Answer:
See Below
Explanation:
Okay, I thinkkk what it is asking by what you summarzied for me issss:
They split the total time into four quarters. They then took (for the first quarter) the start time. Then when the first quarter ends and the second quarter starts is the "end" time.
They then subtract the start time of the second quarter from the end time of the first quarter.
I hope this helps, good luck! :D
Answer:
C hope it helps
call the client and inform her that she was incorrectly charged.
Answer:
a) v = 0.7071 v₀, b) v= v₀, c) v = 0.577 v₀, d) v = 1.41 v₀, e) v = 0.447 v₀
Explanation:
The speed of a wave along an eta string given by the expression
v = 
where T is the tension of the string and μ is linear density
a) the mass of the cable is double
m = 2m₀
let's find the new linear density
μ = m / l
iinitial density
μ₀ = m₀ / l
final density
μ = 2m₀ / lo
μ = 2 μ₀
we substitute in the equation for the velocity
initial v₀ =
with the new dough
v =
v = 1 /√2 \sqrt{ \frac{T_o}{ \mu_o} }
v = 1 /√2 v₀
v = 0.7071 v₀
b) we double the length of the cable
If the cable also increases its mass, the relationship is maintained
μ = μ₀
in this case the speed does not change
c) the cable l = l₀ and m = 3m₀
we look for the density
μ = 3m₀ / l₀
μ = 3 m₀/l₀
μ = 3 μ₀
v =
v = 1 /√3 v₀
v = 0.577 v₀
d) l = 2l₀
μ = m₀ / 2l₀
μ = μ₀/ 2
v =
v = √2 v₀
v = 1.41 v₀
e) m = 10m₀ and l = 2l₀
we look for the density
μ = 10 m₀/2l₀
μ = 5 μ₀
we look for speed
v =
v = 1 /√5 v₀
v = 0.447 v₀
<span>Nothing happens to the pitch of a cell phone ring when the amplitude
of a sound wave increases.
Pitch and amplitude are both characteristics of a wave, but they're not
connected, and they don't influence each other.</span>
Step 1: Identify the variables. ...Step 2: Determine the variable range. ...Step 3: Determine the scale of the graph. ...Step 4: Number and label each axis and title the graph.Step 5: Determine the data points and plot on the graph. ...Step 6: Draw the graph.