PHYSICS HIGH SCHOOL

A demographic tool used to predict a population's future growth isChoose one answer.
a. a hairline growth curve.
b. demographic transition.
c. age structures.
d. maximum sustained growth.
e. population momentum

Answers

Answer 1
Answer: C. Age structure. Hope I helped :) 

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One component of a magnetic field has a magnitude of 0.0404 T and points along the x axis while the other component has a magnitude of 0.0739 T and points along the y axis A particle carrying a charge of 2.80 10 5 C is moving along the z axis at a speed of 4.46 103 m s a Find the magnitude of the net magnetic force that acts on the particle b Determine the angle that the net force makes with respect to the x axis
A moving car has momentum. If it moves twice as fast what is it's momentum
Magnetic reversals have helped to supporta.the theory of Pangaea.c.sea-floor spreading.b.the age of the Earth.d.the theory of Gondwana.help please
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In the water cycle what state of matter is rain

Will amplitude change if the pitch gets louder

Answers


The wording of the question is fundamentally defective,
and the question is therefore unanswerable. (But I'll
gladly accept your points anyway.)

"Amplitude" is the origin of the subjective impression of loudness.
"Pitch" is not.

There is no such concept as the 'loudness' of a pitch.

What is 54 km = _____ cm
Im working on physical science and I need help ):

Answers

1 km = 100,000 cm
54 km = 54 x 100,000 cm = 5,400,000 km
It is 5400000 cm long

The total energy of a 0.050 kg object travelling at 0.70 c is:A) 2.10 × 107 J

B) 3.06 × 107 J

C) 2.46 × 1015 J

D) 6.30 × 1015 J

E) 8.82 × 1015 J

Answers

Answer:

1.1025×10^15Joules

No correct option

Explanation:

The type of energy possessed by the object is kinetic energy. Kinetic energy is the energy due to virtue of an object motion.

KE = 1/2MV² where;

M is the mass of the car = 0.05kg

V is the velocity of the car

Since the car is traveling at 0.7c (c is the speed of light)

speed = 0.7c { 0.7(3×10^8)}

Speed = 2.1×10^8

Substituting this values in the formula given we have;

KE = 1/2×0.05×(2.1×10^8)²

KE = 1.1025×10^15Joules

No correct option.
What is the unit c denotes here

You need to get to class, 200 meters away, and you can only walk in the hallways at about 1.5 m/s. (if you run any faster, you’ll be caught for running). How much time will it take to get to your class?

Answers

200/1.5 = 133.33 seconds. 

= approximately 2 mins and 13 secs. 
Time = distance / velocity

= 200/1.5

= 133.3333 s.

Thus, the time will be around 133 seconds or 2 minutes 13 seconds.

An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is = 2.5 x 10⁻⁷ C/m², and the plates are separated by a distance of 1.7 x 10⁻² m. How fast is the electron moving just before it reaches the positive plate?

Answers

The speed of the electron before reaching the positive plate is 1.30 * 10^(7)\ m / s

Explanation:

As per Gauss law of electro statistics, the electric field generated by a capacitor is directly proportional to the surface charge density of the plate and inversely proportional to the dielectric constant. In simple words, the electric field is proportional to the surface charge density.  So,  

    \text {Electric field}=(\sigma)/(\varepsilon_(0))

And then from the second law of motion, F=m * acceleration

So acceleration exerted by the electrons will be directly proportional to the force exerted on them and inversely proportional to the mass of the electron.

        Acceleration =(F)/(m)

Since force is also calculated as product of charge with electric field in electrostatic force,

       \text {Acceleration}=(q E)/(m)=(q \sigma)/(m \varepsilon_(0))

So, the charge of electronq=1.6 * 10^(-19)\ \mathrm{C}, \sigma=\text { Charge per unit area }=2.5 * 10^(-7)\ \mathrm{C} / \mathrm{m}^(2)

m is the mass of electron which is equal to 9.11 * 10^(-31)\ \mathrm{kg}

\varepsilon_(0)=8.85 * 10^(-12)\ \mathrm{Nm}^(2) \mathrm{C}^(-2)

Then,

    \text { Acceleration }=(1.6 * 2.5 * 10^(-19) * 10^(-7))/(9.11 * 8.85 * 10^(-31) * 10^(-12))=(4 * 10^(-19-7))/(80.62 * 10^(-31-12))

   \text { Acceleration }=0.0496 * 10^(-19-7+31+12)=0.0496 * 10^(17)\ \mathrm{m} / \mathrm{s}^(2)

So the acceleration of the electron in the capacitor will be 4.96 * 10^(15) m / s^(2)

Then, the velocity can be observed from the third equation of motion.

    v^(2)=u^(2)+2 a s

As u = 0 and s is the distance of separation between two plates.

   \begin{array}{c}v^(2)=0+\left(2 * 4.96 * 10^(15) * 1.7 * 10^(-2)\right) \nv^(2)=16.864 * 10^(15-2)=16.864 * 10^(13)=1.684 * 10^(14)\end{array}

Thus, v=\sqrt{\left(1.68 * 10^(14)\right)}=1.30 * 10^(7)\ m/s

So, the speed of the electron before reaching the positive plate is 1.30 * 10^(7) \mathrm{m} / \mathrm{s}.

What is the new weight of a 10kg object on a planet that has twice the earths mass. (assume that the radius of the planet is the same as earths)

Answers


Weight = (mass) x (acceleration of gravity) .

Acceleration of gravity on a planet's surface is proportional to

                         (planet's mass) / (planet's radius)² .

On Earth's surface, acceleration of gravity is 9.8 m/s² .

Weight of 10 kg mass = (10 kg) x (9.8 m/s²) = 98 newtons.

On the surface of a planet with twice the Earth's mass
all stuffed into the same radius, the acceleration of gravity
would be 19.6 m/s² ... double what it is on Earth.

Weight of 10 kg would be (10 kg) x (19.6 m/s²) = 196 newtons ...
double its Earth weight.

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