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3 years ago

How do you move the decimal from megameters to meters?

1 answer:

8 0

The prefix "mega" means million.
Therefore
1 megameter = 10⁶ meters

That is,
1.0 megameter = 1,000,000.0 meters.

Answer:
The decimal is moved right by 6 places to convert a megameter to meters.

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What object best represents a true scale model of the shape of earth?

BabaBlast [244]

Any object that is spherical in shape would best represent a true scale model of the shape of the Earth. Examples are ping pong balls, billiard balls, marble and other smooth spherical objects. The shape of the Earth is called the oblate spheroid. The "oblate" would refer to an oblong shape and "spheroid" would refer to an almost spherical shape. The earth has on almost spherical shape and has a slightly oblong appearance. The diameter from the South pole to the north pole was measured to have a value of 12714 km while the diameter of the equator is approximately 12756 km. As you can see, the values are not equal. This makes the earth not a perfect sphere.

6 0

3 years ago

What is the maximum mass that can hang without sinking from a 20-cm diameter Styrofoam sphere in water? Assume the volume of the

alex41 [277]

Answer:

the maximum mass that can hang without sinking is 2.93 kg

Explanation:

Given: details:

sphere diameter d = 20 cm

so, radius r = 10 cm = 0.10 m

density of the Styrofoam sphere D = 300 kg/m3

sphere volume $V = \frac{4}{3} \pi r^3$

$=\frac{4}{3} \pi 0.10^3$

=4.18*10^{-3} m^3

we know that

$Density = \frac{Mass}{Volume}$

mass M = Density * Volume

= (300)(4.18*10^{-3} m3)

=1.25 kg

mass of the water displace = volume *density of water

= 4.18*10^{-3} m3 * 1000

= 4.18 kg

The difference between the mass of water and mass of styrofoam is the amount of mass that the sphere can support

=4.18 kg -1.25 kg

= 2.93 kg

3 0

3 years ago

the density of dry air at 20 degrees celsius is 1.20 g/L. What is the mass of air, in kilograms, of a room that measures 24.0m b

horsena [70]

<h2>Mass of air in a room that measures 24.0 m by 15.0 m by 4.0 m is 1728 kg.</h2>

Explanation:

Density of air = 1.20 g/L = $\frac{1.20\times 10^{-3}kg}{10^{-3}m^3}=1.2kg/m^3$

Size of room is 24.0m by 15.0 m by 4.0 m

Volume of room = 24 x 15 x 4 = 1440 m³

We know the equation

Mass = Volume x Density

Mass = 1440 x 1.2

Mass = 1728 kg

Mass of air in a room that measures 24.0 m by 15.0 m by 4.0 m is 1728 kg.

8 0

3 years ago

Stars are called blackbody radiators because they

scoundrel [369]

The answer is D) absorb most of the light that falls on them

A blackbody is an object that absorbs most (theoretically all) of the light that falls on it. Stars are called black bodies because they demonstrate something close to this property. Why do they appear white and not black? As black body radiators heat up, they slowly turn from black to grey to white, according to the human eye. Since stars are incredibly hot, they appear white to us.

8 0

3 years ago

he fan blades on a jet engine make one thousand revolutions in a time of 54.9 ms. What is the angular frequency of the blades?

Gnesinka [82]

So, the angular frequency of the blades approximately <u>36.43π rad/s</u>.

<h3>Introduction</h3>

Hi ! Here I will discuss about the angular frequency or what is also often called the angular velocity because it has the same unit dimensions. <u>Angular frequency occurs, when an object vibrates (either moving harmoniously / oscillating or moving in a circle)</u>. Angular frequency can be roughly interpreted as the magnitude of the change in angle (in units of rad) per unit time. So, based on this understanding, the angular frequency can be calculated using the equation :

$\boxed{\sf{\bold{\omega = \frac{\theta}{t}}}}$

With the following condition :

$\sf{\omega}$ = angular frequency (rad/s)
$\sf{\theta}$ = change of angle value (rad)
t = interval of the time (s)

<h3>Problem Solving</h3>

We know that :

$\sf{\theta}$ = change of angle value = 1,000 revolution = 1,000 × 2π rad = 2,000π rad/s >> Remember 1 rev = 2π rad/s.
t = interval of the time = 54.9 s.

What was asked :

$\sf{\omega}$ = angular frequency = ... rad/s

Step by step :

$\sf{\omega = \frac{\theta}{t}}$

$\sf{\omega = \frac{2,000 \pi}{54.9}}$

$\boxed{\sf{\omega \approx 36.43 \pi \: rad/s}}$

<h3>Conclusion :</h3>

So, the angular frequency of the blades approximately 36.43π rad/s.

8 0

2 years ago