Are these a Chemical or Physical Change??

1. Sodium Bicarbonate breaks down into sodium carbonate and water.
A. Chemical Change
B. Physical Change

2. Mercury is cooled until it is in the solid state.
A. Chemical Change
B. Physical Change

Answers

Answer 1
1. chemical change, as it is the breaking of bonds

2. physical change, as it is a change in temperature
Answer 2

The change of sodium bicarbonate breaking down into sodium carbonate and water is a chemical change. The cooling of mercury until it is in the solid state is physical change. The correct option for 1 is A and for 2 is B.

What is chemical and physical change?

A chemical change occurs as a consequence of a reaction, whereas a physical change occurs when transformation occurs forms but retains its chemical identity.

Chemical changes include burning, cooking, rusting, and rotting. Physical changes include boiling, melting, freezing, and shredding.

A physical change is required to a specimen of matter in which some of the material's properties change but the matter's identity does not.

Physical changes can be classified as either reversible or irreversible. Melting is a reversible physical change because the melted ice cube can be refrozen.

In the first case, sodium bicarbonate is breaking down and forming sodium carbonate and water, this implies a chemical change.

In the second case, mercury is just cooling down, changing its physical shape. So it is physical change.

Thus, the correct option for 1 is A and for 2 is B.

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Related Questions

A circular disk of radius 0.200 m rotates at a constant angular speed of 2.50 rev/s. What is the centripetal acceleration (in m/s2) of a point on the edge of the disk?

Answers

[tex]a_c = 3.14\:\text{m/s}^2[/tex]

Explanation:

First, we need to convert the given angular speed [tex]\omega[/tex] from rev/s to rad/s:

[tex]2.50\:\dfrac{\text{rev}}{\text{s}}×\dfrac{2\pi\:\text{rad}}{1\:\text{rev}} = 15.7\:\text{rad/s}[/tex]

The centripetal acceleration [tex]a_c[/tex] is defined as

[tex]a_c = \dfrac{v^2}{r}[/tex]

Recall that [tex]v = r\omega[/tex] so we can write [tex]a_c[/tex] as

[tex]a_c = \dfrac{(r\omega)^2}{r} = \omega^2r[/tex]

[tex]\;\;\;\;\;=(15.7\:\text{rad/s})^2(0.200\:\text{m}) = 3.14\:\text{m/s}^2[/tex]

A student catches a can of soda dropped from some unknown height by Mr. Fineman. If

the can was dropped from rest, and it is traveling at 8.9 m/s before it arrives in the

student's hand...

Answers

Answer:

v² = u² + 2gh

8.9² = 0² + 2(9.8)h

h = 4.0 m

v = at

t = 8.9/9.8

t = 0.91 s

Which of the following is the equation for acceleration?

Answers

Answer:

A

Explanation:

acceleration = velocity / time

Hope that helps

Answer:

D

Explanation:

Average acceleration = final velocity- initial velocity, divided by elapsed time.

ā = V - Vo/t

= change in velocity ÷ change in time

A car travels a certain distance from A to B with a speed of 60km/hr and then returns along the same path to the starting point with a speed of 40km/hr. Find the average speed and average velocity.
a) Km/hr
b) m/s

wrong answers will be reported!

Answers

Answer:

Explanation:

Speed is total distance traveled over time taken to do so.

If AB is measured in kilometers, time (t) for the whole trip is

t = AB/60 + AB/40  

t = 2AB/120 + 3AB/120

t = 5AB/120 hrs

Average speed is distance over time

s = 2AB / (5AB/120)

s = 2(120)/5

s = 48 km/hr

s = 48(1000 m/km / 3600 s/hr) = 13.333333.... 13 m/s

Velocity is displacement over time.

As displacement is zero, velocity is zero

v = 0 km/hr = 0 m/s

Pretty harsh reporting answers just because they are wrong.

37. The progressive loss of material from a surface by the mechanical action of a fluid on a surface is called

Answers

Answer:

The progressive loss of material from a surface by the mechanical action of a fluid on a surface is called erosion.

I need help. please look at the image below and let me know I need this by 7:20 am pst. ​

Answers

Answer:

3(1.5) = 4.5 V

Explanation:

A mars surface exploration vehicle drops a rock off a 1.00 I'm high vertical Cliff. The sound of the rock landing at the base of the cliff is recorded by instruments on the vehicle 27.1 seconds later. Calculate the acceleration due to gravity on Mars given that the speed of sound on Mars is 320 m/s

Answers

The acceleration due to gravity on Mars is 11.81 m/s².

The given parameters:

Height of the cliff, h = 1 mTime of motion of the sound wave, t = 27.1 sSpeed of sound in mass, v = 320 s

The equation of motion to determine the acceleration due to gravity on the moon is calculated as follows;

[tex]s = vt + \frac{1}{2} gt^2[/tex]

where;

s is the distance traveledt is the time of motion

Since the time measured is two way time, the new equation for the total distance traveled is calculated as;

[tex]v = \frac{2d}{t} \\\\2d = vt\\\\d = \frac{vt}{2} \\\\d = \frac{320 \times 27.1}{2} \\\\d = 4,336 \ m[/tex]

The acceleration due to gravity is calculated as follows;

[tex]s = vt + \frac{1}{2} gt^2\\\\4,336 = 0 \ + \ \frac{1}{2} \times g \times (27.1)^2\\\\4,336 = 367.21g\\\\g = \frac{4,336}{367.21} \\\\g = 11.8 1 \ m/s^2[/tex]

Thus, the acceleration due to gravity on Mars is 11.81 m/s².

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I need a short answer ?

Answers

Answer:

Explanation:

7a) t = d/v = 100/45cos14.5 = 2.29533...= 2.30 s

7b) h = ½(9.81)(2.29533/2)² = 6.46056... = 6.45 m

  or

  h = (45sin14.5)² / (2(9.81)) = 6.47 m

which rounds to the same 6.5 m when limiting to the two significant digits of the initial velocity.

A solid, uniform sphere with a mass of 2.5 kg rolls without slipping down an incline plane starting from rest at a vertical height of 19 m. If the sphere has a radius of 0.60 m, what is the angular speed of the sphere at the bottom of the incline plane

Answers

Answer:

1/2 m v^2 + 1/2 I ω^2 = m g h       conservation of energy

I = 2/5 m R^2     inertia of solid sphere

1/2 m v^2 + 1/5 m ω^2 R^2 = m g h

1/2 v^2 + 1/5 v^2 = g h

v^2 = 10 g h / 7 = 1.43 * 9.80 * 19 m^2/s^2 = 266 m^2/s^2

v = 16.3 m/s

v = R ω

ω = 16.3 / .6 = 27.2 / sec

This is an image of a satellite traveling around Earth. Explain what are the two forces that are keeping the satelite around Earth without flying off or hitting the ground.​

Answers

Answer:

One force will be gravity & inertia.

Explanation:

Bioth are combine to keep Earth in orbit around the sun, and the moon in orbit around Earth

A body is moving along a circular path with variable speed, it has both radial and tangential acceleration.

Select one:
True
False

Answers

Answer:

True;   ar = v^2 / R      Radial acceleration because it moves in a circular path

    at = α R = tangential acceleration because its speed changes

a = at + ar    total acceleration equals sum of radial and tangential

A 5 kilograms bowling ball is dropped out a window. It hits the ground, and bounces upward. The velocity change of the ball is noted to be 15 m/s downward and 12 m/s upward. What is the contact time for the ball if the force applied on the ball from the ground is equal to 10 N?

Answers

Answer:

13.5

Explanation:

Mass: 5kg

Initial Velocity: -15

Final Velocity: 12

Force: 10

We can use the equation: Vf = Vi + at

We need to find acceleration, and we can use the equation, F=ma,

We have mass and the force so it would look like this, 10=5a, and 5 times 2 would equal 10, so acceleration would be 2.

Now we have all the variables to find time.

Back to Vf = Vi + at, plug the numbers in, 12 = -15 + 2(t)

Plugging them in into desmos gives 13.5 for time.

3. A car travelling at 12 m/s into a stationary truck of about 10 times the cars mass. a. If the collision was completely inelastic, what velocity would the two travel at if the stuck together? b. If the collision was completely elastic, what would be the velocities of the car and truck after the collision? c. In order to exert a force of only 3500N on the truck during the collision, how much time would the collision have to take?

Answers

(a) The final velocity of the two vehicles if the collision was inelastic is 1.1 m/s.

(b)  For the elastic collision, the final velocity of the car is 9.81 m/s backwards and the final velocity of the truck is 2.19 m/s forward.

(c) The time taken to exert the given force is 0.00625 m (s).

The given parameters;

Initial velocity of the car, u₁ = 12 m/sInitial velocity of the truck, u₂ = 0Mass of the car, = mMass of the truck, = 10m

(a) The final velocity of the two vehicles if the collision was inelastic is calculated as follows;

[tex]m_1 u_1 + m_2u_2 = v(m_1+ m_2)\\\\12m + 10m(0) = v(m + 10m)\\\\12m = v(11m)\\\\v = \frac{12m}{11m} \\\\v = 1.1 \ m/s[/tex]

(b) The final velocity of the two vehicles if the collision was elastic is calculated as follows;

[tex]m_1 u_1 + m_2u_2 = m_1v_1 + m_2v_2\\\\\12m \ + \ 10m(0) = mv_1 + 10mv_2\\\\12m = m(v_1 + 10v_2)\\\\12 = v_1 + 10 v_2\ \ - --(1)[/tex]

Apply one-directional velocity equation:

[tex]u_1 +v_1 = u_2 + v_2\\\\12 + v_1 = 0 + v_2\\\\12+ v_1 = v_2 \ \ --- (2)[/tex]

Substitute the value of [tex]v_2[/tex] into equation (1);

[tex]12 = v_1 + 10(12 + v_1)\\\\12= v_1 + 120 + 10v_1\\\\12- 120 = 11v_1\\\\-108 = 11v_1\\\\v_1 = \frac{-108}{11} \\\\v_1 = -9.81 \ m/s\\\\[/tex]

Solve for [tex]v_2[/tex];

[tex]v_2 = 12 + v_1\\\\v_2 = 12 - 9.81\\\\v_2 = 2.19 \ m/s[/tex]

Thus, for the elastic collision, the final velocity of the car is 9.81 m/s backwards and the final velocity of the truck is 2.19 m/s forward.

(c)

The change in the momentum of the truck is calculated as;

[tex]\Delta P = m_2(v_2 - u_2)\\\\\Delta P = 10m(2.19)\\\\\Delta P = 21.9m[/tex]

The time taken to exert the given force is calculated as follows;

[tex]Ft = \Delta P\\\\t = \frac{\Delta P}{F} \\\\t = \frac{21.9 \ m}{3500} \\\\t = 0.00625 \ m (seconds)[/tex]

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Four men are pushing on a broken car. One man pushes on the car with 345 N, another with 203 N, another
with 291 N and 101 N. The friction between the car and road pushes in the opposite direction with a force of 940 N.
1.) Make a claim: What is the net force on the car?
2.) Cite evidence: What evidence supports your claim?
3.) Make a claim: Will the car move?
4.) Provide reasoning: Why do you say that?

Answers

Answer:

1. 3 N

2. The total force being put on the car by the men is 943 while the friction is going in the opposite direction with a force of 940.

3. Yes

4. The net force is not equal to zero

Explanation:

1. 345 + 203 + 291 + 101 = 943 + -940 = 3

An object, initially traveling at a velocity of 73 m/s, experiences an acceleration of -9.8 m/s^2. How much time will it take it to come to rest?

Answers

7.4 s

Explanation:

Given:

[tex]v_0 = 73\:\text{m/s}[/tex]

[tex]v = 0[/tex]

[tex]a = -9.8\:\text{m/s}^2[/tex]

[tex]t = ?[/tex]

To solve the time it takes for the object to come to a stop, we are going to use the equation below:

[tex]v = v_0 + at \Rightarrow t = \dfrac{v - v_0}{a}[/tex]

Using the given values above, we get

[tex]t = \dfrac{0 - 73\:\text{m/s}}{-9.8\:\text{m/s}^2}[/tex]

[tex]\;\;\;\;= 7.4\:\text{s}[/tex]

the distance between two charges q a and q b is r and the force between them is F. What is the force between them if the distance between them is doubled?

Answers

The force will be reduced to 1/4 of the original

Explanation:

According to Coulomb's law, the force between two charges [tex]q_a\:\text{and}\:q_b,[/tex] separated by a distance r is given by

[tex]F = k\dfrac{q_aq_b}{r^2}[/tex]

where k is the Coulomb constant.

Now let F' be the force between the two charges when their separation distance is doubled. We can write this force as

[tex]F' = k\dfrac{q_aq_b}{(2r)^2} = k\dfrac{q_aq_b}{4r^2}[/tex]

[tex]\;\;\;\;\;= \frac{1}{4}\left(k\dfrac{q_aq_b}{r^2}\right) = \frac{1}{4}F[/tex]

Therefore, the force will be reduced to a quarter of its original value.

Hope you could understand.

If you have any query, feel free to ask.

A 30-cm-tall, 4.0-cm-diameter plastic tube has a sealed bottom. 250 g of lead pellets are poured into the bottom of the tube, whose mass is 30 g, then the tube is lowered into a liquid. The tube floats with 5.0 cm extending above the surface. What is the density of the liquid

Answers

The density of the liquid will be equal to   [tex]\rho=0.892 \ \dfrac{g}{cm^3}[/tex]

What is density?

The density of an object is defined as the ratio of the mass of an object to the volume of the object.

Volume of tube = 2^2 * pi * 30 = 377 cm^3

Volume of tube submerged = 25* 377 / 30 = 314 cm^3

Buoyancy = weight of liquid displaced

Volume of liquid displaced = 314 cm^3

Mass of tube and lead = 250 + 30 = 280 g

Now from the mass density by definition

[tex]\rho = \dfrac{m}{v}[/tex]

[tex]m=\rho \times v[/tex]

Mass of liquid displaced = Mass being supported

[tex]314 \times \rho = 280 g[/tex]

[tex]\rho= \dfrac{280}{ 314 } = .892 \frac{g}{cm^3}[/tex]

Thus the density of the liquid will be equal to   [tex]\rho=0.892 \ \dfrac{g}{cm^3}[/tex]

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The density of the liquid is 1.67 g/[tex]cm^3[/tex].

The volume of the tube is

30 * 4 * 3.14 * 0.25 = 94.2 [tex]cm^3[/tex].

The mass of the lead pellets and the plastic tube is

30 + 250 = 280 g.

The volume of the lead pellets is

250 / 11.34 = 22 [tex]cm^3[/tex].

The volume of the liquid that the tube displaces is

94.2 - 22 = 72 [tex]cm^3[/tex].

The density of the liquid is

280 / 72 = 1.67 g/ [tex]cm^3[/tex].

Therefore, the density of the liquid is 1.67 g/ [tex]cm^3[/tex].

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Find the required angular speed, ω, of an ultracentrifuge for the radial acceleration of a point 1.60 cm from the axis to equal 300000 g (where g is the acceleration due to gravity).
Express your answer numerically in revolutions per minute.

Answers

Answer:

Explanation:

300000(9.8) = ω²(0.0160)

ω = 13555 rad/s

13555 rad/s (60 s/min/ 2π rad/rev) = 129,445 rev/min

How to fix this? I need to save both

Answers

Should be able to pull it off the same way you got it on

Answer:

Id say beat it with a hammer and light it on fire. Then if it doesn't work, sacrifice it.

Explanation:

9.The force of gravity between two asteroids is 10,000 newtons (N).

When the distance between the asteroids is halved, what will be the force of gravity between them?

Answers

Answer:

F1 = G m1 m2 / R^2        force of attraction

F2 = G m1 m2 / (R/2)^2

F2 / F1 = 4       the force of gravity will be quadrupled

Which of the following describes the motion of a block while it is in equilibrium? The block:
A. moves at a constant speed
B. slows down gradually to stop
C. speeds up for a bit, then moves at a constant speed
D. Accelerates constantly

Answers

The statement that describes the motion of a block while it is in equilibrium is: The block moves at a constant speed.

EQUILIBRIUM:

A state of equilibrium in physics refers to a state of rest or the forces exerted on the object is in a balanced state.

In dynamic equilibrium, the acceleration of a body is zero. This means that the body is moving at a uniform speed.

Therefore, the statement that describes the motion of a block while it is in equilibrium is: The block moves at a constant speed.

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Pls answer ASAP pls bc I’m tryna get my grade up please

Answers

Answer:

The right answer for this question is 85%.

(I had the same question.)

what is the pressure exerted by a force of 25 N on an area of 5m square

Answers

Answer:

pressure = force / area

then pressure = 25 / 5 = "5" N/m^2


When you put a pot of water on the stove, the stove transfers thermal energy to the water. As the water gains large
amounts of energy, how does it transfer this energy to its surrounding environment?

Answers

Answer:

It releases some of the energy into the atmosphere as hot steam.

Explanation:

Light and Reflection

Diagram Skills

E

STI

500

Mirrot

Flat Mirrors

1. The point of a 20.0 cm

D

pencil is placed 25.0 cm

from a flat mirror. Its

eraser is 15.0 cm from

the mirror. Three of the

light rays from the

pencil's point hit the

mirror with incident

angles of 0°, 20°, and

50° at points A, B, and C as shown.

a. Use a protractor to draw the reflected rays from points A, B, and C.

b. Where do reflected rays or their extensions intersect?

Mirror

B

c. What is the distance between the pencil's head and its image?

d. Would a person's eye located at point D perceive one of the reflected rays

drew? Will the person be able to see the image? Explain.

e. What if the eye is located at point E?

f. Draw incident rays from the eraser of the pencil to point A and to poin

Answers

The law of reflection allows to find the results for the questions about ray reflection in a plane mirror are:

    a) Attachment we see a diagram of the incident and reflected rays, incident and reflected angles are equal.

    b)  The extension of the reflected rays is what forms the image.

    c)  The image's distance  is 20 cm behind the flat mirror.

    d) The point D (normal for an angle of 50º) cannot perceive the rays coming from point A, B, C  

    e) the Rays at points A, B, C cannot perceive in the point E.

    f) attachment we see the rays that come out from the pencil eraser.

    g) The image is behind the mirror at 15 cm.

The geometric interaction describes the interaction of light rays with surfaces, looking for where the rays are directed, it is described by two phenomenological laws:

Refraction. Establishes a relationship between incident rays and those transmitted by material means. Reflection. It establishes that the angle of incidence and reflection of the rays is the same.

             [tex]\theta_i = \theta_{r}[/tex]  

From these two general laws, geometric optics establishes a relationship for the formation of the image, called the constructor's equation.

            [tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]

Where f is the focal length, p and q are the distance to the object and the image, respectively.

 

In this exercise, the medium is a mirror, which is why it must comply with the law of reflection.

a) In the attachment we see a diagram of the incident and reflected rays for the three points.

According to the law of reflection, the incident and reflected angles are equal.

b) From the diagram we can see that the extension of the reflected rays is what forms the image, which is called virtual and is located behind the mirror.

c) In the diagram we see two rays to form the image, we see that the distance to the object is equal to the distance to the image.

From the constructor's equation a plane mirror has an infinite radius.

      p = -q

Therefore the image's distance  is 20 cm behind the flat mirror.  Therefore the distance to the object and the image are the same, the negative sign indicates that the image is behind the mirror.

d) A person located at point D (normal for an angle of 50º) cannot perceive the rays coming from point A, B, C since their angle of reflection is not equal to the incident angle.

To perceive a ray it must have an angle of incidence of 25º.

e) Point E is located very far from the pencil, so the incident angle increases as does the reflected angle.

the Rays at points A, B, C cannot perceive.

f) In the attachment we see the rays that come out from the pencil eraser, they indicate that the distance to the plane mirror is 15.0 cm,

g) The image is behind the mirror at 15 cm.

In conclusion using the law of reflection we can find the results for the questions are:

     a) Attachment we see a diagram of the incident and reflected rays, incident and reflected angles are equal.

     b)  The extension of the reflected rays is what forms the image.

     c)  The image's distance  is 20 cm behind the flat mirror.

     d) The point D (normal for an angle of 50º) cannot perceive the rays coming from point A, B, C  

    e) the Rays at points A, B, C cannot perceive in the point E.

    f) attachment we see the rays that come out from the pencil eraser.

   g) The image is behind the mirror at 15 cm.

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What is Acceleration?​

Answers

Answer:

Depends on what are you looking at. All of the following are valid definitions, it just depends on which way you are analizing the problem.

From a kinematics point of view:

Acceleration is, by definition, is a vector quantity that measures the rate of change of the rate of change in position (add brackets if it helps visualizing the idea). This leads to the following different definitions - which are more like means of calculating it

the second derivative of position with respect to time [tex]a = \ddot x= \frac{d}{dt}({\frac d{dt}} x)[/tex];the first derivative of velocity with respect to time [tex]a = \dot v = \frac d{dt} v[/tex].

From a dinamics point of view

Acceleration is the effect of a force applied to a body, and measures the ratio of the force applied to a body of mass m and the mass itself (which is another formulation of Newton second law):

[tex]a = \frac Fm[/tex]

Acceleration is the rate of change of velocity

 

To define acceleration, We need to know more about motion.

Motion: This can be defined as the change in position of a body from one point to another. When an object accelerates, it undergoes motion.

Definition

Acceleration can be defined as the rate of change of velocity. The S.I unit of acceleration is meter-per-squared seconds. (m/s²)

The formula of acceleration is

a = (v-u)/t................. Equation 1

⇒ Where:

a = accelerationu = initial velocityv = final velocityt = time

Hence, Acceleration is the rate of change of velocity

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Please help me.............................

Answers

Answer:

[tex]a[/tex]

Explanation:

is a corect anser

Answer This!!!!!!
I'll give brainliest to whoever gets it right.
And 100 Points!

Answers

Answer: Time (days) = 88, the (mass) = 0.2180

Explanation:

Time (days) = 88, the (mass) = 0.2180

Cody hits up food king and uses a scale to weigh the mass of an apple. if the spring potential energy in the scale is .09 j and is spring is stretched 0.6 meters, calculate the spring constant

Answers

Answer:

oK so  here's  what you should do is  add .09 and 0.6

Explanation:

When laser light of wavelength 632.8 nm passes through a diffraction grating, the first bright spots occur at ± 17.8 ∘ from the central maximum. What is the line density

Answers

Answer:

Depends on the weight i would be 7.120

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