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Hitman42 [59]
3 years ago
10

Evidence suggests that there may be _______ momentum and ________ reversal patterns in stock price behavior.

Physics
1 answer:
bearhunter [10]3 years ago
8 0

CORRECT ANSWER:

D. short-run; long run

STEP-BY-STEP EXPLANATION:

The whole question from book is

Evidence suggests that there may be _______ momentum and ________ reversal patterns in stock price behavior.

A. short-run; short-run

B. long-run; long-run

C. long-run; short-run

D. short-run; long run

Evidence suggests that there may be <em><u>short-run</u></em> momentum and <em><u>long run </u></em>reversal patterns in stock price behavior.

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What is Density? (3-5 sentences or your own word don't copy the<br> notes)
lara31 [8.8K]

Answer:

the density makes something float or sink below water its the "water weight "

Explanation:

8 0
4 years ago
Consider a cylindrical segment of a blood vessel 2.70 cm long and 3.10 mm in diameter. What additional outward force would such
Lyrx [107]

This question is incomplete, the complete question is;

- Calculate the difference in blood pressure between the feet and top of the head of a person who is 1.80m Tall

- Consider a cylindrical segment of a blood vessel 2.70 cm long and 3.10 mm in diameter. What additional outward force would such a vessel need to withstand in the person's feet compared to a similar vessel in her head

Answer:

- the difference in blood pressure is 18698.4 Pa

- the additional outward force F is 4.86 N

Explanation:

Given the data in the question;

we know that the expression for difference in blood pressure is;

ΔP = pgh

where p is density = 1060 kg/m³

g is acceleration due to gravity  = 9.8 m/s²

and h is height = 1.80 m

now we substitute

ΔP = 1060 × 9.8 × 1.80

ΔP = 18698.4 Pa

therefore the difference in blood pressure is 18698.4 Pa

Also given that;

diameter of blood vessel d = 3.10 mm

radius r = 3.10 mm / 2 = 1.55 mm = 0.00155 m

length l = 2.70 cm = 0.027 m

Surface area of the cylindrical segment of a blood vessel is

A = 2πrl

we substitute

A = 2 × π × 0.00155 × 0.027

A = 2.6 × 10⁻⁴ m²

so

the required for will be;

F = PA

we substitute

F = 18698.4 Pa × 2.6 × 10⁻⁴ m²

F = 4.86 N

Therefore, the additional outward force F is 4.86 N

5 0
3 years ago
Kevin is refinishing his rusty wheelbarrow. He moves his sandpaper back and forth 45 times over a rusty area, each time moving a
dmitriy555 [2]
W = _|....F*dx*cos(a)........With F=force, x=distance over which force acts on object,
.......0.............................and a=angle between force and direction of travel.

Since the force is constant in this case we don't need the equation to be an integral expression, and since the force in question - the force of friction - is always precisely opposite the direction of travel (which makes (a) equal to 180 deg, and cos(a) equal to -1) the equation can be rewritted like so:

W = F*x*(-1) ............ or ............. W = -F*x

The force of friction is given by the equation: Ffriction = Fnormal*(coeff of friction)

Also, note that the total work is the sum of all 45 passes by the sandpaper. So our final equation, when Ffriction is substituted, is:

W = (-45)(Fnormal)(coeff of friction)(distance)
W = (-45)...(1.8N).........(0.92).........(0.15m)
W = ................-11.178 Joules
5 0
3 years ago
Two waves can share the same space and pass through one another. What type of interference occurs when the crests of one wave ov
bulgar [2K]

Question 25 Answer: Destructive interference occurs.

<span>Question 26Answer: The waves are closer together (as they move) because the object is moving toward you.</span>

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4 0
3 years ago
At the same moment, one rock is dropped and one is thrown downward with an initial velocity of 29m/s from the top of a building
Inessa [10]

Answer:

The thrown rock strike 2.42 seconds earlier.

Explanation:

This is an uniformly accelerated motion problem, so in order to find the arrival time we will use the following formula:

x=vo*t+\frac{1}{2} a*t^2\\where\\x=distance\\vo=initial velocity\\a=acceleration

So now we have an equation and unkown value.

for the thrown rock

\frac{1}{2}(9.8)*t^2+29*t-300=0

for the dropped rock

\frac{1}{2}(9.8)*t^2+0*t-300=0

solving both equation with the quadratic formula:

\frac{-b\±\sqrt{b^2-4*a*c} }{2*a}

we have:

the thrown rock arrives on t=5.4 sec

the dropped rock arrives on t=7.82 sec

so the thrown rock arrives 2.42 seconds earlier (7.82-5.4=2.42)

6 0
3 years ago
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