Answer:
P = d g h pressure due to density of liquid of height h
d1 g h1 = d2 g h2 since P1 equals P2
d1 h1 = d2 h2 and the density of Hg is 13.6 times that of water
h2 = (d1 / d2) * h1 = 13.6 * 70 cm = 952 cm or 9.52 m
Answer:
a) L = 33.369 m
, b) 21
Explanation:
The analysis of the ocean depth can be performed assuming that at the bottom of the ocean there is a node and the surface must have a belly, so the expression for resonance is
λ = 4 L / n
n = 1, 3, 5, ...
The speed of the wave is
v = λ f
v = 4L / n f
L = n v / 4f
Let's write the expression for the two frequencies
L = n₁ 343/4 53.95
L = n₁ 1,589
L = n₂ 343/4 59
L = n₂ 1.4539
Let's solve the two equations
n₁ 1,589 = n₂ 1,459
n₁ / n₂ = 1.4539 / 1.589
n₁ / n2 = 0.91498
Since the two frequencies are very close the whole numbers must be of consecutive resonances, let's test what values give this value
n₁ n₂ n₁ / n₂
1 3 0.3
3 5 0.6
5 7 0.7
7 9 0.77
9 11 0.8
17 19 0.89
19 21 0.905
21 23 0.913
23 25 0.92
Therefore the relation of the nodes is n₁ = 21 and n₂ = 23
Let's calculate
L = n₁ 1,589
L = 21 1,589
L = 33.369 m
b) the number of node and nodes is equal therefore there are 21 antinode
Answer:
53.895 m.
Explanation:
Using the equation of motion,
v² = u² + 2as .............. Equation 1
Where v = final velocity of the swan, u = initial velocity of the swan, a = acceleration of the swan, s = distance covered by the swan.
make s the subject of the equation,
s = (v² - u²)/2a----------- Equation 2
Given: v = 6.4 m/s, u = 0 m/s ( from rest) a = 0.380 m/s².
Substitute into equation 2
s = (6.4²-0²)/(2×0.380)
s = 40.96/0.76
s = 53.895 m.
Hence the swan will travel 53.895 m before becoming airborne.
Answer: C. -1.16 meters/second2
Explanation:
A= v/t (velocity/time)
in this case: v=7 and t=6
So, A= 7/6
A=1.16
The graph is decreasing so accelleration would be negative
A= <u>-1.16 meters/second2</u>
<u>Option C!</u> ; )
<u></u>
Answer:
a - As long as the time between 2 events is reconcilable with a light signal, the time between the events, in that frame, can be determined.
