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valentina_108 [34]
3 years ago
13

15. What is one way in which scientists study dark

Physics
1 answer:
Kay [80]3 years ago
8 0

Answer:

It's d

Explanation:

Scientists study dark matter by looking at how it affects visible matter.

You might be interested in
The rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3 × 10-11 e-250/T and 2
Vlada [557]

Answer:

Calculate the ratio of the rates of ozone destruction by these catalysts at 20 km, given that at this altitude the average concentration of OH is about 100 times that of Cl and that the temperature is about -50 °C

Knowing

Rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3x10^{-11} e^{-255/T}  and 2x10^{-12} e^{-940/T}  

T = -50 °C = 223 K

The reaction rate will be given by [Cl] [O3] 3x10^{-11} e^{-255/223} = 9.78^{-12} [Cl] [O3]  

Than, the reaction rate of OH with O3 is

Rate = [OH] [O3] 2x10^{-12} e^{-940/223} = 2.95^{-14} [OH] [O3]

Considering these 2 rates we can realize the ratio of the reaction with Cl to the reaction with OH is 330 * [Cl] / [OH]

Than, the concentration of OH is approximately 100 times of Cl, and the result will be that the reaction with Cl is 3.3 times faster than the  reaction with OH

Calculate the rate constant for ozone destruction by chlorine under conditions in the Antarctic ozone hole, when the temperature is about -80 °C and the concentration of atomic chlorine increases by a factor of one hundred to about 4 × 105 molecules cm-3

Knowing

Rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3x10^{-11} e^{-255/T}  and 2x10^{-12} e^{-940/T}  

T = -80 °C = 193 K

The reaction rate will be given by [Cl] [O3] 3x10^{-11} e^{-255/193} = 8.21^{-12} [Cl] [O3]  

Than, the reaction rate of OH with O3 is

Rate = [OH] [O3] 2x10^{-12} e^{-940/193} = 1.53^{-14} [OH] [O3]

Considering these 2 rates we can realize the ratio of the reaction with Cl to the reaction with OH is 535 * [Cl] / [OH]

Than, considering the concentration of Cl increases by a factor of 100 to about 4 × 10^{5} molecules cm^{-3}, the result will be that the reaction with OH will be 535 + (100 to about 4 × 10^{5} molecules cm^{-3}) times faster than the  reaction with Cl

Explanation:

4 0
3 years ago
Function of a simple pendulum​
Misha Larkins [42]

Answer:

A pendulum is a mechanical machine that creates a repeating, oscillating motion. A pendulum of fixed length and mass (neglecting loss mechanisms like friction and assuming only small angles of oscillation) has a single, constant frequency. This can be useful for a great many things.

From a historical point of view, pendulums became important for time measurement. Simply counting the oscillations of the pendulum, or attaching the pendulum to a clockwork can help you track time. Making the pendulum in such a way that it holds its shape and dimensions (in changing temperature etc.) and using mechanisms that counteract damping due to friction led to the creation of some of the first very accurate all-weather clocks.

Pendulums were/are also important for musicians, where mechanical metronomes are used to provide a notion of rhythm by clicking at a set frequency.

The Foucault pendulum demonstrated that the Earth is, indeed, spinning around its axis. It is a pendulum that is free to swing in any planar angle. The initial swing impacts an angular momentum in a given angle to the pendulum. Due to the conservation of angular momentum, even though the Earth is spinning underneath the pendulum during the day-night cycle, the pendulum will keep its original plane of oscillation. For us, observers on Earth, it will appear that the plane of oscillation of the pendulum slowly revolves during the day.

Apart from that, in physics a pendulum is one of the most, if not the most important physical system. The reason is this - a mathematical pendulum, when swung under small angles, can be reasonably well approximated by a harmonic oscillator. A harmonic oscillator is a physical system with a returning force present that scales linearly with the displacement. Or, in other words, it is a physical system that exhibits a parabolic potential energy.

A physical system will always try to minimize its potential energy (you can accept this as a definition, or think about it and arrive at the same conclusion). So, in the low-energy world around us, nearly everything is very close to the local minimum of the potential energy. Given any shape of the potential energy ‘landscape’, close to the minima we can use Taylor expansion to approximate the real potential energy by a sum of polynomial functions or powers of the displacement. The 0th power of anything is a constant and due to the free choice of zero point energy it doesn’t affect the physical evolution of the system. The 1st power term is, near the minimum, zero from definition. Imagine a marble in a bowl. It doesn’t matter if the bowl is on the ground or on the table, or even on top of a building (0th term of the Taylor expansion is irrelevant). The 1st order term corresponds to a slanted plane. The bottom of the bowl is symmetric, though. If you could find a slanted plane at the bottom of the bowl that would approximate the shape of the bowl well, then simply moving in the direction of the slanted plane down would lead you even deeper, which would mean that the true bottom of the bowl is in that direction, which is a contradiction since we started at the bottom of the bowl already. In other words, in the vicinity of the minimum we can set the linear, 1st order term to be equal to zero. The next term in the expansion is the 2nd order or harmonic term, a quadratic polynomial. This is the harmonic potential. Every higher term will be smaller than this quadratic term, since we are very close to the minimum and thus the displacement is a small number and taking increasingly higher powers of a small number leads to an even smaller number.

This means that most of the physical phenomena around us can be, reasonable well, described by using the same approach as is needed to describe a pendulum! And if this is not enough, we simply need to look at the next term in the expansion of the potential of a pendulum and use that! That’s why each and every physics students solves dozens of variations of pendulums, oscillators, oscillating circuits, vibrating strings, quantum harmonic oscillators, etc.; and why most of undergraduate physics revolves in one way or another around pendulums.

Explanation:

7 0
3 years ago
The force that opposes motion to moving parts is _____
Colt1911 [192]
The force that opposes motion to moving parts is F<span>riction</span><span>

Hope this helped!
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6 0
4 years ago
A beam of monochromatic light with a wavelength of 400 nm in air travels into water. what is the wavelength of the light in wate
slava [35]
The refractive index of water is n=1.33. This means that the speed of the light in the water is:
v= \frac{c}{n}= \frac{3 \cdot 10^8 m/s}{1.33 }=2.26 \cdot 10^8 m/s

The relationship between frequency f and wavelength \lambda of a wave is given by:
\lambda= \frac{v}{f}
where v is the speed of the wave in the medium. The frequency of the light does not change when it moves from one medium to the other one, so we can compute the ratio between the wavelength of the light in water \lambda_w to that in air \lambda as
\frac{\lambda_w}{\lambda}= \frac{ \frac{v}{f} }{ \frac{c}{f} } = \frac{v}{c}
where v is the speed of light in water and c is the speed of light in air. Re-arranging this formula and by using \lambda=400 nm, we find
\lambda_w = \lambda \frac{v}{c}=(400 nm) \frac{2.26 \cdot 10^8 m/s}{3 \cdot 10^8 m/s}=301 nm
which is the wavelength of light in water.
5 0
3 years ago
When 1,250^3/4 is written in simplest radical form, which value remains under the radical?
GaryK [48]

Answer:

125\sqrt[4]{8}

Explanation:

A number of the form

a^{\frac{m}{n}}

can be re-written in the radical form as follows:

\sqrt[n]{a^m}

In this problem, we have:

a = 1,250

m = 3

n = 4

So, if we apply the formula, we get

1,250^{\frac{3}{4}}=\sqrt[4]{(1,250)^3}

Then, we can rewrite 1250 as

1250 = 2\cdot 5^4

So we can rewrite the expression as

=\sqrt[4]{(2\cdot 5^4)^3}=5^3 \sqrt[4]{2^3}=125\sqrt[4]{8}

7 0
3 years ago
Read 2 more answers
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