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dimulka [17.4K]
3 years ago
9

An agriculturalist working with Australian pine trees wanted to investigate the relationship between the age and the height of t

he Australian pine. A random sample of Australian pine trees was selected, and the age, in years, and the height, in meters, was recorded for each tree in the sample. Based on the recorded data, the agriculturalist created the following regression equation to predict the height, in meters, of the Australian pine based on the age, in years, of the tree. predicted height 0.29+048 (age) Which of the following is the best interpretation of the slope of the regression line?
A. The height increases, on average, by 1 meter each 0.48 year.
B. The height increases, on average, by 0.48 meter each year.
C. The height increases, on average, by 0.29 meter each year.
D. The height increases, on average, by 0.29 meter each 0.48 year.
E. The difference between the actual height and the predicted height is, on average, 0.48 meter for each year.
Physics
1 answer:
SVETLANKA909090 [29]3 years ago
6 0

Answer:

The correct answer is B.

Explanation:

Step 1:

The available regression equation is: Predict height= 0.29 + 0.48 (age).

Here, the predict height is dependent variable and the  age is in-dependent variable.

Intercept = 0.29

Slope      = 0.48

The given regression equation indicates the y on x model and the intercept coefficients of the regression equation is 0.29 and the slope is 0.48.

Step 2:

The height increases, an average, by 0.48 m per year.

Because co-efficient of slope variable indicate the positive sign and we increase 1 year in age then automatically height increased is 0.48 m.

<h3></h3><h3>The height increases, on average, by 0.48 meter each year.</h3>
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<em>b) 61.908 J</em>

<em>c)  </em>\frac{4492.18}{v_{car} ^{2} }

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Explanation:

The complete question is

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.

Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?

Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?

Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar. Enter an expression for the mass of the car, in terms of the quantities defined here.

moment of inertia is given as

I = \frac{1}{2}mr^{2}

where m is the mass of the flywheel,

and r is the radius of the flywheel

for the flywheel with radius 1.1 m

and mass 11 kg

moment of inertia will be

I =  \frac{1}{2}*11*1.1^{2} = 6.655 kg-m^2

The maximum speed of the flywheel = 35 m/s

we know that v = ωr

where v is the linear speed = 35 m/s

ω = angular speed

r = radius

therefore,

ω = v/r = 35/1.1 = 31.82 rad/s

maximum rotational energy of the flywheel will be

E = Iw^{2} = 6.655 x 31.82^{2} = <em>6738.27 J</em>

<em></em>

b) second flywheel  has

radius = 2.8 m

mass = 16 kg

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I = \frac{1}{2}mr^{2} =  \frac{1}{2}*16*2.8^{2} = 62.72 kg-m^2

According to conservation of angular momentum, the total initial angular momentum of the first flywheel, must be equal to the total final angular momentum of the combination two flywheels

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E = Iw^{2} = 6.655 x 3.05^{2} = <em>61.908 J</em>

<em></em>

<em></em>

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6738.27/3 = 2246.09 J

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Equating the energy

2246.09 =  \frac{1}{2}mv_{car} ^{2}

making m the subject of the formula

mass of the car m = \frac{4492.18}{v_{car} ^{2} }

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