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Dominik [7]
3 years ago
9

Why are there variations of traits within a population?

Physics
2 answers:
nydimaria [60]3 years ago
3 0

Answer:

The variation and distribution of traits in a population depend on genetic and environmental factors. Genetic variation can result from mutations caused by environmental factors or errors in DNA replication, or from chromosomes swapping sections during meiosis.

Explanation:

Hope this helps!

docker41 [41]3 years ago
3 0

Answer:

The variation and distribution of traits in a population depend on genetic and environmental factors. Genetic variation can result from mutations caused by environmental factors or errors in DNA replication, or from chromosomes swapping sections during meiosis.

Explanation:

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in the circuit diagram, what does the symbol made of two long lines and two short likes with a positive and a negative sign at e
IrinaVladis [17]
It is a battery (the long and short lines represent 2 cells)
7 0
3 years ago
A 1500 kg car traveling at 15.0 m/s to the south collides with a 4500 kg truck that is at rest at a stopligt. The car comes to a
Arlecino [84]

Answer:

<em><u>M</u></em><em><u>a</u></em><em><u>t</u></em><em><u>h</u></em><em><u>e</u></em><em><u>m</u></em><em><u>a</u></em><em><u>t</u></em><em><u>i</u></em><em><u>c</u></em><em><u>a</u></em><em><u>l</u></em><em><u>l</u></em><em><u>y</u></em><em><u>:</u></em>

That will be

<em>=</em><em> </em><em>1</em><em>5</em><em>0</em><em>0</em><em> </em><em>x</em><em> </em><em>1</em><em>5</em><em> </em><em>x</em><em> </em><em>4</em><em>5</em><em>0</em><em>0</em>

<em>=</em><em> </em><em><u>1</u></em><em><u>0</u></em><em><u>1</u></em><em><u>,</u></em><em><u>2</u></em><em><u>5</u></em><em><u>0</u></em><em><u>,</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em>

5 0
3 years ago
If a stone is dropped from a height of 400 feet, its height after t seconds is given by s = 400 − 16t2. Find its instantaneous v
soldier1979 [14.2K]

Answer:

v = -32 t

Explanation:

given,

s = 400- 16 t²

we know,

Velocity of an object is defined as the change in displacement per unit change in time.

velocity an also be return as

v = \dfrac{ds}{dt}

v = \dfrac{d}{dt}(400-16t^2)

v= 0 -2\times 16 t

v = -32 t

Hence, instantaneous velocity function given by v = -32 t

To calculate instantaneous velocity, you need to insert value of time.

ex, instantaneous velocity at t = 4 s

       v = -32 x 4 = -128 m/s.

5 0
3 years ago
A lead ball is dropped into a lake from a diving board 5.0 m above the water. After entering the water, it sinks to the bottom w
nirvana33 [79]

Answer:

|D_{depth} |=19.697m

Explanation:

To find Depth D of lake we must need to find the time taken to hit the water.So we use equation of simple motion as:

Δx=vit+(1/2)at²

x_{f}-x_{i}=v_{i}t+(1/2)at^{2}\\  -5.0m=(o)t+(1/2)(-9.8m/s^{2} )t^{2}\\ -4.9t^{2}=-5.0\\ t^{2}=5/4.9\\t=\sqrt{1.02} \\t=1.01s

As we have find the time taken now we need to find the final velocity vf from below equation as

v_{f}=v_{i}+at\\v_{f}=0+(-9.8m/s^{2} )(1.01s) \\v_{f}=-9.898m/s

So the depth of lake is given by:

first we need to find total time as

t=3.0-1.01 =1.99 s

|D_{depth} |=|vt|\\|D_{depth} |=|(-9.898m/s)(1.99s)|\\|D_{depth} |=19.697m

6 0
3 years ago
A train starts at rest in a station and accelerates at a constant 0.987 m/s2 for 182 seconds. Then the train decelerates at a co
algol [13]

Answer:

\displaystyle X_T=66.6\ km

Explanation:

<u>Accelerated Motion </u>

When a body changes its speed at a constant rate, i.e. same changes take same times, then it has a constant acceleration. The acceleration can be positive or negative. In the first case, the speed increases, and in the second time, the speed lowers until it eventually stops. The equation for the speed vf at any time t is given by

\displaystyle V_f=V_o+a\ t

where a is the acceleration, and vo is the initial speed .

The train has two different types of motion. It first starts from rest and has a constant acceleration of 0.987 m/s^2 for 182 seconds. Then it brakes with a constant acceleration of -0.321 m/s^2 until it comes to a stop. We need to find the total distance traveled.

The equation for the distance is

\displaystyle X=V_o\ t+\frac{a\ t^2}{2}

Our data is

\displaystyle V_o=0,a=0.987m/s^2,\ t=182\ sec

Let's compute the first distance X1

\displaystyle X_1=0+\frac{0.987\times 182^2}{2}

\displaystyle X_1=16,346.7\ m

Now, we find the speed at the end of the first period of time

\displaystyle V_{f1}=0+0.987\times 182

\displaystyle V_{f1}=179.6\ m/s

That is the speed the train is at the moment it starts to brake. We need to compute the time needed to stop the train, that is, to make vf=0

\displaystyle V_o=179.6,a=-0.321\ m/s^2\ ,V_f=0

\displaystyle t=\frac{v_f-v_o}{a}=\frac{0-179.6}{-0.321}

\displaystyle t=559.5\ sec

Computing the second distance

\displaystyle X_2=179.6\times559.5\ \frac{-0.321\times 559.5^2}{2}

\displaystyle X_2=50,243.2\ m

The total distance is

\displaystyle X_t=x_1+x_2=16,346.7+50,243.2

\displaystyle X_t=66,589.9\ m

\displaystyle \boxed{X_T=66.6\ km}

8 0
3 years ago
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