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)?
<em>number of waves that pass a given point in one second is called <u>frequency..</u></em>
Answer:
<h2>FUNDAMENTAL UNITS INVOLVED ARE : NEWTON AND SECOND .</h2>
<h2>FORMULA OF PRESSURE = </h2>
<h2>P=F/A </h2>
Answer:
The weight of measuring stick is 9.8 N
Explanation:
given information:
the mass of the rock, 
= 1 kg
measuring stick, x =1 m
d = 0.25 m
to find the weight of measuring stick, we can use the following equation:
τ = Fd
τ = 0
- 
= 0
F_{r} = the force of the rock
F_{s} = the force of measuring stick
= m g
= 1 kg x 9.8 m/s
= 9.8 N
thus, the weight of measuring stick is 9.8 N
Answer:
2,500 Joules (J) or Newton Meter (N M)
Explanation:
Work = Force x Distance
The force in this equation is 500 Newtons. The distance (displacement) is 5 meters. Plug it into the equation above.
Work = 5m x 500n
Work = 2,500 Joules or Newton-Meters.
Therefore 2,500 Joules or Newton Meters of work is done on an object.
