How scientists decide if an exoplanet might be like Earth?
1 answer:
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The angular speed of the merry-go-round reduced more as the sandbag is
placed further from the axis than increasing the mass of the sandbag.
The rank from largest to smallest angular speed is presented as follows;
[m = 10 kg, r = 0.25·R]
⇩
[m = 20 kg, r = 0.25·R]
⇩
[m = 10 kg, r = 0.5·R]
⇩
[m = 10 kg, r = 0.5·R] = [m = 40 kg, r = 0.25·R]
⇩
[m = 10 kg, r = 1.0·R]
Reasons:
The given combination in the question as obtained from a similar question online are;
<em>1: m = 20 kg, r = 0.25·R</em>
<em>2: m = 10 kg, r = 1.0·R</em>
<em>3: m = 10 kg, r = 0.25·R</em>
<em>4: m = 15 kg, r = 0.75·R</em>
<em>5: m = 10 kg, r = 0.5·R</em>
<em>6: m = 40 kg, r = 0.25·R</em>
According to the principle of conservation of angular momentum, we have;
The moment of inertia of the merry-go-round, = 0.5·M·R²
Moment of inertia of the sandbag = m·r²
Therefore;
0.5·M·R²· = (0.5·M·R² + m·r²)·
Given that 0.5·M·R²· is constant, as the value of m·r² increases, the value of
decreases.
The values of m·r² for each combination are;
Combination 1: m = 20 kg, r = 0.25·R; m·r² = 1.25·R²
Combination 2: m = 10 kg, r = 1.0·R; m·r² = 10·R²
Combination 3: m = 10 kg, r = 0.25·R; m·r² = 0.625·R²
Combination 4: m = 15 kg, r = 0.75·R; m·r² = 8.4375·R²
Combination 5: m = 10 kg, r = 0.5·R; m·r² = 2.5·R²
Combination 6: m = 40 kg, r = 0.25·R; m·r² = 2.5·R²
Therefore, the rank from largest to smallest angular speed is as follows;
Combination 3 > Combination 1 > Combination 5 = Combination 6 >
Combination 2
Which gives;
[<u>m = 10 kg, r = 0.25·R</u>] > [<u>m = 20 kg, r = 0.25·R</u>] > [<u>m = 10 kg, r = 0.5·R</u>] > [<u>m = </u>
<u>10 kg, r = 0.5·R</u>] = [<u>m = 40 kg, r = 0.25·R</u>] > [<u>m = 10 kg, r = 1.0·R</u>].
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Answer:
Explanation:
Using Kepler's third law, we can relate the orbital periods of the planets and their average distances from the Sun, as follows:
Where and
are the orbital periods of Mercury and Earth respectively. We have
and
. Replacing this and solving for
Explanation:
It is given that,
The separation between two parallel wires, r = 5.6 cm = 0.056 m
Current in both the wires is 2.65 A
(a) We need to find the magnitude of the force per unit length between the wires. It can be given by :
(b) As the current is in same direction, the wires will attract each other.