No because there must be an even # if their is an even amount one of the forces isn’t being cancelled
To measure the strength of an earthquake, you can use either a Richter scale or Mercalli scale. Richter scale uses the amplitued of the wave and the distance from the source. Mercalli scale uses observations of people and is not considered to be scientific as Richter scale.
Answer:
The portfolio should invest 48.94% in equity while 51.05% in the T-bills.
Explanation:
As the complete question is not given here ,the table of data is missing which is as attached herewith.
From the maximized equation of the utility function it is evident that
For the equity, here as
is percentage of the equity which is to be calculated
is the Risk premium whose value as seen from the attached data for the period 1926-2015 is 8.30%
is the risk aversion factor which is given as 4.
is the standard deviation of the portfolio which from the data for the period 1926-2015 is 20.59
By substituting values.
So the weight of equity is 48.94%.
Now the weight of T bills is given as
So the weight of T-bills is 51.05%.
The portfolio should invest 48.94% in equity while 51.05% in the T-bills.
Answer:
- Fx = -9.15 N
- Fy = 1.72 N
- F∠γ ≈ 9.31∠-10.6°
Explanation:
You apparently want the sum of forces ...
F = 8.80∠-56° +7.00∠52.8°
Your angle reference is a bit unconventional, so we'll compute the components of the forces as ...
f∠α = (-f·cos(α), -f·sin(α))
This way, the 2nd quadrant angle that has a negative angle measure will have a positive y component.
= -8.80(cos(-56°), sin(-56°)) -7.00(cos(52.8°), sin(52.8°))
≈ (-4.92090, 7.29553) +(-4.23219, -5.57571)
≈ (-9.15309, 1.71982)
The resultant component forces are ...
Then the magnitude and direction of the resultant are
F∠γ = (√(9.15309² +1.71982²))∠arctan(-1.71982/9.15309)
F∠γ ≈ 9.31∠-10.6°
Answer:
t = 39.60 s
Explanation:
Let's take a careful look at this interesting exercise.
In the first case the two motors apply the force in the same direction
F = m a₀
a₀ = F / m
with this acceleration it takes t = 28s to travel a distance, starting from rest
x = v₀ t + ½ a t²
x = ½ a₀ t²
t² = 2x / a₀
28² = 2x /a₀ (1)
in a second case the two motors apply perpendicular forces
we can analyze this situation as two independent movements, one in each direction
in the direction of axis a, there is a motor so its force is F/2
the acceleration on this axis is
a = F/2m
a = a₀ / 2
so if we use the distance equation
x = v₀ t + ½ a t²
as part of rest v₀ = 0
x = ½ (a₀ / 2) t²
let's clear the time
t² = (2x / a₀) 2
we substitute the let of equation 1
t² = 28² 2
t = 28 √2
t = 39.60 s
