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sergey [27]
2 years ago
8

Mars’ axial tilt is 25. 19°. Earth’s axial tilt is 23. 5°. What does this most likely indicate about mars?.

Physics
1 answer:
sweet-ann [11.9K]2 years ago
4 0

In comparison to Earth, Mars will have more extreme weather/climatic conditions

<h3>what is Axial tilt?</h3>

Axial tilt is the angular measure between the axis of rotation and the axis of revolution. it affects the extents of varying climatic conditions in a direct proportional relationship. This means that the more the axial tilt the more the difference in extremes of the climatic condition.

Therefore comparing the two axial tilts one can arrive at. Mars having more axial tilt will have more extreme seasonal conditions than Earth.

Read more on axial tilt here:

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Question 1<br> 2 pts<br> Explain what causes a solution to be a strong acid.
lubasha [3.4K]

Answer:

Cuanto más fuerte es el ácido, más rápido se disocia para generar H +start superscript, plus, end superscript. Por ejemplo, el ácido clorhídrico (HCl) se disocia completamente en iones hidrógeno y cloruro cuando se mezcla con agua, por lo que se considera un ácido fuerte.

5 0
3 years ago
Which carbon reservoirs did not receive any carbon?
Allisa [31]
Need a picture of the problem
3 0
3 years ago
A 2 000-kg car is slowed down uniformly from 20.0 m/s to 5.00 m/s in 4.00 s. (a) What average force acted on the car during that
slava [35]

Answer:

the answer is c

Explanation:

3 0
3 years ago
If a sinusoidal electromagnetic wave with intensity 18 W/m2 has an electric field of amplitude E, then a 36 W/m2 wave of the sam
Neporo4naja [7]

Answer:

The  correct option is D

Explanation:

From the question we are told that

  The intensity of the first  electromagnetic wave is  I =  18 \  W/m^2

  The amplitude of the electric field is  E_{max}_1 =A

   The intensity of the second electromagnetic wave is  I =  36 \  W/m^2

Generally the an electromagnetic wave intensity is mathematically represented as

       I  =  \frac{1}{2} *  \epsilon_o  * c  * E_{max}^2

Looking at this equation we see that

     I \ \ \alpha  \ \ E^2_{max}

=>  \frac{I_1}{I_2}  =  [ \frac{ E_{max}_1}{ E_{max}_2} ] ^2

=>   E_{max}_2 = \sqrt{\frac{x}{y} }  *  E_{max}_1

=>  E_{max}_2 = \sqrt{\frac{36}{18} }  * E        

=>  E_{max}_2 = \sqrt{2 }  E        

5 0
3 years ago
A uniformly charged ring of radius 10.0 cm has a total charge of 50.0 μC Find the electric field on the axis of the ring at 30.0
Grace [21]

Answer: 4.27 *10^6 N/C

Explanation: In order to calculate the electric field along the axis of charged ring we have to use the following expression:

E=k*x/(a^2+x^2)^3/2    where a is the ring radius and x the distance to the point measured from the center of the ring.

Replacing the data we have:

E= (9* 10^9* 0.3* 50 * 10^-6)/(0.1^2+0.3^2)^3/2

then

E=4.27 * 10^6 N/C

8 0
3 years ago
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