Answer:
Explanation:
Let the intensity of the noise be represented by I
Given that
40dB = 10 log 10 ( I /I•) ........ 1
I• is the lowest or threshold intensity of sound made.
I represents the intensity of the sound/ noise
The intensity of noise of 1000flies will be
β = 10 log 10 (1000I/I•)
Open up the bracket
β = 10 log 10(1000)+ 10 log 10(I/I•)
10 log 10(10^3)+10 log 10(I/I•)
3×10(10 log 10) +10 log 10(I/I•)
Recall, 10 log 10 = 1
30×1 + 10 log 10(I/I•).........2
Put equation 1 into 2
β =30+40
= 70db
Momentum should be conserved. The momentum of both
objects must balance with their initial and final momentum.
Let m1 and v1 be the mass and velocity of the
bowling ball
And m2 and v2 be the mass and velocity of the
bowling pin
(m1v1)i + (m2v2)i = (m1v1)f + (m2v2)f
30 kg m/s + (1.5 kg)(0 m/s) = 13kg m/s + 1.5v2f
V2f = 11.33 m/s
<span>So the momentum = 1.5 kg(11.33 m/s) = 17 kg m/s</span>
Mass= volume x density
Mass= 90kg/m^3 x 2.3m^3
Therefore, Mass= 207 kg
The first opiton is the answer A)<span>Rahul’s weight
</span>
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
Read more on waves here
brainly.com/question/25699025
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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)?
