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
We know that speed is a scalar unit while velocity is a vector. The signs are mostly concerned with the magnitude that the car should travel in, the direction is not as important.
Hope I helped :)
Answer:
Explanation:
Given that,
Number of extra electrons, n = 21749
We need to find the net charge on the metal ball. Let Q is the net charge.
We know that the charge on an electron is 
To find the net charge if there are n number of extra electrons is :
Q = n × q
So, the net charge on the metal ball is 
. Hence, this is the required solution.
Answer:
0.37sec
Explanation:
Period of oscillation of a simple pendulum of length L is:
T
=
2
π
×
√
(L
/g)
L=length of string 0.54m
g=acceleration due to gravity
T-period
T = 2 x 3.14 x √[0.54/9.8]
T = 1.47sec
An oscillating pendulum, or anything else in nature that involves "simple harmonic" (sinusoidal) motion, spends 1/4 of its period going from zero speed to maximum speed, and another 1/4 going from maximum speed to zero speed again, etc. After four quarter-periods it is back where it started.
The ball will first have V(max) at T/4,
=>V(max) = 1.47/4 = 0.37 sec
The answer is B frequency. When frequency increases more wave crests pass a fixed point each second. That means the wavelength shortens. So, as frequency increases, wavelength decreases
