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3 years ago

8

What are some possible examples of genetic characteristics that can be passed down to child??????

2 answers:

elena-s [515]3 years ago

8 0

They can have a close similar appearance to the parents, have close relation of child reactions.

for example, everyone born in my father's side of the family had the tendency to bump their head on something as they fall asleep up to the point when you are a toddler.

attashe74 [19]3 years ago

6 0

Curly hair, freckles, cleft chin, dimples, the way you clasp your hands,<span>from eye color to the length of their toes and shapes of their earlobes.and there hight and body shape</span>

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What volume of .2500 m cobalt iii chloride is required to react completely with 25 ml of .0315 m calcium hydroxide?

Lynna [10]

<span>f 0.0315 M calcium hydroxide?<span>Co2(SO4)3 + 3Ca(OH)2---> 2Co(OH)3+ 3CaSO4 Co Sulfate and Ca hydroxide react

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8 0

3 years ago

!! Offering 50 points !!

Evgesh-ka [11]

Answeryes they are the same

Explanation:

0.5-0.5=10 is gotta be sorry if its wrong

8 0

2 years ago

Read 2 more answers

A 0.50 kg toy is attached to the end of a 1.0 m very light string. The toy is whirled in a horizontal circular path on a frictio

xenn [34]

Answer:

The maximum speed will be 26.475 m/sec

Explanation:

We have given mass of the toy m = 0.50 kg

radius of the light string r = 1 m

Tension on the string T = 350 N

We have to find the maximum speed without breaking the string

For without breaking the string tension must be equal to the centripetal force

So $T=\frac{mv^2}{r}$

So $350=\frac{0.5\times v^2}{1}$

$v^2=700$

v = 26.475 m /sec

So the maximum speed will be 26.475 m/sec

6 0

2 years ago

g In a certain binary-star system, each star has the same mass which is 8.2 times of that of the Sun, and they revolve about the

Mademuasel [1]

To solve this problem it is necessary to apply the concepts related to the Third Law of Kepler.

Kepler's third law tells us that the period is defined as

$T^2 = \frac{4\pi^2 d^3}{2GM}$

The given data are given with respect to known constants, for example the mass of the sun is

$m_s = 1.989*10^{30}$

The radius between the earth and the sun is given by

$r = 149.6*10^9m$

From the mentioned star it is known that this is 8.2 time mass of sun and it is 6.2 times the distance between earth and the sun

Therefore:

$m = 8.2*1.989*10^{30}$

$d = 6.2*149.6*10^6$

Substituting in Kepler's third law:

$T^2 = \frac{4\pi^2 d^3}{2}$

$T^2 = \frac{4\pi^2(6.2*149.6*10^9)^3}{2(6.674*10^{-11} )(8.2*1.989*10^30 )}$

$T=\sqrt{\frac{4\pi^2(6.2*149.6*10^9)^3}{2(6.674*10^{-11} )(8.2*1.989*10^30)}}$

$T = 120290789.7s$

$T = 120290789.7s(\frac{1year}{31536000s})$

$T \approx 3.8143 years$

Therefore the period of this star is 3.8years

7 0

2 years ago

An undiscovered planet, many lightyears from Earth, has one moon in a periodic orbit. This moon takes 23.0 days on average to co

Gelneren [198K]

Answer:

See it in the pic

Explanation:

See it in the pic

7 0

2 years ago