The time interval that is between the first two instants when the element has a position of 0.175 is 0.0683.
<h3>How to solve for the time interval</h3>
We have y = 0.175
y(x, t) = 0.350 sin (1.25x + 99.6t) = 0.175
sin (1.25x + 99.6t) = 0.175
sin (1.25x + 99.6t) = 0.5
99.62 = pi/6
t1 = 5.257 x 10⁻³
99.6t = pi/6 + 2pi
= 0.0683
The time interval that is between the first two instants when the element has a position of 0.175 is 0.0683.
b. we have k = 1.25, w = 99.6t
v = w/k
99.6/1.25 = 79.68
s = vt
= 79.68 * 0.0683
= 5.02
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complete question
A transverse wave on a string is described by the wave function y(x, t) = 0.350 sin (1.25x + 99.6t) where x and y are in meters and t is in seconds. Consider the element of the string at x=0. (a) What is the time interval between the first two instants when this element has a position of y= 0.175 m? (b) What distance does the wave travel during the time interval found in part (a)?
Answer: The height of the fluid rise is 0.01m
Explanation:
Using the equation
h = (2TcosѲ )/rpg
h= height of the fluid rise
diameter of the tube =3mm
radius of the tube= 3/2 =1.5mm=0.0015
T= surface tension = 600mN/m=0.6N/m
Ѳ = contact angle =
C
p= density =3.7g/cm3= 3700kg/m3
g= acceleration due to gravity =9.8m/s2
h = ( 2*0.6*0.5)/(0.0015*3700*9.8)
h = 0.6/54.39
h= 0.01m
Therefore,the height of the fluid rise is 0.01m
Answer:

Explanation:
We are given that
Diameter of wire=d=4.12 mm
Radius of wire=r

Current=I=8 A
Drift velocity=
We have to find the density of free electrons in the metal
We know that
Density of electron=
Using the formula
Density of free electrons=
By using Area of wire=

Density of free electrons=