All categories

1 year ago

In which atmosphere layer does 80 percent of the gas in the atmosphere reside?

Physics

2 answers:

VladimirAG [237]1 year ago

7 0

Answer: TheTroposphere contains 80% of the total gas in the atmosphere

tatiyna1 year ago

5 0

Answer:

The correct answer is Troposphere.

Explanation:

The troposphere is the closest atmospheric layer of the earth's surface and 80% of the total mass of the atmosphere and 99% of the water vapor are located here.

In this layer, the higher the altitude, the temperature and the steam will be lower. This steam is responsible for regulating air temperatures thanks to the absorption of energy from the sun. <u>Here is the air we breathe and all the climatic processes of our planet. </u>

80% of the mass of the atmosphere is located in this layer because in it the air is much denser than in the other layers.

You might be interested in

A uniform meterstick of mass 0.20 kg is pivoted at the 40 cm mark. where should one hang a mass of 0.50 kg to balance the stick?

Tcecarenko [31]

The weight of the meterstick is:
$W=mg=0.20 kg \cdot 9.81 m/s^2 = 1.97 N$
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
$d_1 = 0.50 m - 0.40 m=0.10 m$
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
$M_w = W d_1 = (1.97 N)(0.10 m)=0.20 Nm$

To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
$(mg) d_2 = 0.20 Nm$
from which we find the value of d2:
$d_2 = \frac{0.20 Nm}{mg}= \frac{0.20 Nm}{(0.5 kg)(9.81 m/s^2)}=0.04 m$

So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.

4 0

2 years ago

When a test charge q0 = 2 nC is placed at the origin, it experiences a force of 8 times 10-4 N in the positive y direction. What

ser-zykov [4K]

Answer:

Electric field, $E=4\times 10^5\ N/C$

Explanation:

It is given that,

Magnitude of charge, $q_o=2\ nC=2\times 10^{-9}\ C$

Force experienced, $F=8\times 10^{-4}\ N$

We need to find the electric field at the origin. It is given by :

$F=q_o\times E$

$E=\dfrac{F}{q_o}$

$E=\dfrac{8\times 10^{-4}}{2\times 10^{-9}}$

$E=4\times 10^5\ N/C$

So, the electric field at the origin is $4\times 10^5\ N/C$ . Hence, this is the required solution.

3 0

1 year ago

Whennes

rodikova [14]

Answer:

See the explanation below.

Explanation:

12) When an object is falling, how does the objects velocity change? what formula do you use?

The speed of a falling object is increased by a value of 9.81 meters per second per second. That is if we throw any body regardless of mass from a considerable height, its speed in the first second will be 9.81[ m/ s] , in the next second will be equal to 19.62 [m/s] in the next will be equal to 29.43 [m/ s].

The formula is:

$v=v_{0}+g*t$

where:

vo = initial velocity = 0

g = gravity = 9.81[m/s^2]

t = time [s]

13)

what is a falling speed at 6s, 9s, 112s?

v = 0 + (9.81*6) = 58.86[m/s]

v = 0 + (9.81*9) = 88.29 [m/s]

v = 0 + (9*112) = 1098.72 [m/s]

14)

If an object is falling at 65 [m/s]. How long has it been falling ? what is the formula that you use?

The formula is the same:

$v=v_{o}+g*t$

65 = 0 + 9.81*t

t = 65/9.81

t = 6.62[s]

15)

What formula is used to determine the distance an object is falling ?

$y = y_{o}+v_{o}*t + 0.5*9.81*t^{2}$

where:

y = distance [m]

yo = initial distance, in most of the cases and depending of the reference point it will be eqaul to zero

vo = initial velocity, if it is free fall, then = 0

t = time [s]

g = gravity = 9.81[m/s^2]

This equation will be reduce to:

$y = 0.5*g*t^{2}$

16)

using the times given in problem 13. Determine the distance fallen for each.

y = 0.5*9,81*(6)^2 = 176.58 [m]

y = 0.5*9,81*(9)^2 = 397.3 [m]

y = 0.5*9,81*(112)^2 = 61528.3 [m]

17)

If an object has fallen a distance of 87.3 [m]. How long was it falling?

87.3 = 0.5*9.81*t^2

$t=\sqrt{\frac{87.3}{0.5*9.81} }\\ t=4.21[s]$

4 0

2 years ago

Most workers in nanotechnology are actively monitored for excess static charge buildup. the human body acts like an insulator as

irga5000 [103]

Sometimes arithmetic problems can be solved much more easily using the dimensional analysis approach. You focus on the units of the given information. Then, you manipulate them applying the laws of algebra where like units cancel, in order to end up with the unit of the unknown.

Given:
-50 nc/step
31 steps
Unknown: charge

Thus,
Charge = -50 nc/step * 31 steps =<em> -1550 nc</em>

7 0

1 year ago

What is the frequency of radiation whose wavelength is 11.5 a0 ?

irakobra [83]

Answer:

The frequency of radiation is $2.61 \times 10^{17} s^{-1}$