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liraira [26]
2 years ago
7

The local newspaper has published the passage below. Last week, scientists released incredible images of aurora australis, known

as the northern lights, as seen from the Hubble telescope. The Hubble telescope was placed in orbit in 1990 by the space shuttle and has since delivered the sharpest images we have of our planet, our galaxy, and the universe. The images delivered last week show the beautiful display of lights in the southern hemisphere. These lights are the result of the solar activity known as solar wind. Traces of solar wind reach well beyond Earth and have been detected as far as Saturn. Which statement in the news passage above is inaccurate
Physics
2 answers:
OverLord2011 [107]2 years ago
6 0

Answer:

The statement as aurora australis known as northern lights is incorrect. As the designation of aurora australis is for the southern lights i.e. which occur in the southern hemisphere.

Explanation:

Aurora or natural lights is a phenomenon that occurs at the poles of the Earth due to interaction between the Earth's magnetic field and cosmic rays. This interaction results in the beautiful display of colors on both poles. These are named, aurora borealis or aurora australis depending on their geographical location. If they occur on the northern pole they are termed as aurora borealis while those occurring on the southern pole are named aurora australis.

defon2 years ago
5 0

Answer:

C. The aurora australis is known as the northern lights.

Explanation

The answer on edge

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A trumpet player is tuning his instrument by playing an A note simultaneously with the first-chair trumpeter, who has perfect pi
KIM [24]

To develop this problem, it is necessary to apply the concepts related to Beat

The Beat is an acoustic phenomenon that is generated by two sine waves interfering with slightly different frequencies. The beat frequency is equal to the difference in the frequencies of the two original waves:

f_{bat} = |(f_1 \pm f_B)|

Our values are given as

f_1 = 440Hz

f_B = 2.1Hz

For the particular case we have two possible frequencies:

1) f_{bat} = |(f_1 - f_B)|

f_{bat} = |(440 - 2.1)|

f_{bat} = 437.9Hz

2) f_{bat} = |(f_B + f_1)|

f_{bat} = |(440 +2.1)|

f_{bat} = 442.1Hz

Therefore the two possibles frequencies of the other players note are 437.9Hz and 442.1Hz

4 0
3 years ago
How do i calculate this?
Lesechka [4]

CAR 1

Momentum = Mass/Velocity

M = 2100/20

M = 105 m/s^2

CAR 2

Momentum = Mass/Velocity

M = 2100/30

M = 70 m/s^2

8 0
3 years ago
Suppose a net force of 7900 N is applied to slide a 950 kg crate across the floor. Compute the resulting acceleration of the cra
LiRa [457]

Explanation:

F=ma

7900=950a

a=7900/950=8.58m/s^2

4 0
2 years ago
Dry air is primarily composed of nitrogen. In a classroom demonstration, a physics instructor pours 3.6 L of liquid nitrogen int
neonofarm [45]

Answer:

The  value is  V_n  =  2.2498 \  m^3

Explanation:

From the question we are told that

   The volume of  liquid nitrogen is  V_n  =  3.6 \  L=  3.6 *10^{-3} \ m^3

   The  density of  nitrogen at gaseous form   is  \rho_n =  1.2929 \  kg/m^3  =  The dry air at sea level

   

Generally the density of nitrogen at liquid form is  

         \rho _l = 808 \  kg/m^3

And this is mathematically represented as

      \rho_l  =  \frac{m}{V_l }

=>   m  =  \rho_l  *  V_l

Now the density of  gaseous nitrogen is

       \rho_n  =  \frac{m}{V_n }

=>   m  =  \rho_n  *  V_n

Given that the mass is constant

       \rho_n  *  V_n  =   \rho_l  *  V_l

        1.2929*  V_n  =   808  *  3.6*10^{-3}

=>   V_n  =  2.2498 \  m^3

       

3 0
2 years ago
How long must a 0.70-mm-diameter aluminum wire be to have a 0.40 a current when connected to the terminals of a 1.5 v flashlight
Natalka [10]
By using Ohm's law, we can find what should be the resistance of the wire, R:
R= \frac{V}{I}= \frac{1.5 V}{0.40 A} =3.75 \Omega

Then, let's find the cross-sectional area of the wire. Its radius is half the diameter,
r=35 mm=0.35 \cdot 10^{-3} m
So the area is
A=\pi r^2 = \pi (0.35 \cdot 10^{-3} m)^2=3.85 \cdot 10^{-7} m^2

And by using the resistivity  of the Aluminum, \rho=2.65 \cdot 10^{-8} \Omega m, we can use the relationship between resistance R and resistivity:
R= \frac{\rho L}{A}
to find L, the length of the wire:
L= \frac{RA}{\rho}= \frac{(3.75 \Omega)(3,85 \cdot 10^{-7} m^2)}{2.65 \cdot 10^{-8} \Omega m}=54.48 m
4 0
3 years ago
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