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3 years ago

During development, the cells in these organisms become specialized.

Physics

1 answer:

Y_Kistochka [10]3 years ago

7 0

Answer:

A- multicellular

Explanation:

fertilized egg I think that is the answer

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What are examples of government action? Please HELP ME!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

alexira [117]

The answer is <span>Encourage private ownership Profit motives and promote competition and monopoly regulate companies sell food and regulate pollution</span>

6 0

3 years ago

Please help!

julsineya [31]

Answer:

Answer:

3 g/cm

Explanation:

Mass of stone = 30 g

Initial volume of cylinder = 50 cm

Increased volume = 60 cm

Volume = 60-50

= 10 cm

Density of stone = Mass/Volume

= 30/10

= 3 g/ml as 1 ml = 1 cm

= 3 g/ml

Explanation:

6 0

3 years ago

Micah knows that a car had a change in velocity of 15 m/s.To determine acceleration ,Micah also needs the blank of the total tri

Fittoniya [83]

Answer: To determine acceleration ,Micah also needs the Time of the total trip in seconds.

Explanation:

Acceleration can be defined as rate of change of velocity.

$a = \frac{dv}{dt}$

for calculating acceleration, initial and final velocity are required in meter per second and the total time of the trip in seconds. Then acceleration is measured in meter per second square.

Thus, Micah knows that a car had a change in velocity of 15 m/s.To determine acceleration ,Micah also needs the <u>Time</u> of the total trip in seconds.

7 0

3 years ago

Read 2 more answers

Saturn has an orbital period of 29.46 years. In two or more complete sentences, explain how to calculate the average distance fr

vivado [14]

This question can be solved from the Kepler's law of planetary motion.

As per this law the square of time period of a planet is proportional to the cube of semi major axis.

Mathematically it can be written as $T^{2} \alpha R^{3}$

⇒ $T^{2} = KR^{3}$

Here K is the proportionality constant.

If $T_{1}$ and $T_{2}$ are the orbital periods of the planets and

$R_{1}$ and $R_{2}$ are the distance of the planets from the sun, then Kepler's law can be written as-

$\frac{T_{1} ^{2} }{T_{2} ^{2} } =\frac{R_{1} ^{3} }{R_{2} ^{2} }$

⇒ $R_{1} ^{3} =R_{2} ^{3} *\frac{T_{1} ^{2} }{T_{2} ^{2} }$

Here we are asked to calculate the the distance of Saturn from sun.It can solved by comparing it with earth.

Let the distance from sun and orbital period of Saturn is denoted as $R_{1}$ and $T_{1}$ respectively.

Let the distance from sun and orbital period of earth is denoted as $R_{2}$ and $T_{2}$ respectively.

we are given that $T_{1} =29.46 years$

we know that $R_{2} =$ 1 AU and $T_{2} =$ 1 year.

1 AU is the mean distance of earth from the sun which is equal to 150 million kilometre.

Hence distance of Saturn from sun is calculated as -

From Kepler's law as mentioned above-

$R_{1} ^{3} =R_{2} ^{3} *\frac{T_{1} ^{2} }{T_{2} ^{2} }$

= $[1 ]^{3} *\frac{[29.46]^{2} }{[1]^{2} } AU$

$=867.8916 AU^{3}$

⇒ $R_{1} =\sqrt[3]{867.8916}$

=9.5386 AU [ans]

5 0

3 years ago

a force of 30 N is exerted on cart there is no friction the acceleration of the cart is 6 m/s^2 find the mass of the cart

Greeley [361]

Here we would use the equation
$f = m \times a$
When we sub in we would would write it as
$30 = m \times 6$
you would the solve the equation for an answer of 5

8 0

3 years ago