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GrogVix [38]
3 years ago
6

Which term refers to a variable that a scientist adjusts during an experiment?

Physics
1 answer:
azamat3 years ago
5 0

Answer:

it’s c

Explanation:

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The big bang produced an imprint of leftover heat called _____. hydrogen cosmic heat CMB radiation redshift
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I believe the answer you are looking for is CMB Radiation

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D) it explodes as a super nova

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Which phrase best describes a semidiurnal tide pattern?
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A 32.5 g cube of aluminum initially at 45.8 °C is submerged into 105.3 g of water at 15.4 °C. What is the final temperature of b
lbvjy [14]

Answer:

T = 17.26 ^oC

Explanation:

At thermal equilibrium we have heat given by aluminium must be equal to the heat absorbed by the water

so we will have

Q_1 = Q_2

m_1s_1\Delta T_1 = m_2s_2\Delta T_2

so we will have

32.5(900)(45.8 - T) = 105.3(4186)(T - 15.4)

so we have

(45.8 - T) = 15.1(T - 15.4)

so we have

16.1 T = 277.87

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3 0
3 years ago
A dwarf planet discovered out beyond the orbit of Pluto is known to have an orbital period of 619.36 years. What is its average
Maksim231197 [3]

Answer: 72.66 AU=1.089(10)^{10} km

Explanation:

Let's begin by explaining that according to Kepler’s Third Law of Planetary motion “The square of the orbital period T of a planet is proportional to the cube of the semi-major axis a of its orbit”:

T^{2}\propto a^{3} (1)  

Now, if T is measured in years (Earth years), and a is measured in astronomical units (equivalent to the distance between the Sun and the Earth: 1AU=1.5(10)^{8}km), equation (1) becomes:  

T^{2}=a^{3} (2)  

So, knowing T=619.36 years and isolating a from (2) we have:  

a=\sqrt[3]{T^{2}} (3)  

a=\sqrt[3]{(619.36 years)^{2}} (4)  

Finally:

a=72.66 AU T his is the distance between the dwarf planet and the Sun in astronomical units

Converting this to kilometers, we have:

a=72.66 AU \frac{1.5(10)^{8}km}{1 AU}=1.089(10)^{10} km

4 0
3 years ago
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