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1 year ago

Determine whether the underlined numerical value is a parameter or a statistic. explain your reasoning. a certain zoo

Physics

1 answer:

irinina [24]1 year ago

6 0

Parameter, because the data set of all 843 animals in a zoo is a population.

The population of all animals in the Zoo is 843

Proportion = 8%, which is the ratio regarding the population. So 8% is populaton proporiton, i..e Parameter .

A zoo, short for the zoological garden, is a place where animals are kept in enclosures, cared for, presented to the public, and occasionally bred for conservation efforts. The phrase zoological garden refers to zoology, the study of animals. It is also used to refer to an animal park or menagerie. In the United States alone, over 181 million people visit zoos every year. The goal of the modern zoo, which first appeared in the 19th century in the United Kingdom was to entertain and inspire the public by offering scientific inquiry and later educational exhibits. The London Zoological Gardens, which was opened to the public in 1847 after being opened for scientific study in 1828, is where the abbreviation "zoo" first appeared.

To learn more about the Zoo, visit:

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The lowest possible temperature in outer space is 3.13 K. What is the rms speed of hydrogen molecules at this temperature? (The

Umnica [9.8K]

Answer:

$v_{rms} =196.59 m/s$

Given:

Temperature, T = 3.13 K

molar mass of molecular hydrogen, m = 2.02 g/mol = $2.02\times 10^{-3}kg/mol$

Solution:

To calculate the root mean squarer or rms speed of hydrogen molecule, we use the given formula:

$v_{rms} = \sqrt{\frac{3TR}{m}}$

where

R = rydberg's constant = 8.314 J/mol-K

Putting the values in the above formula:

$v_{rms} = \sqrt{\frac{3\times 3.13\times 8.314}{2.02\times 10^{-3}}}$

$v_{rms} =196.59 m/s$

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3 years ago

An object, initially at rest, is subject to an acceleration of 34 m/s^2. How long will it take for that object to reach 3400m ?

Norma-Jean [14]

Vf^2 = Vi^2 + 2ad
a= 34 m/s^2
Vi = 0 m/s
d = 3400m

Vf = 480.83 m/s

a=v/t
t=v/a
t=480.83/34
t=14.142 s

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2 years ago

An observation of the red shift of galaxies suggests that the universe is ______ a.. expanding.. b.. contracting.. c.. reversing

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3 years ago

Read 2 more answers

A 25 kg mass is accelerated by a force at a rate of 5 m/s2. What is the magnitude of the force that accelerates the man?

svlad2 [7]

Force=mass*acceleration
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3 years ago

A solid sphere, a solid disk, and a thin hoop are all released from rest at the top of the incline (h0 = 20.0 cm).

Ede4ka [16]

Answer:

a. The object with the smallest rotational inertia, the thin hoop

b. The object with the smallest rotational inertia, the thin hoop

c. The rotational speed of the sphere is 55.8 rad/s and Its translational speed is 1.67 m/s

Explanation:

a. Without doing any calculations, decide which object would be spinning the fastest when it gets to the bottom. Explain.

Since the thin has the smallest rotational inertia. This is because, since kinetic energy of a rotating object K = 1/2Iω² where I = rotational inertia and ω = angular speed.

ω = √2K/I

ω ∝ 1/√I

since their kinetic energy is the same, so, the thin hoop which has the smallest rotational inertia spins fastest at the bottom.

b. Again, without doing any calculations, decide which object would get to the bottom first.

Since the acceleration of a rolling object a = gsinФ/(1 + I/MR²), and all three objects have the same kinetic energy, the object with the smallest rotational inertia has the largest acceleration.

This is because a ∝ 1/(1 + I/MR²) and the object with the smallest rotational inertia has the smallest ratio for I/MR² and conversely small 1 + I/MR² and thus largest acceleration.

So, the object with the smallest rotational inertia gets to the bottom first.

c. Assuming all objects are rolling without slipping, have a mass of 2.00 kg and a radius of 3.00 cm, find the rotational and translational speed at the bottom of the incline of any one of these three objects.

We know the kinetic energy of a rolling object K = 1/2Iω² + 1/2mv² where I = rotational inertia and ω = angular speed, m = mass and v = velocity of center of mass = rω where r = radius of object

The kinetic energy K = potential energy lost = mgh where h = 20.0 cm = 0.20 m and g = acceleration due to gravity = 9.8 m/s²

So, mgh = 1/2Iω² + 1/2mv² = 1/2Iω² + 1/2mr²ω²

Let I = moment of inertia of sphere = 2mr²/5 where r = radius of sphere = 3.00 cm = 0.03 m and m = mass of sphere = 2.00 kg

So, mgh = 1/2Iω² + 1/2mr²ω²

mgh = 1/2(2mr²/5 )ω² + 1/2mr²ω²

mgh = mr²ω²/5 + 1/2mr²ω²

mgh = 7mr²ω²/10

gh = 7r²ω²/10

ω² = 10gh/7r²

ω = √(10gh/7) ÷ r

substituting the values of the variables, we have

ω = √(10 × 9.8 m/s² × 0.20 m/7) ÷ 0.03 m

= 1.673 m/s ÷ 0.03 m

= 55.77 rad/s

≅ 55.8 rad/s

So, its rotational speed is 55.8 rad/s

Its translational speed v = rω

= 0.03 m × 55.8 rad/s

= 1.67 m/s

So, its rotational speed is of the sphere is 55.8 rad/s and Its translational speed is 1.67 m/s

6 0

2 years ago