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Nataly_w [17]
3 years ago
5

Why is it important to have a control group in a experiment?

Physics
2 answers:
goldenfox [79]3 years ago
5 0
You would compare the results from the experimental group with the results of the control group to see what happens when you change the variable you want to examine. A control group is an essential part of an experiment because it allows you to eliminate and isolate these variables.
Hopefully this helped.
zysi [14]3 years ago
4 0

Answer:it allows you to eliminate and isolate variables.

Explanation:

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explain how a mutation in an individual's DNA occurs, and describe the effect it may have on the individual's traits.
zysi [14]
A mutation in the DNA ca happen during growth, when the DNA splits and becomes RNA. Mutations can also occur from other invasive cells, such as bacteria or viruses. A mutation in your DNA can also be there from when you were born. These mutations can cause variations in your appearance, such as eye color, ability to see, taste, smell and the complexion of your face. The mutations can also result in variations in your growth, causing immature or late growth. These mutations are dangerous, and can result in failure of the organisms cells , to complete and utter death of an organism.
8 0
3 years ago
Read 2 more answers
Desde una altura de 120 m se deja caer un cuerpo. Calcular a los 2,5 s i) la rapidez que lleva; ii) cuánto ha descendido; iii) c
stira [4]

Answer:

i) 24.5 m/s

ii) 30,656 m

iii) 89,344 m

Explanation:

Desde una altura de 120 m se deja caer un cuerpo. Calcule a 2.5 s i) la velocidad que toma; ii) cuánto ha disminuido; iii) cuánto queda por hacer

i) Los parámetros dados son;

Altura inicial, s = 120 m

El tiempo en caída libre = 2.5 s

De la ecuación de caída libre, tenemos;

v = u + gt

Dónde:

u = Velocidad inicial = 0 m / s

g = Aceleración debida a la gravedad = 9.81 m / s²

t = Tiempo de caída libre = 2.5 s

Por lo tanto;

v = 0 + 9.8 × 2.5 = 24.5 m / s

ii) El nivel que el cuerpo ha alcanzado en 2.5 segundos está dado por la relación

s = u · t + 1/2 · g · t²

= 0 × 2.5 + 1/2 × 9.81 × 2.5² = 30.656 m

iii) La altura restante = 120 - 30.656 = 89.344 m.

6 0
3 years ago
an Olympic runner leaps over a hurdle.if the runner initial vertical speed is 2.2m/s how much will the runners center of mass be
Snowcat [4.5K]
Supposing the runner is condensed to a point and moves upward at 2.2 m/s.
It takes a time = 2.2/g = 2.2/9.8 = 0.22 seconds to increase to max height. 
Now looking at this condition in opposite - that is the runner is at max height and drops back to earth in 0.22 s (symmetry of this kind of motion). 
From what height does any object take 0.22 s to fall to earth (supposing there is no air friction)? 
d = 1/2gt²= (0.5)(9.8)(0.22)²= 0.24 m
7 0
3 years ago
A particle initially located at the origin has an acceleration of vector a = 2.00ĵ m/s2 and an initial velocity of vector v i =
natali 33 [55]

With acceleration

\mathbf a=\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)\,\mathbf j

and initial velocity

\mathbf v(0)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i

the velocity at time <em>t</em> (b) is given by

\mathbf v(t)=\mathbf v(0)+\displaystyle\int_0^t\mathbf a\,\mathrm du

\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\displaystyle\int_0^t\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)\,\mathbf j\,\mathrm du

\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\bigg|_{u=0}^{u=t}

\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf j

We can get the position at time <em>t</em> (a) by integrating the velocity:

\mathbf x(t)=\mathbf x(0)+\displaystyle\int_0^t\mathbf v(u)\,\mathrm du

The particle starts at the origin, so \mathbf x(0)=\mathbf0.

\mathbf x(t)=\displaystyle\int_0^t\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\,\mathrm du

\mathbf x(t)=\left(\left(8.00\dfrac{\rm m}{\rm s}\right)u\,\mathbf i+\dfrac12\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf j\right)\bigg|_{u=0}^{u=t}

\mathbf x(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)t\,\mathbf i+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)t^2\,\mathbf j

Get the coordinates at <em>t</em> = 8.00 s by evaluating \mathbf x(t) at this time:

\mathbf x(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)(8.00\,\mathrm s)\,\mathbf i+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)(8.00\,\mathrm s)^2\,\mathbf j

\mathbf x(8.00\,\mathrm s)=(64.0\,\mathrm m)\,\mathbf i+(64.0\,\mathrm m)\,\mathbf j

so the particle is located at (<em>x</em>, <em>y</em>) = (64.0, 64.0).

Get the speed at <em>t</em> = 8.00 s by evaluating \mathbf v(t) at the same time:

\mathbf v(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)(8.00\,\mathrm s)\,\mathbf j

\mathbf v(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(16.0\dfrac{\rm m}{\rm s}\right)\,\mathbf j

This is the <em>velocity</em> at <em>t</em> = 8.00 s. Get the <em>speed</em> by computing the magnitude of this vector:

\|\mathbf v(8.00\,\mathrm s)\|=\sqrt{\left(8.00\dfrac{\rm m}{\rm s}\right)^2+\left(16.0\dfrac{\rm m}{\rm s}\right)^2}=8\sqrt5\dfrac{\rm m}{\rm s}\approx17.9\dfrac{\rm m}{\rm s}

5 0
3 years ago
Which Of The Following Statements Are True?
aksik [14]

Answer:

A) The north pole of a bar magnet will attract the south pole of another bar magnet.

B) Earth's geographic north pole is actually a magnetic south pole.

E) The south poles of two bar magnets will repel each other.

Explanation:

<u>According to </u><u>classical physics</u>, a magnetic field always has two associated magnetic poles (north and south), the same happens with magnets. This means that if we break a magnet in half, we will have two magnets, where each new magnet will have a new south pole, and a new north pole.

This is because <u>for classical physics, naturally, magnetic monopoles can not exist. </u>

In this context,  Earth is similar to a magnetic bar with a north pole and a south pole. This means, the axis that crosses the Earth from pole to pole is like a big magnet.  

Now, by convention, on all magnets the north pole is where the magnetic lines of force leave the magnet and the south pole is where the magnetic lines of force enter the magnet.  

Then, for the case of the Earth, the north pole of the magnet is located towards the geographic south pole and the south pole of the magnet is near the geographic north pole.  

And it is for this reason, moreover, that the magnetic field lines enter the Earth through its magnetic south pole (which is the geographic north pole).

5 0
3 years ago
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