Answer:
1.566 x 10^2
Move the decimal to where the number being multiplied by 10^x is greater than 1 but less than 10. Then multiply it by 10^x
X is the number of times you moved the decimal, so in this case it would be 10^2
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)?
Complete Question
The complete question is shown on the first uploaded image
Answer:
The workdone is 
Explanation:
From the question we are told that
The initial Volume is 
The final volume is 
The external pressure is
Generally the change in volume is

Substituting values we have


Generally workdone is mathematically represented as

W is negative because the working is done on the environment by the system which is indicated by volume increase
Substituting values


Now 
Therefore 

Answer:
Latent heatnof fusion = 417.5 J
Explanation:
Specific latent heat of fusion of water is 334kJ.kg-1.
The heat required to melt water when it's ice I called latent heat because there is no temperature change, the only change observed is change in physical structure.
The amount of heat required to change 1 kg of solid to its liquid state (at its melting point) at atmospheric pressure is called Latent heat of Fusion.
Latent heat = ML
Latent heat= 1.25 kg * 334kJ.kg-1
Latent heat = 1.25*334 *(J/kg)*kg
Latent heat = 417.5 J