The sun is the mother star of the solar system, which only emits light to half of the planet, while the other part is always dark.
The sun emits light towards the earth, which dominates all life on earth. The movements of the Moon around the Earth and of the Earth around the Sun are complex. Movements of rotation around their own axes are superimposed on movements of orbital translation. The Earth and the Moon rotate around their own axes: This is rotation.
Answer:
0.661 s, 5.29 m
Explanation:
In the y direction:
Δy = 2.14 m
v₀ = 0 m/s
a = 9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
(2.14 m) = (0 m/s) t + ½ (9.8 m/s²) t²
t = 0.661 s
In the x direction:
v₀ = 8 m/s
a = 0 m/s²
t = 0.661 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (8 m/s) (0.661 s) + ½ (0 m/s²) (0.661 s)²
Δx = 5.29 m
Round as needed.
No.
Since repeated measurements are taken and the average and 95% confidence interval are calculated, the possibility of the lack of agreement being a random error has been minimized or even eliminated.
<h3>What is a random error?</h3>
Random error is defined as the deviation of the total error from its mean value due to chance.
Random errors can result from the instrument not being precise or from mistakes by the researcher.
Random errors can be minimized by taking multiple readings and averaging the results.
Since repeated measurements are taken and the average and 95% confidence interval are calculated, the possibility of the lack of agreement being a ransom error has been minimized.
Learn more about random errors at: brainly.com/question/22041172
<span>The angular momentum of a particle in orbit is
l = m v r
Assuming that no torques act and that angular momentum is conserved then if we compare two epochs "1" and "2"
m_1 v_1 r_1 = m_2 v_2 r_2
Assuming that the mass did not change, conservation of angular momentum demands that
v_1 r_1 = v_2 r_2
or
v1 = v_2 (r_2/r_1)
Setting r_1 = 40,000 AU and v_2 = 5 km/s and r_2 = 39 AU (appropriate for Pluto's orbit) we have
v_2 = 5 km/s (39 AU /40,000 AU) = 4.875E-3 km/s
Therefore, </span> the orbital speed of this material when it was 40,000 AU from the sun is <span>4.875E-3 km/s.
I hope my answer has come to your help. Thank you for posting your question here in Brainly.
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