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:
<u>Given</u><em> </em><em>-</em><em> </em><u>M</u><u> </u><u>=</u><u> </u>20 kg
k = 0.4
F = 200 N
<u>To </u><u>find </u><u>-</u><u> </u> acceleration
<u>Solution </u><u>-</u><u> </u>
F= kMA
200 = 0.4 * 20 * acceleration
200 = 8 * a
a = 8/200
a = 0.04 m s²
<h3>a = 0.04 m s²</h3>
Answer:
0.15625 grams
Explanation:
Half life: It is related to the decay of radioactive material. The duration in which half of the material will be degraded/decayed. That means after half life 50% of the radioactive material will be left. Here the half life is 28 years.
Initial quantity of the sample: 2.5 grams.
After 28 years, the leftover quantity = 1.25 grams
After 56 years, the leftover quantity = 0.625 grams
After 84 Years, the leftover quantity = 0.3125 grams
After 112 years, the leftover quantity = 0.15625 grams
Kinetic energy depends on the mass and the speed of a moving object.
If the speeds are equal, then the rick with more mass has more kinetic energy.