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3 years ago

Define alpha and beta

Physics

1 answer:

elena55 [62]3 years ago

4 0

alpha is the excess return on an investment after adjusting for market related volatility and random fluctuations.

beta is a measure of volatility relative to a benchmark ,such as the S&P 500.

Explanation:

alpha and beta are two different parts of an equation used to explain the performance of stocks and investments funds. But in maths alpha and beta is the Greek alphabet

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At NASA's Zero Gravity Research Facility in Cleveland, Ohio, experimental payloads fall freely from rest in an evacuated vertica

Diano4ka-milaya [45]

Answer:

(a). Energy is 64,680 J

(b) velocity is 51.43m/s

Explanation:

(a).

The potential energy $P$ of the payload of mass $m$ is at a vertical distance $h$ is

$P =mgh$ .

Therefore, for the payload of mass $m = 50kg$ at a vertical distance of $h = 132 m$ , the potential energy is

$P = (50kg)(9.8m/s^2)(132m)$

$\boxed{P = 64,680J}$

(b).

When the payload reaches the bottom of the shaft, all of its potential energy is converted into its kinetic energy; therefore,

$mgh= \dfrac{1}{2}mv^2$

$v= \sqrt{2gh}$

$v = \sqrt{2*9.8*135}$

$\boxed{v = 51.43m/s}$

(c).

The velocity in mph is

$\dfrac{51.43m}{s} * \dfrac{3600s}{hr} * \dfrac{1mile}{1609.34m}$

$\boxed{v= 115.0mph}$

5 0

4 years ago

The compressed-air tank ab has a 250-mm outside diameter and an 8-mm wall thickness. it is fitted with a collar by which a 40-kn

valentinak56 [21]

Assume: neglect of the collar dimensions. Ď_h=(P*r)/t=(5*125)/8=78.125 MPa ,Ď_a=Ď_h/2=39 MPa Ď„=(S*Q)/(I*b)=(40*ă€–10ă€—^3*Ď€(ă€–0.125ă€—^2-ă€–0.117ă€—^2 )*121*ă€–10ă€—^(-3))/(Ď€/2 (ă€–0.125ă€—^4-ă€–0.117ă€—^4 )*8*ă€–10ă€—^(-3) )=41.277 MPa @ Point K: Ď_z=(+M*c)/I=(40*0.6*121*ă€–10ă€—^(-3))/(8.914*ă€–10ă€—^(-5) )=32.6 MPa Using Mohr Circle: Ď_max=(Ď_h+Ď_a)/2+âš(Ď„^2+((Ď_h-Ď_a)/2)^2 ) Ď_max=104.2 MPa, Ď„_max=45.62 MPa

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4 years ago

Pablo wants to enter the science fair. Which question could be answered through scientific investigation?

FinnZ [79.3K]

D.What is the best angle to throw a ball for maximum distance? Because this is the only question in above all of them which has some scientific investigation like angle that is use in many of the physics relations.

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4 years ago

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A baseball diamond is a square (don’t be fooled that it is usually shown rotated by 45°). Each side of the square is 90 feet lon

gayaneshka [121]

A line from second to home creates a right triangle with 45 degree angles.

This makes the diagonal line the side length of the square multiplied by the sqrt(2)

Second to home = 90sqrt(2) = 127.279 feet ( round answer as needed)

8 0

3 years ago

g While hauling a log in the back of a flatbed truck, a driver is pulled over by the state police. Although the log cannot roll

irga5000 [103]

Answer:

The minimum coefficient of static friction required, µ = 0.10

Note. The question is incomplete. The complete question is given below:

While hauling a log in the back of a flatbed truck, a driver is pulled over by the state police. Although the log cannot roll sideways, the police claim that the log could have slid out the back of the truck when accelerating from rest. The driver claims that the truck could not possibly accelerate at the level needed to achieve such an effect. Regardless, the police write a ticket anyway and now the driver court date is approaching.

The log has a mass of m = 929 kg; the truck has a mass of M = 8850 kg. According to the truck manufacturer, the truck can accelerate from 0 to 55 mph in 23.0 seconds, but this does not account for the additional mass of the log. Calculate the minimum coefficient of static friction μs needed to keep the log in the back of the truck.

Explanation:

First, velocity in mph is converted to m/s

1 mph = 0.447 m/s

55 mph ≈ 24.6 m/s

The acceleration of the empty truck is a = v/t = 24.6 / 23 = 1.07 m/s²

Force that can be generated by the truck, F = ma

F = 8850kg * 1.07 m/s² = 9469.5 N

However, with the added mass of the log on it, the acceleration of the truck will become;

a = F / m = 9469.5 N / (8550+929)kg = 0.97 m/s²

Frictional force between the log and the truck = 0.97 m/s² * 929 kg = 901.13 N

Normal reaction on the truck due to the weight of the log, R = mg

R = 929 kg * 9.8m/s² = 9104.2 N

Coefficient of static friction, µ = F/R

µ = 901.13/9104.2

µ = 0.098 ≈ 0.10

Therefore, the minimum static friction required is µ = 0.10

8 0

3 years ago

Define alpha and beta​

Define alpha and beta