Answer:

Explanation:
Given:
- length of a steel-string,

- area of the string,

- Young's modulus of the steel,

- force of tension on the string,

We have the relation for change in length:




Answer:
The frictional force acting on the bear during the slide is 207.5 N
Explanation:
Given;
mass of beam, m = 25-kg
vertical height, h = 12 m
speed of fall, v = 6 m/s
Change in potential energy of the beam:
ΔP.E = -mgh = - 25 x 9.8 x 12 = -2940 J
Change in kinetic energy of the beam:
Δ K.E = ¹/₂mv² = ¹/₂ x 25 x (6)² = 450 J
Change in thermal energy of the system due to friction:
ΔE = - (ΔP.E + Δ K.E)
ΔE = - (-2940 J + 450 J)
ΔE = 2940 J - 450 J = 2490 J
Frictional force (in N) acting on the bear during the slide:
F x d = Fk x h = ΔE
Where;
Fk is the frictional force
Fk = ΔE/h
Fk = 2490J / 12m
Fk = 207.5 N
Therefore, the frictional force acting on the bear during the slide is 207.5 N
The beat frequency produced by the two standing waves is 13 Hz.
<h3>The wavelength of the shorter string</h3>
The wavelength of the shorter string is calculated as follows;

<h3>The length of the longer string</h3>

The frequency of the longer string is calculated as follows;

<h3>Beat frequency</h3>
The beat frequency produced by the two standing waves is calculated as follows;

Learn more about beat frequency here: brainly.com/question/3086912
To solve this problem it is necessary to apply the concepts related to destructive interference.
The concept refers to an overlap of two or more waves of identical or similar frequency that, when interfering, creates a new pattern of waves of lower intensity
By definition destructive interference is given by

Where,

n=integer (1,2,3,4,5,6...etc)
t = thickness
M= Index of refrqaction
For minimum thickness to satisfy this condition n will be minimum there,
n=0
Therefore

Solving to find M,



Therefore the correct answer is B. 1.38
Answer:
they share electrons between them.
Explanation:
taking the test rn lol i think its right