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irga5000 [103]
3 years ago
7

What measurement can be used to determine the stability of the atmosphere?

Physics
1 answer:
agasfer [191]3 years ago
5 0
I would say pressure, but it might depend on what you mean by stability. Pressure will tell you how many particles and how energized those particles are, the more particles the higher the pressure, the harder it is to change or remove the atmosphere due to the larger mass of it. Also consistent pressure measurements can tell you how stable the atmosphere in a particular region is. So if stability means how consistent are the weather conditions, pressure can be a good indicator too, differences and changing pressures can cause weather catastrophes like tornados and hurricanes.
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Determine the kinetic energy of 1000-kg roller coaster car that is moving with speed of 20.0m/s
nevsk [136]
B, i got the same question
7 0
3 years ago
Calculate the electric field at one corner of a square 50 cm on a side if the other corners are occupied by 250x10-7C (charges)
SIZIF [17.4K]

The electric field at one corner of a square is 1614217 N/C.

Explanation:

The distance between x and y direction diagonals.

As per the given details the distance between diagonals is calculated as

0.5² + 0.5² = c²  =>  c = 0.707 m

Charge to the right:  In x direction

In order to find the electric charge towards x direction

we use e = kq/r² formula

As 'k' is coulomb's constant it's value is 9 x 10^{9} N m²/C²

e = (9 x 10^{9})(250 x 10^{-7}) / (0.5)²

e = 9 x 10^{5} N/C

Charge diagonal:

e = kq/r²

e = [(9 x 10^{9})(250 x 10^{-7}) / (0.707)²] cos 45

e = 225000√2 N/C

X direction sum = 1218198 N/C.

Similarly as shown in x direction the charge is same for y direction also

Charge below:  For y direction

e = kq/r²

e = (9 x 10^{9})(250 x 10^{-7}) / (0.5)²

e = 9 x 10^{5} N/C

Charge diagonal:

e = kq/r²

e = [(9 x 10^{9})(250 x 10^{-7}) / (0.5)²] sin 45

e = 159099 N/C

Y direction sum = 1059099 N/C

Resultant electric field strength:

1218198 ² + 1059099² = e²

e = 1614217 N/C [45 degrees below the horizontal]

4 0
3 years ago
A thin 1.5 mm coating of glycerine has been placed between two microscope slides of width 0.8 cm and length 3.9 cm . Find the fo
Radda [10]

The  force required to pull one of the microscope sliding at a constant speed of 0.28 m/s relative to the other is zero.

<h3>Force required to pull one end at a constant speed</h3>

The force required to pull one of the microscope sliding at a constant speed of 0.28 m/s relative to the other is determined by applying Newton's second law of motion as shown below;

F = ma

where;

  • m is mass
  • a is acceleration

At a constant speed, the acceleration of the object will be zero.

F = m x 0

F = 0

Thus, the  force required to pull one of the microscope sliding at a constant speed of 0.28 m/s relative to the other is zero.

Learn more about constant speed here: brainly.com/question/2681210

3 0
2 years ago
Determine the rate at which the electric field changes between the round plates of a capacitor, 8.0 cm in diameter, if the plate
Daniel [21]

Solution:

The relation between the potential difference and the electric field between the plates of the parallel plate capacitor is given by :

$E=\frac{V}{D}$

Differentiating on both the sides with respect to time, we get

$\frac{dE}{dt}=\frac{1}{D}\frac{dV}{dt}$

Therefore, the rate of the electric field changes between the plates of the parallel plate capacitor is given by :

$\frac{dE}{dt}=\frac{1}{D}\frac{dV}{dt}$

      $=\frac{1}{1.4 \times 10^{-3}} \times 110$

      $=7.85 \times 10^4$  V/m-s

8 0
3 years ago
Which of these does NOT result in the acceleration of an object?
Zanzabum

I guess the correct answer is the first one.

6 0
3 years ago
Read 2 more answers
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