = height of the mass above the reference line or equilibrium position = 0.50 m
x = compression of the spring = 0.50 m
v = speed of mass = 2.4 m/s
A = maximum amplitude of the oscillation
v' = speed of mass at the maximum amplitude location = 0 m/s
using conservation of energy between the point where the speed is 2.4 m/s and the highest point at which displacement is maximum from equilibrium
kinetic energy + spring potential energy + gravitational potential energy = kinetic energy at maximum amplitude + spring potential energy at maximum amplitude + gravitational potential energy at maximum amplitude
(0.5) m v² + m g h + (0.5) k x² = (0.5) m v'² + m g A + (0.5) k A²
inserting the values
(0.5) (10) (2.4)² + (10) (9.8) (0.50) + (0.5) (250) (0.50)² = (0.5) (10) (0)² + (10) (9.8) A + (0.5) (250) A²
109.05 = (98) A + (125) A²
A = 0.62 m
Answer:
t = (ti)ln(Ai/At)/ln(2)
t = 14ln(16)/ln(2)
Solving for t
t = 14×4 = 56 seconds
Explanation:
Let Ai represent the initial amount and At represent the final amount of beryllium-11 remaining after time t
At = Ai/2^n ..... 1
Where n is the number of half-life that have passed.
n = t/half-life
Half life = 14
n = t/14
At = Ai/2^(t/14)
From equation 1.
2^n = Ai/At
Taking the natural logarithm of both sides;
nln(2) = ln(Ai/At)
n = ln(Ai/At)/ln(2)
Since n = t/14
t/14 = ln(Ai/At)/ln(2)
t = 14ln(Ai/At)/ln(2)
Ai = 800
At = 50
t = 14ln(800/50)/ln(2)
t = 14ln(16)/ln(2)
Solving for t
t = 14×4 = 56 seconds
Let half life = ti
t = (ti)ln(Ai/At)/ln(2)
