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Sergeu [11.5K]
3 years ago
5

Consider formula A to be v = and formula B to be v2 = G. Write the letter of the appropriate formula to use in each scenario. De

termine the tangential speed of the moon given the mass of Earth and the distance from Earth to the moon. Determine the tangential speed of a satellite that takes 90 minutes to complete an orbit 150 km above Earth’s surface.
Biology
2 answers:
suter [353]3 years ago
6 0
The formulas that will be used are:

Situation one:
The moon will be experiencing gravitational force in the form of centripetal force, so we equate the two formulas.

Gravitational force = GMm /r²
Centripetal force = mv²/r

Equating,
GMm/r² = mv²/r
v² = GM/r

The first scenario will use the formula v² = GM/r


Situation 2:
The second situation will use the simple distance over time formula for velocity, where the distance will be the circumference and the time will be in seconds.

v = 2rπ/t
DedPeter [7]3 years ago
6 0

B is the first one and A is the second one

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After an exceptionally rainy spring, there is no longer a barrier between the populations, and some individuals migrate inland f
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Answer

After inland population after migration allele frequency is 0.62 or 62%

Explanation:

Given,  

Coastal striped phenotype freq. = 0.22

ss = 0.22

q_{coastal} \times q_{coastal} = 0.22

q_{2 coastal}= 0.22

Similarly, inland striped phenotype freq. = 0.43

q_{2inland} = 0.43

q_{coastal} = \sqrt{q_{2coastal}}

= \sqrt{0.22}

= 0.4690

q_{coastal}= 0.47 i.e. 47%

q_{inland} = \sqrt{0.43}

= 0.655

q_{inland} = 0.66 i.e. 66%

the migration range (m) is given as 20%

m= 0.2

allele freq. after migration = pre migration + ∆q

here,                  

∆q = change in the allele frequency

or                    

migration of allele freq. from coastal to inland

=m(q_{coastal} - q_{inland})

= 0.2 (0.47 – 0.66)  

=0.2 \times (- 0.191)

= - 0.0382    

∆q = -0.04   i.e. 4%

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