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4 years ago

Which layer of the atmosphere is directly above the troposhpere? A.troposhpere B.stratosphere C.mesosphere D.exoshpere

Physics

2 answers:

nordsb [41]4 years ago

5 0

It is is B. Stratosphere

vivado [14]4 years ago

4 0

Its B. i hope i helped!!!

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Coherent light of wavelength 525 nm passes through two thin slits that are 4.15Ã—10^(âˆ’2) mm apart and then falls on a screen 7

IRINA_888 [86]

A) 4.7 cm

The formula for the angular spread of the nth-maximum from the central bright fringe for a diffraction from two slits is

$sin \theta=\frac{n \lambda}{d}$

where

n is the order of the maximum

$\lambda$ is the wavelength

$a$ is the distance between the slits

In this problem,

n = 5

$\lambda=525 nm =5.25\cdot 10^{-7} m$

$a=4.15\cdot 10^{-2} mm=4.15\cdot 10^{-5} m$

So we find

$\theta=sin^{-1} (\frac{(5)(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=3.62^{\circ}$

And given the distance of the screen from the slits,

$D=75.0 cm = 0.75 m$

The distance of the 5th bright fringe from the central bright fringe will be given by

$y=D tan \theta = (0.75 m)tan 3.62^{\circ}=0.047 m = 4.7 cm$

B) 8.1 cm

The formula to find the nth-minimum (dark fringe) in a diffraction pattern from double slit is a bit differente from the previous one:

$sin \theta=\frac{(n+\frac{1}{2}) \lambda}{d}$

To find the angle corresponding to the 8th dark fringe, we substitute n=8:

$\theta=sin^{-1} (\frac{(8+\frac{1}{2})(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=6.17^{\circ}$

And the distance of the 8th dark fringe from the central bright fringe will be given by

$y=D tan \theta = (0.75 m)tan 6.17^{\circ}=0.081 m = 8.1 cm$

5 0

3 years ago

You and Mae are in free-float frames. Mae has just passed you, traveling at a speed of 60 km/hr. You throw a ball toward her at

Zina [86]

The ball will be stationary to her

4 0

3 years ago

One ounce of a well-known breakfast cereal contains 104 Calories (1 food Calorie = 4186 J). If 2.5% of this energy could be conv

GalinKa [24]

Answer:

482.4 kg

Explanation:

104 calorie = 435344 J

The 2.5% of this energy equals to

435344 * 0.0025 = 10883.6 J

If this energy is converted to work done on lifting a barbell by a distance of 2.3. By the formula of Work W = force * distance then the force he exerts on the barbell is

$F = \frac{W}{D} = \frac{10883.6
}{2.3} = 4732 N$

Assuming gravitational constant g = 9.81 m/s2 then the mass of the barbell is

m = 4732 / 9.81 = 482.4 kg

5 0

3 years ago

A car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of 1.85 m/s2

Tasya [4]

Answer:

Coefficient of friction = 0.836

Explanation:

If v be the speed after one quarter of the circular path

v² = 2as = 2 x 1.85 x 2πr/4 ; v²/r = 1.85 x 3.14 = 5.8

tangential acceleration = 5.8 m/s²

radial acceleration = v² /r = 5.8

total acceleration = √2 x 5.8

m x√2 x 5.8 = m x g xμ

μ = √2 x 5.8 / 9.8 = 0.836

7 0

3 years ago

In multicellular organisms, which choice ranks the level of organization from simplest to increasingly more complex?

Andreyy89

Answer:

<em>The answer is C.</em>

Explanation:

Cells are the smallest organisms and makes up all living thing or being. The tissues are made of cells. The organs are made of tissues. The organ systems are made of organs. Therefore the order is C. cells, tissues, organs, and organ systems.

3 0

3 years ago