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3 years ago

Which vocabulary term is survival of the fittest

1 answer:

6 0

Natural selection also means survival of the fittest. This is where organisms who are better adapted to their environment survive and produce more offspring compared to the ones who aren’t as adapted to their environment.

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Mass (kg) 4.0

densk [106]

Answer:

25 m/s

Explanation:

First of all, we can find the acceleration the object by using Newton's second law of motion:

$F=ma$

where

F = 20.0 N is the net force applied on the object

m = 4.0 kg is the mass of the object

a is its acceleration

Solving for a, we find

$a=\frac{F}{m}=\frac{20}{4}=5.0 m/s^2$

Now we know that the motion of the object is a uniformly accelerated motion, so we can find its final velocity by using the following suvat equation:

$v=u+at$

where

v is the final velocity

u = 0 is the initial velocity

$a=5.0 m/s^2$ is the acceleration

t = 5 s is the time

By substituting,

$v=0+(5.0)(5)=25 m/s$

8 0

2 years ago

The type of bond formed by shared electrons is covalent, ionic, or metallic

loris [4]

I believe the answer in Covalent Bond.

7 0

3 years ago

Read 2 more answers

if you dig a hole through the earth, from one hemisphere to the other, and drop a tennis ball into it, would it get stuck at the

tia_tia [17]

Answer:

Depending on which hemisphere it is, like western to eastern, It would most likely get stuck at the center. You would also have to put more things into thought like acceleration, velocity, and speed.

BUT since the question asked "would it pop out the other side?", I'm assuming it's talking about northern to southern hemisphere. so in that case it would pop out the other side since gravity makes things go downwards.

7 0

2 years ago

Read 2 more answers

Consider a block on a spring oscillating on a frictionless surface. The amplitude of the oscillation is 11 cm, and the speed of

IRISSAK [1]

Answer:

The angular frequency of the block is ω = 5.64 rad/s

Explanation:

The speed of the block v = rω where r = amplitude of the oscillation and ω = angular frequency of the oscillation.

Now ω = v/r since v = speed of the block = 62 cm/s and r = the amplitude of the oscillation = 11 cm.

The angular frequency of the oscillation ω is

ω = v/r

ω = 62 cm/s ÷ 11 cm

ω = 5.64 rad/s

So, the angular frequency of the block is ω = 5.64 rad/s

6 0

3 years ago

Three positive charges A, B, and C, and a negative charge D are placed in a line as shown in the diagram. All four charges are o

polet [3.4K]

Answer:

a. charge C experiences the greatest net force, and charge B receives the smallest net force

b. ratio=9

Explanation:

<u>Electrostatic Force</u>

Two point-charges $q_1$ and $q_2$ separated a distance d will exert a force on each other of a magnitude given by the Coulomb's formula

$\displaystyle F=\frac{k\ q_1\ q_2}{r^2}$

Where k is the proportional constant of value

$k=9*10^9\ N.m^2/c^2$

The diagram provided in the question shows four identical charges (let's assume their value is Q) separated by identical distance (of value d). The force between the charges next to others is

$\displaystyle F_1=\frac{k\ Q\ Q}{d^2}$

$\displaystyle F_1=\frac{k\ Q^2}{d^2}$

The force between charges separated 2d is

$\displaystyle F_2=\frac{k\ Q^2}{(2d)^2}$

$\displaystyle F_2=\frac{k\ Q^2}{4d^2}$

And the force between the charges A and D is

$\displaystyle F_3=\frac{k\ Q^2}{(3d)^2}$

$\displaystyle F_3=\frac{k\ Q^2}{9d^2}$

Now, let's analyze each charge and the force applied to them by the others

Let's recall equally signed charges repel each other and differently signed charges attrach each other

Charge A. It receives force to the left from B and C and to the right from D

$\displaystyle F_A=-F_1-F_2+F_3=-\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{9d^2}$

$\displaystyle F_A=\frac{k\ Q^2}{d^2}(-1-\frac{1}{4}+\frac{1}{9})$

$\displaystyle F_A=-\frac{41}{36}F_1$

Charge B. It receives force to the right from A and D and to the left from C

$\displaystyle F_B=F_1-F_1+F_2=\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{d^2}+\frac{k\ Q^2}{4d^2}$

$\displaystyle F_B=\frac{1}{4}F_1$

Charge C. It receives forces to the right from all charges.

$\displaystyle F_C=F_2+F_1+F_1=\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{d^2}+\frac{k\ Q^2}{d^2}$

$\displaystyle F_C=\frac{9}{4}F_1$

Charge D. It receives forces to the left from all charges

$\displaystyle F_D=-F_3-F_2-F_1=-\frac{k\ Q^2}{9d^2}-\frac{k\ Q^2}{4d^2}-\frac{k\ Q^2}{d^2}$

$\displaystyle F_D=-\frac{49}{36}F_1$

Comparing the magnitudes of each force is just a matter of computing the fractions

$\displaystyle \frac{41}{36}=1.13,\ \frac{1}{4}=0.25,\ \frac{9}{4}=2.25,\ \frac{49}{36}=1.36$

We can see the charge C experiences the greatest net force, and charge B receives the smallest net force

The ratio of the greatest to the smallest net force is

$\displaystyle \frac{\frac{9}{4}}{\frac{1}{4}}=9$

The greatest force is 9 times the smallest net force

7 0

3 years ago