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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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1. Is the relationship between velocity and centripetal force a direct, linear relationship or is it a nonlinear square relation
Svetllana [295]

Answer:

non linear square relationship

Explanation:

formula for centripetal force is given as

a = mv^2/r

here a ic centripetal acceleration , m is mass of body moving in circle of radius r and v is velocity of body . If m ,and r are constant we have

a = constant × v^2

a α v^2

hence non linear square relationship

7 0
2 years ago
Which statements about acceleration are true?
bonufazy [111]
It's definitely not B or C. There are things missing from A and D so we can't narrow it down any farther.
4 0
3 years ago
Describe the importance of SI units?​
anyanavicka [17]
SI unit is an international system of measurements that are used universally in technical and scientific research to avoid the confusion with the units. Having a standard unit system is important because it helps the entire world to understand the measurements in one set of unit system.
5 0
2 years ago
Read 2 more answers
In a nuclear physics experiment, a proton (mass 1.67×10^(−27)kg, charge +e=+1.60×10^(−19)C) is fired directly at a target nucleu
Arte-miy333 [17]

The given question is incomplete. The complete question is as follows.

In a nuclear physics experiment, a proton (mass 1.67 \times 10^(-27)kg, charge +e = +1.60 \times 10^(-19) C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed 2.50 \times 10^6 m/s. The proton comes momentarily to rest at a distance 5.31 \times 10^(-13) m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are 5.31 \times 10^{-13} m apart?

Explanation:

The given data is as follows.

Mass of proton = 1.67 \times 10^{-27} kg

Charge of proton = 1.6 \times 10^{-19} C

Speed of proton = 2.50 \times 10^{6} m/s

Distance traveled = 5.31 \times 10^{-13} m

We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.

  (K.E + P.E)_{initial} = (K.E + P.E)_{final}

 (\frac{1}{2} m_{p}v^{2}_{p}) = (\frac{kq_{p}q_{t}}{r} + 0)

where,    \frac{kq_{p}q_{t}}{r} = U = Electric potential energy

     U = (\frac{1}{2}m_{p}v^{2}_{p})

Putting the given values into the above formula as follows.

        U = (\frac{1}{2}m_{p}v^{2}_{p})

            = (\frac{1}{2} \times 1.67 \times 10^{-27} \times (2.5 \times 10^{6})^{2})

            = 5.218 \times 10^{-15} J

Therefore, we can conclude that the electric potential energy of the proton and nucleus is 5.218 \times 10^{-15} J.

4 0
3 years ago
Estimate the smallest possible period of a satellite in a circular orbit around earth. (mass and radius of earth is 5.98 x 1024
myrzilka [38]
Given: Mass of earth Me = 5.98 x 10²⁴ Kg

           Radius of earth r = 6.37 x 10⁶ m

            G = 6.67 x 10⁻¹¹ N.m²/Kg²

Required: Smallest possible period T = ?

Formula: F = ma;  F = GMeMsat/r²     Centripetal acceleration ac = V²/r

               but V = 2πr/T

equate T from all equation.

F = ma

GMeMsat/r² = Msat4π²/rT²    

GMe = 4π²r³/T²

T² = 4π²r³/GMe  

T² = 39.48(6.37 x 10⁶ m)³/6.67 x 10⁻¹¹ N.m²/Kg²)(5.98 x 10²⁴ Kg)

T² = 1.02 x 10²² m³/3.99 x 10¹⁴ m³/s²

T² = 25,563,909.77 s²

T = 5,056.08 seconds    or around   1.4 Hour

3 0
3 years ago
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