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)?
The gravity on Neptune is 11.15 m/s²
the gravity on earth is 9.81 m/s²
divide the Neptune and earth gravity we get 1.13
which means object on neptune is 1.13 heavier than earth
yield, weigh of the object on neptune is 1.13×15=17.04kg
Answer:
a = 0.7267
, acceleration is positive therefore the speed is increasing
Explanation:
The definition of acceleration is
a = dv / dt
they give us the function of speed
v = - (t-1) sin (t² / 2)
a = - sin (t²/2) - (t-1) cos (t²/2) 2t / 2
a = - sin (t²/2) - t (t-1) cos (t²/2)
the acceleration for t = 4 s
a = - sin (4²/2) - 4 (4-1) cos (4²/2)
a = -sin 8 - 12 cos 8
remember that the angles are in radians
a = 0.7267
the problem does not indicate the units, but to be correct they must be m/s²
We see that the acceleration is positive therefore the speed is increasing
The igneous rocks which were deposited on the surface and then cooled are known as extrusive. These rocks are a result of a magma reaching the surface of the Earth which cools the magma quickly. Examples are rhyolite, basalt, obsidian and andesite.
