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zheka24 [161]
3 years ago
10

It is important to understand the differences between chemical and physical changes. Chemical changes result in new substances,

but physical changes do not. Which process is an example of a physical change?
F) Water turns to steam when boiled over a Bunsen burner.

G) Carbon combines with oxygen to form carbon dioxide gas.

H) Water breaks down into hydrogen and oxygen gases.

I) Limestone breaks down into lime and carbon dioxide when heated.
Physics
1 answer:
sergiy2304 [10]3 years ago
5 0

Explanation:

Other characteristics of these changes are:

Physical changes are easily reversible but chemical changes are not.

Physical changes involves no change in mass like chemical changes.

Physical changes require little energy unlike chemical changes.

<u>Water turns to steam when boiled over a Bunsen burner   </u><u>Physical change</u>

This is a physical change of state. It is called evaporation. Here, hydrogen bonds between each water molecules are broken and they turn to gases in steam phase.

                    H₂O_{l}    →      H₂O_{g}

<u>Carbon combines with oxygen to form carbon dioxide gas </u><u>Chemical change</u>

This is a typical chemical change":

            C   + O₂   →    CO₂

A new product is formed and this reaction is not easily reversible.

<u>Water breaks down into hydrogen and oxygen gases.   </u><u>Chemical change</u>

This is also a chemical change in which new products are formed:

                      H₂O  →   H₂   +   O₂

<u>Limestone breaks down into lime and carbon dioxide when heated</u><u> Chemical change</u>

This is a chemical change as new product forms:

                    CaCO₃   →  CaO + CO₂

Learn more:

Chemical change brainly.com/question/9388643

#learnwithBrainly

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An ice skater can change the rate of their spin by moving their arms into and away from their body without any other forces or t
pychu [463]

Answer:

Angular momentum is conserved if there are no external forces

P1 = P2

I1 ω1 = I2 ω2

ω2 / ω1 = I1 / I2

If the skater pulls their arms in (I2 < I1) then the angular speed must increase for angular momentum to be conserved.

3 0
3 years ago
Two trucks with equal mass are attracted to each other with a gravitational force of 5.3 x 10 -4 N. The trucks are separated by
HACTEHA [7]

Answer:

7327 kg or 7.3 tons

Explanation:

We have Newton formula for attraction force between 2 objects with mass and a distance between them:

F_G = G\frac{M_1M_2}{R^2}

where G =6.67408*10^{-11} m^3/kgs^2 is the gravitational constant on Earth. M = M_1 = M_2 is the masses of the 2 objects. and R = 2.6m is the distance between them.

F_G = 5.3*10^{-4}N is the attraction force.

5.3*10^{-4} = 6.67408*10^{-11}\frac{M^2}{2.6^2}

7941265 = \frac{M^2}{2.6^2}

M^2 = 53682948.76

M = \sqrt{53682948.76} \approx 7327 kg or 7.3 tons

3 0
4 years ago
You are traveling in a car that is moving at a velocity of 30 m/s. Suddenly, a car 15 meters in front of you slams on its brakes
Artemon [7]

Answer:

The acceleration of car is 6.67 m/s².

Explanation:

Given that,

Initial velocity = 30 m/s

Distance = 15 m

Final velocity = 10 m/s

Time = 3 sec

We need to calculate the acceleration

Using formula of acceleration

a=\dfrac{v_{f}-v_{i}}{t}

a=\dfrac{10-30}{3}

a=\dfrac{-20}{3}

a=-6.67\ m/s^2

Negative sign shows the car is slowing down.

Hence, The acceleration of car is 6.67 m/s².

7 0
4 years ago
The surface of the dock is 6 feet above the water. If you pull the rope in at a rate of 2 ft/sec, how quickly is the boat approa
Oduvanchick [21]

Answer:

The boat is approaching the dock at a rate of <u>2.5 ft/s</u>.

Explanation:

Let the rope length be 'l' at any time 't', the distance of boat from dock be 'b' at any time 't'.

Given:

The height of dock above water (h) = 6 feet

Rate of pull of rope or rate of change of rope is, \frac{dl}{dt}=2\ ft/s

As clear from the question, the height is fixed and only the length 'l' and distance 'b' varies with time 't'.

Now, the above situation represents a right angled triangle as shown below.

Using Pythagoras Theorem, we have:

l^2=h^2+b^2\\\\l^2=6^2+b^2\\\\l^2=36+b^2----------(1)

Now, differentiating the above equation with time 't', we get:

2l\frac{dl}{dt}=0+2b\frac{db}{dt}\\\\l\frac{dl}{dt}=b\frac{db}{dt}\\\\\frac{db}{dt}=\frac{l}{b}\frac{dl}{dt}------(2)

Now, the distance 'b' can be calculated using 'l=10 ft' in equation (1). This gives,

b^2=10^2-36\\\\b=\sqrt{64}=8\ ft

Now, substituting all the given values in equation (2) and solve for \frac{db}{dt}. This gives,

\frac{db}{dt}=\frac{10}{8}\times 2\\\\\frac{db}{dt}=2.5\ ft/s

Therefore, the boat is approaching the dock at a rate of 2.5 ft/s.

7 0
3 years ago
A 1.2 kg block sliding on a horizontal frictionless surface is attached to a horizontal spring with k =480 N/m. Let x be the dis
Angelina_Jolie [31]

Answer:

(a). The frequency is 3.18 Hz.

(b). The amplitude of the block's motion is 0.255 m.

(c). The expression for x as a function of time is x=0.255\cos(19.9 t+\dfrac{\pi}{2})

Explanation:

Given that,

Mass of block = 1.2 kg

Spring constant = 480 N/m

Speed = 5.2 m/s

We need to calculate the frequency

Using formula of frequency

f=\dfrac{1}{2\pi}\sqrt{\dfrac{480}{1.2}}

f=3.18\ Hz

The frequency is 3.18 Hz.

(b). We need to calculate the amplitude of the block's motion

Using relation of equation of amplitude and kinetic energy

\dfrac{1}{2}\times kA^2=\dfrac{1}{2}\times mv^2

Put the value into the formula

\dfrac{1}{2}\times500\times A^2=\dfrac{1}{2}\times1.2\times(5.2)^2

A^2=\dfrac{1.2\times(5.2)^2}{500}

A=\sqrt{\dfrac{1.2\times(5.2)^2}{500}}

A=0.255\ m

The amplitude of the block's motion is 0.255 m.

(c). We need to write the expression for x as a function of time

x=A\cos(\omega t+\phi)

Put the value into the equation

x=0.255\cos(19.9 t+\dfrac{\pi}{2})

The expression for x as a function of time is x=0.255\cos(19.9 t+\dfrac{\pi}{2})

Hence, This is the required solution.

6 0
3 years ago
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