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
Answer:
31.905 ft/s²
Explanation:
Given that
Mass of the pilot, m = 120 lb
Weight of the pilot, w = 119 lbf
Acceleration due to gravity, g = 32.05 ft/s²
Local acceleration of gravity of found by using the relation
Weight in lbf = Mass in lb * (local acceleration/32.174 lbft/s²)
119 = 120 * a/32. 174
119 * 32.174 = 120a
a = 3828.706 / 120
a = 31.905 ft/s²
Therefore, the local acceleration due to gravity at that elevation is 31.905 ft/s²
Answer:
Explanation:
We know that when we don't have air friction on a free fall the mechanical energy (I will symbololize it with ME) is equal everywhere. So we have:
where me(1) is mechanical energy while on h=10m
and me(2) is mechanical energy while on the ground
Ek(1) + DynamicE(1) = Ek(2) + DynamicE(2)
Ek(1) is equal to zero since an object that has reached its max height has a speed equal to zero.
DynamicE(2) is equal to zero since it's touching the ground
Using that info we have
we divide both sides of the equation with mass to make the math easier.
Answer:
The maximum volume is 1417.87 
Explanation:
<u>Optimization Using Derivatives</u>
We have a 24x30 inch piece of metal and we need to make a rectangular box by cutting a square from each corner of the piece and bending up the sides. The width of the piece is 24 inches and its length is 30 inches
When we cut a square of each corner of side x, the base of the box (after bending up the sides) will be (24-2x) and (30-2x), width and length respectively. The volume of the box is
Operating
To find the maximum value of V, we compute the first derivative and equate it to zero
Simplifying by 12
Completing squares
We have two values for x
The first value is not feasible because it will produce a negative width (24-2(13.58))=-6.16
We'll keep only the solution
The width is
The length is
And the height
The maximum volume is
