Describe two 'beneficial' aspects from the last 10,000 years of humanity, as well as two 'dis-advantageous' aspects. Agriculture should be your guiding point here.

Answers

Answer 1

Answer:

Two beneficial aspects of the last 10,000 years of mankind have been the development of writing and the development of civilized communities.

Thus, writing has allowed the recording of the different historical events that have taken place during the development of humanity, setting precedents for subsequent generations, allowing documenting the progress that humanity has been achieving. On the other hand, the development of civilized communities around urban systems has allowed for greater economic and social development on the part of human groups, which through cooperation have been acquiring an increasingly higher quality of life.

On the other hand, two negative aspects of these years have been the world overpopulation and the pollution of the planet.

Thus, the amount of the world's population has been getting closer and closer to the limit of our planet's carrying capacity, which could lead humanity to a point where resources are definitely scarce, causing famines and various catastrophes. On the other hand, the pollution arising from the development of industrial societies has notably undermined the health of individuals and has damaged their standard of living, while causing irreparable damage to the natural environment.


Related Questions

A vacuum gauge connected to a tank reads 30.0 kPa. If the local atmospheric pressure is 13.5 psi, what is the absolute pressure in units of psi, with 3 sig figs

Answers

Answer:

[tex]P_a=17.85psi[/tex]

Explanation:

From the question we are told that:

Tank Pressure [tex]P_t=30.0kpa[/tex]

Atmospheric Pressure [tex]P_a=13.5 psi[/tex]

Where

 [tex]1kpa=0.148psi[/tex]

Therefore

 [tex]30kpa=4.35psi[/tex]

Generally the equation for Absolute pressure [tex]P_a[/tex] is mathematically given by

 [tex]P_a=13.5+4.35[/tex]

 [tex]P_a=17.85psi[/tex]

Forces applied in the opposite direction are 

Added

Subtracted

Multiplied

Divided

Answers

Answer:

its number 2 one but i am not sure hope its right

You want to produce a magnetic field of magnitude 5.50 x 10¹ T at a distance of 0.0 6 m from a long, straight wire's center. (a) What current is required to produce this field? (b) With the current found in part (a), how strong is the magnetic field 8.00 cm from the wire's center?

Answers

Answer:

(a) I = 1650000 A

(b) 4.125 T

Explanation:

Magnetic field, B = 5.5 T

distance, r = 0.06 m

(a) Let the current is I.

The magnetic field due to a long wire is given by

[tex]B =\frac{\mu o}{4\pi }\frac{2 I}{r}\\5.5= 10^{-7}\times \frac{2\times I}{0.06}\\I =1650000 A[/tex]

(b) Let the magnetic field is B' at distance r = 0.08 m.

[tex]B =\frac{\mu o}{4\pi }\frac{2 I}{r}\\B = 10^{-7}\times \frac{2\times 1650000}{0.08}\\B'= 4.125 T[/tex]

a body thrown vertically upwards from grounf with inital vel 40m/s then time taken by it to reach max hieght is?

Answers

Answer:

t = 4.08 s

Explanation:

if the body is thrown upward, it has negative gravity. Knowing through the International System that the earth's gravity is 9.8 m/s²

Data:

Vo = 40 m/sg = -9.8 m/s²t = ?

Use formula:

[tex]\boxed{\bold{t=\frac{-(V_{0})}{g}}}[/tex]

Replace and solve:

[tex]\boxed{\bold{t=\frac{-(40\frac{m}{s})}{-9.8\frac{m}{s^{2}}}}}[/tex][tex]\boxed{\boxed{\bold{t=4.08\ s}}}[/tex]

Time taken by it to reach max height is 4.08 seconds.

Greetings.

A proton is accelerated from rest through a potential difference V0 and gains a speed v0. If it were accelerated instead through a potential difference of 3V0, what speed would it gain? Group of answer choices

Answers

Answer:

[tex]v_{0,new} = v0\sqrt{}2[/tex]

Explanation:

Initial work done on the proton is given by, [tex]\DeltaW0 = q V_o[/tex]

we know that, [tex]\DeltaW = \DeltaK.E[/tex]

[tex]qV0 = (1/2) m v_0^2[/tex]

[tex]v_0 = \sqrt{}2 q V_0 / m[/tex]                                                        { eq.1 }

If it were accelerated instead through a potential difference of 2V0, then it would gain a speed will be given as :

using the above formula, we have

[tex]v_{0,new} = \sqrt{}2 q (2V0) / m[/tex]    

[tex]v_{0,new} = \sqrt{}4 q V0 / m[/tex]    

[tex]v_{0,new} = v0\sqrt{}2[/tex]

A 1000 kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 16 s, then the motor stops. The rocket altitude 20 s after launch is 6600 m. You can ignore any effects of air resistance.

Required:
a. What was the rocket's acceleration during the first 16s?
b. What is the rocket's speed as it passes through acloud 5100 m above the ground?

Answers

Answer:

a)   a = 34.375 m / s²,  b)    v_f = 550 m / s

Explanation:

This problem is the launch of projectiles, they tell us to ignore the effect of the friction force.

a) Let's start with the final part of the movement, which is carried out from t= 16 s with constant speed

             v_f = [tex]\frac{x-x_1}{t}[/tex]

we substitute the values

             v_f = [tex]\frac{ 6600 -x_1}{4}[/tex]  

The initial part of the movement is carried out with acceleration

             v_f = v₀ + a t

             x₁ = x₀ + v₀ t + ½ a t²

the rocket starts from rest v₀ = 0 with an initial height x₀ = 0

             x₁ = ½ a t²

             v_f = a t

we substitute the values

              x₁ = 1/2  a 16²

              x₁ = 128 a

              v_f = 16 a

let's write our system of equations  

               v_f = [tex]\frac{6600 - x_1}{4}[/tex]

               x₁ = 128 a

               v_f = 16 a

we substitute in the first equation  

               16 a = [tex]\frac{6600 -128 a}{4}[/tex]

               16 4 a = 6600 - 128 a

                a (64 + 128) = 6600

                a = 6600/192

                 a = 34.375 m / s²

b) let's find the time to reach this height

                x = ½ to t²

                t² = 2y / a

                t² = 2 5100 / 34.375

                t² = 296.72

                t = 17.2 s

We can see that for this time the acceleration is zero, so the rocket is in the constant velocity part

               v_f = 16 a

               v_f = 16 34.375

               v_f = 550 m / s

g as measured from the earth, a spacecraft is moving at speed .80c toward a second spacecraft moving at speed .60c back toward the first spacecraft. What is the speed of the first spacecraft as viewed from the second spacecraft

Answers

Answer:

the speed of the first spacecraft as viewed from the second spacecraft is 0.95c

Explanation:

Given that;

speed of the first spacecraft from earth v[tex]_a[/tex] = 0.80c

speed of the second spacecraft from earth v[tex]_b[/tex] = -0.60 c

Using the formula for relative motion in relativistic mechanics

u' = ( v[tex]_a[/tex] - v[tex]_b[/tex] ) / ( 1 - (v[tex]_b[/tex]v[tex]_a[/tex] / c²) )

we substitute

u' = ( 0.80c - ( -0.60c)  ) / ( 1 - ( ( 0.80c × -0.60c) / c² ) )

u' = ( 0.80c + 0.60c ) /  ( 1 - ( -0.48c² / c² ) )

u' = 1.4c /  ( 1 - ( -0.48 ) )

u' = 1.4c /  ( 1 + 0.48 )

u' = 1.4c / 1.48

u' = 0.9459c ≈ 0.95c  { two decimal places }

Therefore, the speed of the first spacecraft as viewed from the second spacecraft is 0.95c

as a mercury atom absorbs a photon of energy as electron in the atom changes from energy level B to energy level E. calculate the frequency of the absorb photon.

Answers

Answer:

2.00x 10 14th Hz

Explanation:

Answer:

2.99 x 10^14 Hz

Explanation:

E photon= hf (you have to solve for f)

f= E photon/h

f= 1.98 x 10^-19 J / 6.63 x 10^-34 J x s

f=2.99 x 10^14 Hz

An electron travels 1.49 m in 7.4 µs (microsecWhat is its speed if 1 inch = 0.0254 m? Answer in units of in/min.

Answers

Explanation:

Write what you know

Speed = Distance / Time

micro- = 10^-6

write your conversions as fractions

1 in / 0.0254 m

1 min / 60 sec

First convert time to regular seconds

7.4 x 10^-6 seconds

Use Velocity

1.49m / (7.4 x 10^-6) s

We've written our conversions in fractions because units cancel out just like numbers

[tex] \frac{1.49m}{7.4 \times {10}^{ - 6} } \times \frac{1in}{0.0254m} \times \frac{60sec}{1min} [/tex]

Multiply all the fractions accross and youll have your answer

30.
the horizontal. The force needed to push the body up the plane is
A body of mass 20kg is pushed up a smooth plane inclined at an angle of 30° to
b. 200N c. 100N
d. 20N
a. ION

Answers

Answer:

b. 200N c. 100N

Explanation:

30.

the horizontal. The force needed to push the body up the plane is

Your friend has been given a laser for her birthday. Unfortunately, she did not receive a manual with it and so she doesn't know the wavelength that it emits. You help her by performing a double-slit experiment, with slits separated by 0.36 mm. You find that the two m n = 2 bright fringes are 5.5 mm apart on a screen 1.6 m from the slits.
a. What is the wavelength the light emits?
b. What is the distance between the two n = 1 dark fringes?

Answers

Answer:

a) the wavelength that the light emits is 6.1875 × 10⁻⁷ m

b) the distance between the two n = 1 dark fringes is 5.5 × 10⁻³ m

Explanation:

Given the data in the question;

separation  between two slits  d = 0.36 mm = 0.00036 m

Separation between two adjacent fringes β = 5.5 mm = 0.0055 m

Distance of screen from slits D = 1.6 m

n = 2

a) the wavelength the light emits;

Using the formula;

β = (nD/d)λ

To find wavelength, we make λ the subject of formula;

βd = nDλ

λ = βd / nD

so we substitute

λ = ( 0.0055 m × 0.00036 m ) / ( 2 × 1.6 m )

λ = 0.00000198 / 3.2

λ = 6.1875 × 10⁻⁷ m

Therefore, the wavelength that the light emits is 6.1875 × 10⁻⁷ m

b) the distance between the two n = 1 dark fringes;

To find the distance between the two n = 1 dark fringes, we use the following formula;

y[tex]_m[/tex] = 2nλD / d

given that n = 1, we substitute

y[tex]_m[/tex] = ( 2 × 1 × ( 6.1875 × 10⁻⁷ m ) × 1.6 m ) / 0.00036 m

y[tex]_m[/tex] = 0.00000198 / 0.00036

y[tex]_m[/tex] = 0.0055 m

y[tex]_m[/tex] = 5.5 × 10⁻³ m

Therefore, the distance between the two n = 1 dark fringes is 5.5 × 10⁻³ m

Two long, straight wires are fixed parallel to one another a distance do apart. The wires carry equal constant currents 1, in the same direction. The attractive magnetic force per unit length between them if f = F/L. What is the force per unit length between the wires if their separation is 2d, and each carries current 2I0?
A. f/4
B. f/2
C. 3f/2
D.) 2f

Answers

Answer:

Option D

Explanation:

From the question we are told that:

The attractive magnetic force per unit length as

 [tex]f = F/L[/tex]

Separation Distance [tex]x=2d[/tex]

Generally the equation for  Magnetic force between two current carrying wire is mathematically given by

[tex]\frac{F}{\triangle l}=\frac{\mu_0I_1I_2}{\mu \pi x}[/tex]

[tex]\frac{F}{\triangle l }=\frac{I_1I_2}{ x}[/tex]

Where

[tex]x=2r[/tex]

And

[tex]I_1=I_2=>2I[/tex]

Then

[tex]\frac{F}{\triangle l}=>\frac{2*2}{2}*f[/tex]

[tex]\frac{F}{\triangle l}=>2f[/tex]

Therefore s the force per unit length between the wires if their separation is 2d

[tex]\frac{F}{\triangle l}=>2f[/tex]

Option D

16. Olympic ice skaters are able to spin at about 5 rev/s.
(a) What is their angular velocity in radians per second?
(b) What is the centripetal acceleration of the skater's nose it
it is 0.120 m from the axis of rotation?​

Answers

Answer:

a)   w = 31.4 rad / s,  b)  a = 118.4 m / s²

Explanation:

a) let's reduce to the SI system

   w = 5 rev / s (2pi rad / 1 rev)

   w = 31.4 rad / s

b) the expression for the centripetal acceleration is

      a = v² / r

linear and angular variables are related

      v = w r

    we substitute

     a = w² r

     a = 31.4² 0.120

     a = 118.4 m / s²

comparison between copper properties and aluminium properties​

Answers

Hopes this helps:

Answer: Aluminum has 61 percent of the conductivity of copper, but has only 30 percent of the weight of copper. That means that a bare wire of aluminum weights half as much as a bare wire of copper that has the same electrical resistance. Aluminum is generally more inexpensive when compared to copper conductors.

Suppose that the position of a particle is given by s=f(t)=5t3+6t+9. (a) Find the velocity at time t.

Answers

This question is incomplete, the complete question is;

Suppose that the position of a particle is given by s=f(t)=5t³ + 6 t+ 9.

(a) Find the velocity at time t.

(b) Find the velocity at time t=3 seconds

Answer:

a) the velocity at time t is ( 15t² + 6 ) m/s

b) Velocity at time t=3 seconds is 141 m/s

Explanation:

Give the data in the question;

position of a particle is given by;

s = f(t) = 5t³ + 6t + 9

Velocity at t;

we differentiate with respect to t

so

V(t) = f'(t) = d/dt ( 5t³ + 6t + 9 )

V(t) = f(t) = 5(3t²) + 6(1) + 0 )

V(t) = f(t) = ( 15t²+6 ) m/s

Therefore, the velocity at time t is 15t²+6 m/s

b) Velocity at t = 3 seconds

V(t) = f(t) = ( 15t²+6 ) m/s

we substitute

V(3) = ( 15(3)² + 6 ) m/s

V(3) = ( (15 × 9) + 6 ) m/s

V(3) = ( 135 + 6 ) m/s

V(3) = 141 m/s

Therefore, Velocity at time t=3 is 141 m/s

A laser of wavelength 720 nm illuminates a double slit where the separation between the slits is 0.22 mm. Fringes are seen on a screen 0.85 m away from the slits. How far apart are the second and third bright fringes

Answers

Answer:

The appropriate solution is "2.78 mm".

Explanation:

Given:

[tex]\lambda = 720 \ nm[/tex]

or,

  [tex]= 720\times 10^{-9} \ m[/tex]

[tex]D=0.85 \ m[/tex]

[tex]d = 0.22 \ mm[/tex]

or,

  [tex]=0.22 \times 10^{-3} \ m[/tex]

As we know,

Fringe width is:

⇒ [tex]\beta=\frac{\lambda D}{d}[/tex]

hence,

Separation between second and third bright fringes will be:

⇒ [tex]\theta=\beta=\frac{\lambda D}{d}[/tex]

       [tex]=\frac{720\times 10^{-9}\times 0.85}{0.22\times 10^{-3}}[/tex]

       [tex]=2.78\times 10^{-3} \ m[/tex]

or,

       [tex]=2.78 \ mm[/tex]

Question 11 of 22
A horse of mass 180 kg gallops at a speed of 8 m/s. What is the momentum
of the horse?

Answers
1440
22.5
845
1955

Answers

Momentum = (mass) x (speed)

If you work the problem in the same units as the given data, then you get the momentum in units of kilogram-meters per second, and your horse has 1,440 of them.

Answer:

A

Explanation:

1440 kg*m/s

List five instruments of mechanical fluid ​

Answers

The Barometer: The barometer is a device meant for measuring the local atmospheric pressure. ...
Piezometer or Pressure Tube: ...
Manometers: ...
The Bourdon Gauge: ...
The Diaphragm Pressure Gauge: ...
Micro Manometer (U-Tube with Enlarged Ends):

[tex]\sf{The~ different~ types~ of~ measuring~ instruments~ are:-}[/tex]

Calipers.Micrometer.Laser Measure.Ruler.Compass.

Please helppppppp!!!!!!!!!!!!!!

Answers

Answer:

circuit breaker

Explanation:

A circuit breaker is a device used for electrical safety. It consists of a switch designed to protect an electrical circuit from damage that may result from heating due to overload in the circuit.

Its basic function is to interrupt current flow through its switch that consists of metal stripe which bends when it gets hot.

Fuse has similar action with circuit breaker, the only difference is that fuse can only be used once because it melts when it gets hot.

Therefore, the correct answer is "circuit breaker"

A 0.50-m long solenoid consists of 500 turns of copper wire wound with a 4.0 cm radius. When the current in the solenoid is 22 A, the magnetic field at a point 1.0 cm from the central axis of the solenoid is

Answers

Answer: The magnetic field at a point 1.0 cm from the central axis of the solenoid is 0.0276 T.

Explanation:

Given: Length = 0.50 m

No. of turns = 500

Current = 22 A

Formula used to calculate magnetic field is as follows.

[tex]B = \mu_{o}(\frac{N}{L})I[/tex]

where,

B = magnetic field

[tex]\mu_{o}[/tex] = permeability constant = [tex]4\pi \times 10^{-7} Tm/A[/tex]

N = no. of turns

L = length

I = current

Substitute the values into above formula as follows.

[tex]B = \mu_{o}(\frac{N}{L})I\\= 4 \pi \times 10^{-7} Tm/A \times (\frac{500}{0.5 m}) \times 22\\= 0.0276 T[/tex]

Thus, we can conclude that magnetic field at a point 1.0 cm from the central axis of the solenoid is 0.0276 T.

What is measured by the change in velocity of a moving object?

Answers

Answer:

acceleration is measured

Acceleration is the change of velocity (speed) or direction.

an object moves clockwise around a circle centered at the origin with radius m beginning at the point ​(0,​). a. find a position function r that describes the motion of the object moves with a constant​ speed, completing 1 lap every s. b. find a position function r that describes the motion if it occurs with speed .

Answers

Answer:

Answer to An object moves clockwise around a circle centered at the origin with radius 6 m beginning at ... 6 M Beginning At The Point (0,6) B. Find A Position Function R That Describes The Motion If It Occurs With Speed E T A. R(t)= S The Motion Of The Object Moves With A Constant Speed, Completing 1 Lap Every 12 S.

Explanation:

How much energy must be added to a 1-kg piece of granite with a specific
heat of 600 J/(kg°C) to increase its temperature from 20° C to 100° C?

A. 48,000 J
B. 4,800 J
C. 1,200,000 J
D. 60,000 J

Answers

Answer: 48,000 J

Explanation: i just did it

PLEASE HELPPP MEEE :((​

Answers

power = work / time --> time = work / power = 3600 J / 275 watts = 13.1 seconds.

Make me brainliest plz

Physics part 2

These the other questions 14 - 17

Answers

Answer:

bvihobonlnohovicjfufufufucvkvkvvjcufufydyfuvi

The loudness of a sound is the wave's _______

Answers

Amplitude:)

Hope this help!

Answer:

amplitude

Explanation:

The loudness of a musical sound is a measure of the sound wave's ?

is amplitude explanation:- The loudness of a sound depends upon the amplitude.Loudness of a sound depends on the amplitude of the vibration producing that sound. Greater is the amplitude of vibration, louder is the sound produced by it. if you find this answer helpful please rate positive thank you so much.

while hunting in a cave a bat emits sounds wave of frequency 39 kilo hartz were moving towards a wall with a constant velocity of 8,32 meters per second take the speed of sound as 340 meters per second calculate frequency​

Answers

Complete question:

while hunting in a cave a bat emits sounds wave of frequency 39 kilo hartz were moving towards a wall with a constant velocity of 8.32 meters per second take the speed of sound as 340 meters per second. calculate the frequency​ reflected off the wall to the bat?

Answer:

The frequency reflected by the stationary wall to the bat is 41 kHz

Explanation:

Given;

frequency emitted by the bat, = 39 kHz

velocity of the bat, [tex]v_b[/tex] = 8.32 m/s

speed of sound in air, v = 340 m/s

The apparent frequency of sound striking the wall is calculated as;

[tex]f' = f(\frac{v}{v- v_b} )\\\\f' = 39,000(\frac{340}{340 -8.32} )\\\\f' = 39978.29 \ Hz[/tex]

The frequency reflected by the stationary wall to the bat is calculated as;

[tex]f_s = f'(\frac{v + v_b}{v} )\\\\f_s = 39978.29(\frac{340 + 8.32}{340} )\\\\f_s = 40,956.56 \ Hz[/tex]

[tex]f_s\approx 41 \ kHz[/tex]

Hey guys....
What is the advantage of a capacitor as it stores charge? ​

Answers

First thing capacitor do not store charge, capacitor actually store an imbalance of charge.They are good at delivering ghe stored imbalance of charge.They have extremely low internal resistanceThey are safe to use

A person carries a plank of wood 1.6 m long with one hand pushing down on it at one end with a force F1 and the other hand holding it up at 43 cm from the end of the plank with force F2. If the plank has a mass of 13.7 kg and its center of gravity is at the middle of the plank, what is the force F1

Answers

Answer: [tex]115.52\ N[/tex]

Explanation:

Given

Length of plank is 1.6 m

Force [tex]F_1[/tex] is applied on the left side of plank

Force [tex]F_2[/tex] is applied 43 cm from the left end O.

Mass of the plank is [tex]m=13.7\ kg[/tex]

for equilibrium

Net torque must be zero. Taking torque about left side of the plank

[tex]\Rightarrow mg\times 0.8-F_2\times 0.43=0\\\\\Rightarrow F_2=\dfrac{13.7\times 9.8\times 0.8}{0.43}\\\\\Rightarrow F_2=249.78\ N[/tex]

Net vertical force must be zero on the plank

[tex]\Rightarrow F_1+W-F_2=0\\\Rightarrow F_1=F_2-W\\\Rightarrow F_1=249.78-13.7\times 9.8\\\Rightarrow F_1=115.52\ N[/tex]

If we convert a circuit into a current source with parallel load it is called?​

Answers

Answer:

If we convert a circuit into a current source with parallel load it is called source transformation

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