The two types of vitamins are fat and water soluble?

A classroom is about 3 meters high, 20 meters wide and 30 meters long. If the density of air is 1.29 kg/m3, what is the mass of

Deffense [45]

Answer:

the mass of the air in the classroom = 2322 kg

Explanation:

given:

A classroom is about 3 meters high, 20 meters wide and 30 meters long.

If the density of air is 1.29 kg/m3

find:

what is the mass of the air in the classroom?

density = mass / volume

where mass (m) = 1.29 kg/m³

volume = 3m x 20m x 30m = 1800 m³

plugin values into the formula

1.29 kg/m³ = <u> mass </u>

1800 m³

mass = 1.29 kg/m³ ( 1800 m³ )

mass = 2322 kg

therefore,

the mass of the air in the classroom = 2322 kg

8 0

3 years ago

Read 2 more answers

Many Amtrak trains can travel at a top speed of 42.0 m/s. Assuming a train maintains that speed for several hours, how many kilo

777dan777 [17]

Answer:

605 km

Explanation:

Hello

the same units of measure should be used, then

Step 1

convert 42 m/s ⇒ km/h

1 km =1000 m

1 h = 36000 sec

$42 \frac{m}{s}*\frac{1\ km}{1000\ m}=0.042\ \frac{km}{s}\\ 0.042\ \frac{km}{s}\\$

$0.042\ \frac{km}{s}*\frac{3600\ s}{1\ h} =151.2 \frac{km}{h}\\ \\Velocity =151.2\ \frac{km}{h}$

Step 2

find kilometers traveled after 4 hours

$V=\frac{s}{t}\\ \\$

V,velocity

s, distance traveled

t. time

now, isolating s

$V=\frac{s}{t} \\s=V * t\\$

and replacing

$s=V * t\\s=151.2\frac{km}{h}*4 hours\\ s=604.8 km\\$

S=604.8 Km

Have a great day

4 0

3 years ago

What is the maximum speed when the conditions are mass =450 kg, initial height= 30 m, and the roller coaster is initially at res

Zarrin [17]

Answer:

B. 24.2 m/s

Explanation:

Given;

mass of the roller coaster, m = 450 kg

height of the roller coaster, h = 30 m

The maximum potential energy of the roller coaster due to its height is given by;

$P.E_{max} = mgh\\\\PE_{max} = 450 *9.8*30\\\\PE_{max} = 132,300 \ J$

$P.E_{max} = K.E_{max} \ (law \ of \ conservation\ of \ energy)$

$K.E_{max} = \frac{1}{2}mv_{max}^2\\\\ v_{max}^2 = \frac{2K.E_{max}}{m}\\\\ v_{max}^2 = \frac{2*132300}{450}\\\\ v_{max}^2 =588\\\\v_{max} = \sqrt{588}\\\\ v_{max} = 24.2 \ m/s$

Therefore, the maximum speed of the roller coaster is 24.2 m/s.

3 0

2 years ago

Look at the Grand Canyon cross section. What is the 4th event to take place in the Grand Canyon's geologic history?

mart [117]

Answer:

bass limestone

Explanation:

if I'm right bass limestone is the fourth one because 3 are under it

8 0

3 years ago

A cannon, positioned on a hill, shoots a cannonball horizontally at 23 m/s. The cannonball hits the stone wall 1.96 m below the

irina [24]

Answer: 14. 49 m

Explanation:

We can solve this problem with the following equations:

$x=V_{o} cos \theta t$ (1)

$y-y_{o}=V_{o} sin \theta t-\frac{1}{2}gt^{2}$ (2)

Where:

$x$ is the horizontal distance between the cannon and the ball

$V_{o}=23 m/s$ is the cannonball initial velocity

$\theta=0\°$ since the cannonball was shoot horizontally

$t$ is the time

$y=0$ is the final height of the cannonball

$y_{o}=1.96 m$ is the initial height of the cannonball

$g=9.8 m/s^{2}$ is the acceleration due gravity

Isolating $t$ from (2):

$t=\sqrt{-\frac{2(y-y_{o})}{g}}$ (3)

$t=\sqrt{-\frac{2(0 m-1.96 m)}{9.8 m/s^{2}}}$ (4)

$t=0.63 s$ (5)

Substituting (5) in (1):

$x=(23 m/s) cos(0\°) 0.63 s$ (6)

Finally:

$x=14.49 m$

5 0

3 years ago