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: c
Explanation: it is c because i used my brain to answer it
Answer:
-.457 m/s^2
Explanation:
Actual weight = 60 .3 (9.81) = 591.54 N
Accel of lift changes this to 60.3 ( 9.81 - L) where L - accel of lift
60.3 ( 9.81 - L ) = 564
solve for L = .457 m/s^2 DOWNWARD
so L = - .457 m/s^2
Answer:
-62.45m/s and +62.45m/s
Explanation:
The formula for relativistic speed
This is the speed of A with respect to B
where
will be the velocity of person 1: 39m/s
will be the velocity of person 2: -31m/s (negative because is travelling in opposite direction)
and 
the velocity of light: 100m/s
The velocity of person 1 measured by person 2 is:
The velocity of person 2 measured by person 1 is:
Explanation:
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