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harkovskaia [24]
2 years ago
13

What are the layers of the Earth's atmosphere?

Physics
2 answers:
Mandarinka [93]2 years ago
4 0
<span>Layers of the Earth's Atmosphere: Troposphere, Stratosphere, Mesosphere, Thermosphere, and Exosphere - Windows to the Universe.</span>
borishaifa [10]2 years ago
3 0
<span> <span>Layers of the Earth's Atmosphere: Troposphere, Stratosphere, Mesosphere, Thermosphere, and Exosphere - Windows to the Universe.
</span> </span> Comments
<span>0</span>
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A certain frictionless simple pendulum having a length l and mass m swings with period t. If both l and m are doubled, what is t
iVinArrow [24]

If l and m both are doubled then the period becomes √2*T

what is a simple pendulum?

It is the one which can be considered to be a point mass suspended from a string or rod of negligible mass.

A pendulum is a weight suspended from a pivot so that it can swing freely.

Here,

A certain frictionless simple pendulum having a length l and mass m

mass of pendulum = m

length of the pendulum = l

The period of simple pendulum is:

T = 2\pi \sqrt{\frac{l}{g} }

Where k is the constant.

Now the length and mass are doubled,

m' = 2m

l' = 2l

T' = 2\pi \sqrt{\frac{2l}{g} }

T' = \sqrt{2}* 2\pi \sqrt{\frac{l}{g} }

T' = \sqrt{2} * T

Hence,

If l and m both are doubled then the period becomes √2*T

Learn more about Simple Harmonic Motion here:

<u>brainly.com/question/17315536</u>

#SPJ4

8 0
1 year ago
State two factors that affect the rate of diffusion of a substance
Ronch [10]

Explanation:

<em>Two</em><em> </em><em>factors</em><em> </em><em>that</em><em> </em><em>affect</em><em> </em><em>the</em><em> </em><em>rater</em><em> </em><em>of</em><em> </em><em>diffusion</em><em> </em><em>of</em><em> </em><em>a</em><em> </em><em>substance</em><em> </em><em>are</em><em>:</em><em> </em>

  • <em>Diffusion</em><em> </em><em>of</em><em> </em><em>substance</em><em> </em><em>plays</em><em> </em><em>an</em><em> </em><em>important</em><em> </em><em>role</em><em> </em><em>on</em><em> </em><em>cellular</em><em> </em><em>transport</em><em> </em><em>in</em><em> </em><em>plants</em><em>.</em><em> </em>
  • <em>Diffusion</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>passive</em><em> </em><em>movement</em><em> </em><em>of</em><em> </em><em>substance</em><em> </em><em>from</em><em> </em><em>a</em><em> </em><em>region</em><em> </em><em>of</em><em> </em><em>higher</em><em> </em><em>concentration</em><em> </em><em>to</em><em> </em><em>a</em><em> </em><em>region</em><em> </em><em>of</em><em> </em><em>lower</em><em> </em><em>concentration</em><em>. </em>
4 0
3 years ago
A tank of volume 0.25 ft is designed to contain 50 standard cubic feet of air. The temperature is 80° F.Calculate the pressure i
slavikrds [6]

Answer:

The inside Pressure of the tank is 4499.12 lb/ft^{2}

Solution:

As per the question:

Volume of tank, V = 0.25 ft^{3}

The capacity of tank, V' = 50ft^{3}

Temperature, T' = 80^{\circ}F = 299.8 K

Temperature, T = 59^{\circ}F = 288.2 K

Now, from the eqn:

PV = nRT                      (1)

Volume of the gas in the container is constant.

V = V'

Similarly,

P'V' = n'RT'                       (2)

Also,

The amount of gas is double of the first case in the cylinder then:

n' = 2n

\]frac{n'}{n} = 2

where

n and n' are the no. of moles

Now, from eqn (1) and (2):

\frac{PV}{P'V'} = \frac{nRT}{n'RT'}

P' = 2P\frac{T'}{T} = 2\times 2116\times \frac{299.8}{288.2} = 4499.12 lb/ft^{2}

7 0
3 years ago
A 125-kg astronaut (including space suit) acquires a speed of 2.50 m/s by pushing off with her legs from a 1900-kg space capsule
ryzh [129]

(a) 0.165 m/s

The total initial momentum of the astronaut+capsule system is zero (assuming they are both at rest, if we use the reference frame of the capsule):

p_i = 0

The final total momentum is instead:

p_f = m_a v_a + m_c v_c

where

m_a = 125 kg is the mass of the astronaut

v_a = 2.50 m/s is the velocity of the astronaut

m_c = 1900 kg is the mass of the capsule

v_c is the velocity of the capsule

Since the total momentum must be conserved, we have

p_i = p_f = 0

so

m_a v_a + m_c v_c=0

Solving the equation for v_c, we find

v_c = - \frac{m_a v_a}{m_c}=-\frac{(125 kg)(2.50 m/s)}{1900 kg}=-0.165 m/s

(negative direction means opposite to the astronaut)

So, the change in speed of the capsule is 0.165 m/s.

(b) 520.8 N

We can calculate the average force exerted by the capsule on the man by using the impulse theorem, which states that the product between the average force and the time of the collision is equal to the change in momentum of the astronaut:

F \Delta t = \Delta p

The change in momentum of the astronaut is

\Delta p= m\Delta v = (125 kg)(2.50 m/s)=312.5 kg m/s

And the duration of the push is

\Delta t = 0.600 s

So re-arranging the equation we find the average force exerted by the capsule on the astronaut:

F=\frac{\Delta p}{\Delta t}=\frac{312.5 kg m/s}{0.600 s}=520.8 N

And according to Newton's third law, the astronaut exerts an equal and opposite force on the capsule.

(c) 25.9 J, 390.6 J

The kinetic energy of an object is given by:

K=\frac{1}{2}mv^2

where

m is the mass

v is the speed

For the astronaut, m = 125 kg and v = 2.50 m/s, so its kinetic energy is

K=\frac{1}{2}(125 kg)(2.50 m/s)^2=390.6 J

For the capsule, m = 1900 kg and v = 0.165 m/s, so its kinetic energy is

K=\frac{1}{2}(1900 kg)(0.165 m/s)^2=25.9 J

3 0
3 years ago
The labeled images each represent the wave patterns found in the electromagnetic wave spectrum. which of these images are correc
Phantasy [73]
Energy E of EM radiation is given by the equation E=hf, where h is Planck's constant and f is frequency. It means energy E and frequency f are proportional so as we increase the frequency, energy also increases. Also, the relationship between the wavelength and frequency is c=λ*f where λ is the wavelength and f is frequency and c is the speed of light. This tells us the wavelength and frequency are inversely proportional. So as we increase the frequency the wavelength is getting smaller. So as we go from left to right the frequency increases, energy also increases and the wavelength is decreasing. Or, on the left side we should have low frequency, low radiant energy, and long wavelength. On the right side we should have high frequency, high radiant energy and low wavelength. That is the third graph. 
5 0
3 years ago
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