Answer:
Reading a Graduated Cylinder
Place the graduated cylinder on a flat surface and view the height of the liquid in the cylinder with your eyes directly level with the liquid. The liquid will tend to curve downward. This curve is called the meniscus. Always read the measurement at the bottom of the meniscus.....
hope it helps....
I'm not from that school but I can help you.
Answer:
0.782 s
Explanation:
The water flows horizontally from the hose, so its initial vertical velocity is 0.
Given:
y₀ = 3 m
y = 0 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: t
y = y₀ + v₀ t + ½ at²
0 m = 3 m + (0 m/s) t + ½ (-9.8 m/s²) t²
t = 0.782 s
Round as needed.
Answer:
Since in summer, the eastern side do not face the sunlight and hence the water in eastern pot remain cool in summer.
Answer:
h’ = 1/9 h
Explanation:
This exercise must be solved in parts:
* Let's start by finding the speed of sphere B at the lowest point, let's use the concepts of conservation of energy
starting point. Higher
Em₀ = U = m g h
final point. Lower, just before the crash
Em_f = K = ½ m
energy is conserved
Em₀ = Em_f
m g h = ½ m v²
v_b =
* Now let's analyze the collision of the two spheres. We form a system formed by the two spheres, therefore the forces during the collision are internal and the moment is conserved
initial instant. Just before the crash
p₀ = 2m 0 + m v_b
final instant. Right after the crash
p_f = (2m + m) v
the moment is preserved
p₀ = p_f
m v_b = 3m v
v = v_b / 3
v = ⅓ 
* finally we analyze the movement after the crash. Let's use the conservation of energy to the system formed by the two spheres stuck together
Starting point. Lower
Em₀ = K = ½ 3m v²
Final point. Higher
Em_f = U = (3m) g h'
Em₀ = Em_f
½ 3m v² = 3m g h’
we substitute
h’=
h’ =
h’ = 1/9 h