First, foremost, and most critically, you must look at the graph, and critically
examine its behavior from just before until just after the 5-seconds point.
Without that ability ... since the graph is nowhere to be found ... I am hardly
in a position to assist you in the process.
Answer:
0.000507 kg/m
Explanation:
L = Length of string
T = Tension
= Mass density of string
E denotes the E string
D denotes the D String
Frequency is given by
So
The mass density of the E string is 0.000507 kg/m
Answer:
I BELIEVE THE ANSWER A I REMEMBER THAT QUESTION WHEN I DID IT SO THATS THE BEST ONE OUT OF THE 3
Explanation:
Answer:
Maximum altitude to see(L) = 1.47 × 10⁶ m (Approx)
Explanation:
Given:
wavelength (λ) = 0.12 nm = 0.12 × 10⁻⁹ m
Pupil Diameter (d) = 4.1 mm = 4 × 10⁻³ m
Separation distance (D) = 5.4 cm = 0.054 m
Find:
Maximum altitude to see(L)
Computation:
Resolving power = 1.22(λ / d)
D / L = 1.22(λ / d)
0.054 / L = 1.22 [(0.12 × 10⁻⁹) / (4 × 10⁻³ m)]
0.054 / L = 1.22 [0.03 × 10⁻⁶]
L = 0.054 / 1.22 [0.03 × 10⁻⁶]
L = 0.054 / [0.0366 × 10⁻⁶]
L = 1.47 × 10⁶
Maximum altitude to see(L) = 1.47 × 10⁶ m (Approx)
4x time 10x (I think)
Here, this will help-
http://www.cas.miamioh.edu/mbiws/microscopes/Magnification.html#totalmagnification
