Where does the sun rise at 0 degrees latitude during the winter and summer solstices?

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

A ball is thrown into the air with a vertical velocity of 50 m/s and a horizontal

daser333 [38]

$\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}$

Actually Welcome to the Concept of the Projectile Motion.

Since, here given that, vertical velocity= 50m/s

we know that u*sin(theta) = vertical velocity

so the time taken to reach the maximum height or the time of Ascent is equal to

T = Usin(theta) ÷ g, here g = 9.8 m/s^2

so we get as,

T = 50/9.8

T = 5.10 seconds

thus the time taken to reach max height is 5.10 seconds.

5 0

3 years ago

A train traveling at 25 m/s is blowing its whistle at 440 Hz as it crosses a level crossing. You are waiting at the crossing and

ohaa [14]

Answer:

b) 472HZ, 408HZ

Explanation:

To find the frequencies perceived when the bus approaches and the train departs, you use the Doppler's effect formula for both cases:

$f_o=f\frac{v_s+v_o}{v_s-v}\\\\f_o=f'\frac{v_s-v_o}{v_s+v}\\\\$

fo: frequency of the source = 440Hz

vs: speed of sound = 343m/s

vo: speed of the observer = 0m/s (at rest)

v: sped of the train

f: frequency perceived when the train leaves us.

f': frequency when the train is getTing closer.

Thus, by doing f and f' the subjects of the formulas and replacing the values of v, vo, vs and fo you obtain:

$f=f_o\frac{v_s-v}{v_s+v_o}=(440Hz)\frac{340m/s-25m/s}{340m/s}=408Hz\\\\f'=f_o\frac{v_s+v}{v_s-v_o}=(440Hz)\frac{340m/s+25m/s}{340m/s}=472Hz$

hence, the frequencies for before and after tha train has past are

b) 472HZ, 408HZ

6 0

3 years ago

The refractive index of glass is 1.52

Ksenya-84 [330]

Explanation:

..upper side is glass ..

3 0

3 years ago

You are driving at the speed of 27.7 m/s (61.9764 mph) when suddenly the car in front of you (previously traveling at the same s

densk [106]

Here when car in front of us applied brakes then it is slowing down due to frictional force on it

So here we can say that friction force on the car front of our car is given as

$F_f = \mu m g$

So the acceleration of car due to friction is given as

$F_{net} = - \mu mg$

$a = \frac{F_{net}}{m}$

$a = -\mu g$

now it is given that

$\mu = 0.868$

$g = 9.81 m/s^2$

so here we have

$a = -0.868 * 9.81$

$a = -8.52 m/s^2$

so the car will accelerate due to brakes by a = - 8.52 m/s^2

4 0

4 years ago