5- Clasifica los siguientes cambios de la materia, anotando delante de cada uno cambio físico (F) o cambio químico (Q): • Disolver azúcar en agua • Freir una chuleta • Arrugar un papel • El proceso de la digestión • Secar la ropa al sol • Congelar una paleta de agua • Hacer un avión de papel • Oxidación del cobre • Romper un lápiz • Prender fuegos artificiales • Excavar un hoyo • Quemar basura

Answers

Answer 1

Answer:

1) Disolver azúcar en agua - Cambio químico - Es un caso de una solución en donde el solvente es el azúcar y el soluto es el agua.

2) Freir una chuleta - Cambio físico - Es un proceso de cocinado por un transferencia de calor y transferencia de masa.

3) Arrugar un papel - Cambio físico - Se aplica fuerzas externas para deformar el papel.

4) El proceso de la digestión - Cambio químico - Reducción de los alimentos a desechos y la absorción de nutrientes por el contacto con jugos gástricos o ambientes intestinales.

5) Secar la ropa al sol - Cambio físico - El secado es un fenómeno de transferencia de masa.

6) Congelar una paleta de agua - Cambio físico - Cambio del estado líquido al estado sólido por transferencia de calor.

7) Hacer un avión de papel - Cambio físico - Aplicación de fuerzas externas para plegar y doblar la hoja de papel.

8) Oxidación del cobre - Cambio químico - Proceso de corrosión por contacto con iones que se transportan en el ambiente.

9) Romper un lápiz - Cambio físico - Proceso de ruptura por esfuerzo normal a causa de un momento como consecuencia de una fuerza externa aplicada sobre el lápiz.

10) Prender fuegos artificiales - Cambio químico - Reacción química de reducción-oxidación.

11) Excavar un hoyo - Cambio físico - Remoción de tierra por trabajo físico.

12) Quemar basura - Cambio químico - Reacción de combustión.

Explanation:

A continuación, veremos que representa cada caso:

1) Disolver azúcar en agua - Cambio químico - Es un caso de una solución en donde el solvente es el azúcar y el soluto es el agua.

2) Freir una chuleta - Cambio físico - Es un proceso de cocinado por un transferencia de calor y transferencia de masa.

3) Arrugar un papel - Cambio físico - Se aplica fuerzas externas para deformar el papel.

4) El proceso de la digestión - Cambio químico - Reducción de los alimentos a desechos y la absorción de nutrientes por el contacto con jugos gástricos o ambientes intestinales.

5) Secar la ropa al sol - Cambio físico - El secado es un fenómeno de transferencia de masa.

6) Congelar una paleta de agua - Cambio físico - Cambio del estado líquido al estado sólido por transferencia de calor.

7) Hacer un avión de papel - Cambio físico - Aplicación de fuerzas externas para plegar y doblar la hoja de papel.

8) Oxidación del cobre - Cambio químico - Proceso de corrosión por contacto con iones que se transportan en el ambiente.

9) Romper un lápiz - Cambio físico - Proceso de ruptura por esfuerzo normal a causa de un momento como consecuencia de una fuerza externa aplicada sobre el lápiz.

10) Prender fuegos artificiales - Cambio químico - Reacción química de reducción-oxidación.

11) Excavar un hoyo - Cambio físico - Remoción de tierra por trabajo físico.

12) Quemar basura - Cambio químico - Reacción de combustión.


Related Questions

A penny is dropped into a well. It takes 5 seconds to fall. Calculate the depth of the well in feet.

Answers

Answer:

d=1/2 (a)(t^2)

D = distance

A = acceleration

T = time

acceleration due to gravity is 32 ft/second

so, d=1/2 (32)(5^2)

d=16(25)

d=400

Explanation:

Answer:

400 ft.

Explanation:

D= 1/2 gt^2

=1/2(-32 ft/sec^2)(5 sec^2)

= -(1/2)(32)(25) ft

D= -400 ft, down

E
How does the construction of dams positively affect natural resources?
by providing water for irrigation and restoring trees to areas where forests once existed
by creating reservoirs, preventing flooding, and renewing destroyed ecosystems
by preventing flooding, creating reservoirs, and providing water for irrigation
by renewing destroyed ecosystems and restoring trees to areas where forest once ersted

Answers

Answer:

by preventing flooding, creating reservoirs, and providing water for irrigation

Dams and waterways store and provide water for irrigation so farmers can use the water for growing crops. This idea goes way back into history. Irrigation is an important part of using water. In areas where water and rain are not abundant (like the desert), irrigation canals from rivers and dams are used to carry water.

Dams help in preventing floods. They catch extra water so that it doesn’t run wild downstream. Dam operators can let water out through the dam when needed.

Answer the question plz

Answers

Answer:

b and d

Explanation:

Help Meeeeeeeee. Have a nice day:)

Answers

Answer:

01.

Explanation:

Half the acceleration. Its heavier and moves slower. If it moved the same acceleration, the forces would also have to be doubled since the mass was.

1. Use a micrometer to find the thickness of the foil you used. Do your data confirm the measured thickness? Determine the percentage difference between the measured thickness and the thickness you found in your experiment.

Answers

Answer:

Yes.

Explanation:

Yes, my data confirm the measured thickness because the micrometer measure precise thickness of any object. If the same instrument and right method is used for the measurement of the thickness of the foil then there is zero difference occur between the measured thickness and the thickness in your experiment so using right method can reduce or eliminate the difference between the measured thickness and the thickness in experiment.

An unstrained horizontal spring has a length of 0.40 m and a spring constant of 340 N/m. Two small charged objects are attached to this spring, one at each end. The charges on the objects have equal magnitudes. Because of these charges, the spring stretches by 0.033 m relative to its unstrained length. Determine (a) the possible algebraic signs and (b) the magnitude of the charges.

Answers

Answer:

(a) Both the charges are positive or negative.

(b) Teh value of each charge is 1.53 x 10^-5 C.

Explanation:

Spring constant, K = 340 N/m

Natural length, L = 0.4 m

stretch, y = 0.033 m

(a) Let the charge on each sphere is q and they repel each other so the nature of charge of either sphere may be both positive or both negative.  

(b) The electrostatic force is balanced by the spring force.

[tex]\frac{kq^2}{(L + y)^2}=Ky\\\\\\\frac{9\times 10^9 q^2}{(0.4 +0.033)^2} = 340\times0.033\\\\q= 1.53\times 10^{-5} C[/tex]

Si la rapidez del sonido en el agua es de 1450 m/s. ¿Cuánto tiempo tardará en recorrer 500 m en el fondo del lago?

Answers

Answer:

2.9

Explanation:

es 2 minutos con 9 segundos seria algo así (1450)÷(500)

the bodies in this universe attract one another name the scientist who propounded this statement​

Answers

Answer:

It was proposed by Isaac Newton

Explanation:

The law of universal attraction of expression

          F = [tex]G \ \frac{m_1m_2}{ r^2}[/tex]G m1m2 / r ^ 2

where G is a constant, m₁ and m₂ are the masses of the bodies and r the distance between them.

It was proposed by Isaac Newton

With this law Newton explained that the force that pulls the moon towards the earth is the same as that which attracts an apple towards the earth

The reservoir stores 6 500 000 m3 of water. The density of the water is 998 kg/m3. Calculate the mass of water in the reservoir. Give your answer in standard form.

Answers

Answer:

Mass = 64,870,000,000 kilograms

Explanation:

Given the following data;

Density = 998 kg/m³

Volume = 6,500,000 m³

To find the mass of water in the reservoir;

Density can be defined as mass all over the volume of an object.

Simply stated, density is mass per unit volume of an object.

Mathematically, density is given by the equation;

Density = mass/volume

Making mass the subject of formula, we have;

Mass = density * volume

Mass = 998 * 6,500,000

Mass = 64,870,000,000 kilograms


[tex]what \: is \: reflection \: of \: light \: \: \: \: {?}[/tex]

Answers

Answer:

When a ray of light approaches a smooth polished surface and the light ray bounces back, it is called the reflection of light. The incident light ray which lands upon the surface is said to be reflected off the surface. The ray that bounces back is called the reflected ray.

Liquid to solid with explanation on the basis of kinetic model and freezing point.​

Answers

Answer:

As a liquid is cooled its molecules lose kinetic energy and their motion slows. When they've slowed to where intermolecular attractive forces exceed the collisional forces from random motion, then a phase transition from liquid to solid state takes place and the material freezes.

hope this helps you..!!!!

A 5.0 kg block of ice is at rest at the top of a smooth inclined plane. The block is released and slides 2.0 m down the plane. Assuming there is no friction between the block and the surface, calculate
a) the gravitational potential energy at the top of the plane
b) the component of the weight parallel to the plane
c) the acceleration of the block
d) the velocity of the block at the bottom of the plane
e) the kinetic energy at the bottom of the plane. ​

Answers

Answer:

a) 98.1 Joules

b) 49.05 N × sin(θ)

c) 9.81 × sin(θ)

d) The velocity of the block at the bottom of the plane, v is approximately 6.264 m/s

e) 98.1 Joules

Explanation:

The given parameters of the block are;

The mass of the block, m = 5.0 kg

The distance down the plane the block slides, h = 2.0 m

The friction between the block and the surface = 0

Let θ represent the angle of inclination oof the plane

a) The gravitational potential energy, P.E. = m·g·h

Where;

g = The acceleration due to gravity ≈ 9.81 m/s²

∴ P.E. ≈ 5.0 kg × 9.81 m/s² × 2.0 m = 98.1 Joules

The gravitational potential energy, P.E. ≈ 98.1 Joules

b) The component of the weight of the block parallel to the plane, [tex]w_{\parallel}[/tex], is given as follows;

[tex]w_{\parallel}[/tex] = w × sin(θ) = m·g·sin(θ)

∴ [tex]w_{\parallel}[/tex] ≈ 5.0 kg × 9.81 m/s² × sin(θ) = 49.05 × sin(θ) N

The component of the weight of the block parallel to the plane, [tex]w_{\parallel}[/tex] ≈ 49.05 N × sin(θ)

c) The component of the weight along the inclined plane = The force with which the block moves along the inclined plane, therefore;

[tex]w_{\parallel}[/tex] = m·g·sin(θ) = m·a

Where a represents the acceleration of the block along the plane

Therefore, by comparison, we have;

g·sin(θ) = a

∴ a ≈ 9.81 × sin(θ)

d) Given that the motion of the block is 2.0 m downwards, we have;

The velocity of the block at the bottom of the plane, v² = 2·g·h

Therefore, v² ≈ 2 × 9.81 m/s²× 2.0 m = 39.24 m²/s²

v = √(39.24 m²/s²) ≈ 6.264 m/s

e) The kinetic energy at the bottom of the plane, K.E. = (1/2)·m·v²

∴ K.E. = (1/2) × 5.0 kg × 39.24 m²/s² = 98.1 J

A body of mass 8000 moving with a velocity of 2.5m/s collides with a stationery twice the mass of the first body. After impact the mass moves to 1.0m/s Find the speed of the first body.​

Answers

Answer:

the final speed of the first ball is 0.5 m/s

Explanation:

Given;

mass of the first body, m₁ = 8000 kg

mass of the second body, m₂ = 16,000 kg

initial velocity of the first body, u₁ = 2.5 m/s

initial velocity of the second body, u₂ = 0

final velocity of the stationary mass, v₂ = 1 m/s

let the final velocity of the first body = v₁

Apply the principle of conservation of linear momentum;

m₁u₁  +  m₂u₂  = m₁v₁   +   m₂v₂

8000(2.5)  +   (16,000 x 0)   =   8000v₁   +  16,000 (1)

20,000  +  0 = 8000v₁  +  16,000

8000v₁ = 4000

v₁ = 4000/8000

v₁ = 0.5 m/s

Therefore, the final speed of the first ball is 0.5 m/s

A ball of mass 0.175 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.805 m. What impulse was given to the ball by the floor

Answers

Answer:

The impulse received by the ball is - 1.561 kg.m/s

Explanation:

Given;

mass of the ball, m = 0.175 kg

initial displacement of the ball, h₁ = 1.25 m

final displacement of the ball, h₂ = 0.805 m

Assumptions:

let the downward direction of the ball be positive

let the upward direction of the ball be negative

The following equation of motion will be used to determine the final velocity of the ball at each displacement.

v² = u² ± 2gh

The final velocity of the ball when it is dropped downwards to 1.25 m;

v² = u² + 2gh

v² = 0 + 2gh

v² = 2gh

v = √2gh

v = √(2 x 9.8 x 1.25)

v = 4.95 m/s

The final velocity of the ball when it rebounds from the floor to 0.805 m;

vf² = u² - 2gh

vf² = 0² - 2gh

vf² =  -2gh

vf = -√2gh

vf = - √(2 x 9.8 x 0.805)

vf = -3.97 m/s

The impulse received by the ball is calculated as;

J = ΔP = mΔv = m(vf - v)

                      = 0.175(-3.97 - 4.95)

                       = - 1.561 kg.m/s

The negative sign indicates upward direction of the impulse.

what is chemical change ??​

Answers

Answer:

the combination,decomposition or replacement that occurs in the molecules of matter during chemical change called chemical reaction

Answer:

A chemical change happens when one chemical substance is transformed into one or more different substances, such as when iron becomes rust. ... A chemical change is different from a physical change, which doesn't rearrange atoms or molecules and produce a completely new substance.

Find the sum. Express the answer in scientific notation. (1.54 x 10^6)+(6.15 x 10^6)​

Answers

Answer:

[tex] { \tt{(1.54 \times {10}^{6}) + (6.15 \times {10}^{6}) }} \\ = { \tt{(1.54 + 6.15) \times {10}^{6} }} \\ = { \tt{7.69 \times {10}^{6} }}[/tex]

ive an example of a pair of variables that have negative correlation. A. The quantity of fertilizers used and crop yield. B. The color of a​ person's shirt and the number of meals sold at Chinese restaurants. C. The number of physicists and the speed of sound. D. The number of winter coats sold and the temperature outside.

Answers

Answer:

D. The number of winter coats sold and the temperature outside.

Explanation:

A negative correlation can be described as a relationship existing between two variables such that as one of these variables goes up, the other variable would go down and vice versa.

With this explanation in mind, as the temperature outside increases, less winter coats would be sold because an increase in temperature would mean more heat and hotness. People would be uncomfortable wearing thick clothing.

9. In a __________ collision, 100% of both vehicles' speed is directed towards the point of impact. A. head-on B. rear-end C. side-impact

Answers

Answer:   A

Explanation:

:)

Una grúa eleva un tubo de concreto de
400 kg, con Movimiento rectilíneo uniforme,
con el
cable ABC. Determine la tensión que pueden
soportar los cables AB, BC y BD, sabiendo
que los cables AB=BC y la tensión que
soporta el cable AB es de 150 N.

Answers

Explanation:

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Students are completing a lab in which they let a lab cart roll down a ramp. The students record the mass of the cart, the height of the ramp, and the velocity at the bottom of the ramp. The students then calculate the momentum of the cart at the bottom of the ramp.

A 4 column table with 3 rows. The first column is labeled Trial with entries 1, 2, 3, 4. The second column is labeled Mass of Cart in kilograms with entries 200, 220, 240, 260. The third column is labeled Height of ramp in meters with entries 2.0, 2.1, 1.5, 1.2. The fourth column is labeled Velocity at Bottom in meters per second with entries 6.5, 5.0, 6.4, 4.8.

Which trial’s cart has the greatest momentum at the bottom of the ramp?

Answers

Answer:

second column

Explanation:

Answer:

Trial 3 is the answer.

Explanation:

The electric motor in the car is powered by a battery.

To charge the battery, the car is plugged into the mains supply at 230 V

The power used to charge the battery is 6.9 kW

Calculate the current used to charge the battery.

Answers

Answer:

I = 30 A.

Explanation:

Given that,

The voltage of the battery, V = 230 V

Power used to charge the battery, P = 6.9 kW

We need to find the current used to charge the battery. The formula for the power is given by :

P = VI

Where

I is current

So,

So, the required current is 30 A.

13.Suppose a ball of mass m is thrown vertically upward with an initial speed v, its speed decreases continuously till it becomes zero. Thereafter, the ball begins to fall downward and attains the speed u again before striking the ground. It implies that the magnitude of initial and final momentums of the ball are same. Yet, it is not an example of conservation of momentum. Explain why?​

Answers

Answer:

Law of conservation of momentum is applicable to isolated system (no external force is applied).

In this case, the change in velocity is due to the gravitational force of earth.When the ball is thrown up, negative force of gravity is applied and so the speed becomes zero.After that due to gravitational force the ball falls down with same initial velocity.

What happen to the frequency of transverse vibration of a stretched string if its tension is halved and the area of cross section of the string is doubled?

Answers

Answer:

The fundamental frequency of the stretched string is:

[tex]f=[/tex] [tex]\frac{1}{2} \sqrt{\frac{T}{L} }[/tex] [ T = Tension and μ = mass per unit length]

Here,

 μ = [tex]\frac{m}{L} = \frac{Vp}{L} = Ap[/tex]

[tex]f= \frac{1}{2} \sqrt{\frac{T}{Ap} }[/tex]

If T is halved and A is doubled,

[tex]f= \frac{1}{2} \sqrt{\frac{T'}{A'p} } = \sqrt{\frac{1}{2* 2* A* p} } = \frac{1}{2} (\frac{1}{2} \sqrt{\frac{T}{Ap} } = \frac{1}{2} f[/tex]

Thus, the frequency is reduced to half if its tension is halved and the area of cross-section of the string is doubled.

When the tension is halved and the area of cross section is doubled, the frequency increases by a factor of 1.414.

The frequency of transverse vibration of a stretched string is calculated as follows;

[tex]f = \frac{1}{2l} \sqrt{\frac{T}{\mu} }[/tex]

where;

T is the tension on the stringμ is the mass per unit length

[tex]f = \frac{1}{2l} \sqrt{\frac{T}{\mu} }\\\\f = \frac{1}{2l} \sqrt{\frac{Tl}{m} } \\\\f = \frac{l^2}{2l} \sqrt{\frac{T}{m} } \\\\f = \frac{A}{2l} \sqrt{\frac{T}{m}}[/tex]

when the tension is halved and the area of cross section is doubled, the frequency is calculated as;

[tex]\frac{f_1}{A_1 \sqrt{T_1} } = \frac{f_2}{A_2\sqrt{T_2} } \\\\f_2 = \frac{f_1 A_2\sqrt{T_2}}{A_1 \sqrt{T_1} }\\\\f_2 = \frac{f_1 \times 2A_1\sqrt{0.5T_1} }{\sqrt{T_1} }\\\\f_2 = \frac{f_1 \times 2A_1 \times 0.7071\sqrt{T_1} }{\sqrt{T_1} }\\\\f_2 = 1.414 f_1[/tex]

Thus, when the tension is halved and the area of cross section is doubled, the frequency increases by a factor of 1.414.

Learn more about tension in a string here: https://brainly.com/question/25743940

scientists are seen very busy in designing the solar power equipments, why?

Answers

Answer:

it is because solar energy is the perpetual source of energy and by using it non renewable sources of energy can be conserved for the future

Imagine you see Mars rising in the east at 6:30 pm. Six hours later what direction would you face (look) to see Mars when it is highest in the sky

Answers

Answer:

The Mars appears in the direction of South.

Explanation:

Mars is rising in the east at 6: 30 PM. The period of rotation of earth is 24 hours.

So, 6 hours is the one fourth of the period of rotation of earth. Earth rotates counter clockwise on its axis, so after 6 hours, we see the Mars in the direction of South.

What factors could the skater change to apply the same amount of torque but increase the rate of his spin

Answers

Answer:

The moment of inertia should be decreased.

Explanation:

The torque is given by

Torque = Moment of inertia x angular velocity

To keep the torque constant, the spin rate be increased when the moment of inertia decreases.  

The moment of inertia of the body are the efforts to put the object in rotation.  

A stationary body explodes into four identical fragments such that three of them fly off mutually perpendicular to each other, each with same KE, E 0 ​ . The energy of explosion will be Aa

Answers

Answer:

6Eo

Explanation:

A hot-air balloon is rising upward with a constant speed of 2.85 m/s. When the balloon is 2.50 m above the ground, the balloonist accidentally drops a compass over the side of the balloon. How much time elapses before the compass hits the ground

Answers

Answer:

Explanation:

You have to declare which way is plus -- up or down. I will use down.

vi = - 2.85       The balloon is going up. That is the minus direction.

a = 9.81

d =   2.50 meters  distance in this case is from the object to the ground.

d = vi*t + 1/2 a t^2

-2.50 = -2.85*t + 1/2 * 9.81 * t^2

-2.50 = -2.85*t + 4.905 * t^2           transfer the left to the right.

-4.905 t^2 + 2.85*t + 2.50 = 0

Use the quadratic formula to solve for t.

It turns out that t = 1.06

A stone is dripped into the well 44 meter dip the spalash in heard in 3.12 s find speed of sound in air.

Answers

Answer:

The speed of sound is 366.67 m/s.

Explanation:

height, h = 44 m

total time, T = 3.12 s

Let the time taken by the stone to hit the water is t.

use second equation of motion

[tex]h = u t + 0.5 gt^2\\\\44 = 0.5\times 9.8 t^2\\\\t = 3 s[/tex]

Time taken by sound to g up

t'=T - t =  3.12- 3 = 0.12 s

The speed of sound is

[tex]v = \frac{h}{t'}\\\\v = \frac{44}{0.12}\\\\v = 366.7 m/s[/tex]

You can see stars brighter than magnitude +6 with your naked eye under dark sky conditions, and you can also see the full moon at magnitude –13. (You can glance at the Sun, which is at magnitude –27, but you can't safely look at the Sun.) Given these limits, what is the dynamic range of the naked eye?

Answers

-6 uh 89 0 -+= 62 uwuwhhiwhiuhwiuhwuihhishishoknsk
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