Answer:
The ball will have an upward velocity of 6 m/s at a height of 5.51 m.
Explanation:
Hi there!
The equations of height and velocity of the ball are the following:
y = y0 + v0 · t + 1/2 · g · t²
v = v0 + g · t
Where:
y = height at time t.
y0 = initial height.
v0 = initial velocity.
t = time.
g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).
v = velocity of the ball at time t.
Placing the origin at the throwing point, y0 = 0.
Let´s use the equation of velocity to obtain the time at which the velocity is 12.0 m/s / 2 = 6.00 m/s.
v = v0 + g · t
6.00 m/s = 12.0 m/s -9.81 m/s² · t
(6.00 - 12.0)m/s / -9.81 m/s² = t
t = 0.612 s
Now, let´s calculate the height of the baseball at that time:
y = y0 + v0 · t + 1/2 · g · t² (y0 = 0)
y = 12.0 m/s · 0.612 s - 1/2 · 9.81 m/s² · (0.612 s)²
y = 5.51 m
The ball will have an upward velocity of 6 m/s at a height of 5.51 m.
Have a nice day!
Fog bank formation occurs when the moisture in the cooled air condenses, forming a thick fog.
<h3>How Fog banks formed?</h3>
Fog banks form at sea where cool air moves quickly over the surface of the ocean that is warm. The cool incoming air lowers the temperature of the air just above the water surface and water vapor condenses into fog.
So we can conclude that Fog bank formation occurs when the moisture in the cooled air condenses, forming a thick fog.
Learn more about fog here: brainly.com/question/18943608
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Ultraconservative the future and I have been working with the following questions.
Answer:
Q = 4.52 10¹⁷ J
Explanation:
Thermal energy can be calculated with
Q = m c_{e} ΔT
in this case it indicates that we approximate seawater to pure water with
c_{e} = 4186 J/ kg K
with the density
ρ = m / V
m = ρ V
V = L³
we substitute
m = ρ L³
Q = ρ L3 c_{e} ΔT
calculate
Q = 1000 (3 103) 3 4186 4
Q = 4.52 10¹⁷ J
