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Alinara [238K]
3 years ago
13

HELP ASAP!!

Physics
2 answers:
g100num [7]3 years ago
3 0

Answer:

1.C

2.B

3.1

Explanation:

1.It's a chemical change since the size of the fabric changes

2.Chemical changes involves heat,forms new substance and involves a change in the mass or matter of a substance but it isn't easily reversible.

3.In the law of Conservation of mass matter can neither be created or destroyed.Therefor there won't be a change in the mass under ordinary reaction.

rewona [7]3 years ago
3 0

According to the Law of Conservation of Mass, the total mass of the products and the total mass of the reactant in a chemical reaction are the same

The correct options are;

  • 1. B.) It is a physical change because there is no reaction
  • 2. A.) They form new substances
  • 3. D.) 1

Reason:

1. Physical change is a change in which there is no change in the chemical properties of the substances involved

Therefore, the change is a physical change because there is no reaction

The correct option is option A.)

B.) <u>It is a physical change because there is no reaction</u>

2. A difference between a chemical change and a physical change is that a new substance is formed following chemical change while no new substance is formed in a physical change

Therefore, the correct option is option A.)

A.) <u>They form new substances</u>

3. The Law of Conservation of Mass states that mass can neither be created or destroyed in a closed system

Based on the Law of Conservation of Mass, the observation that contains an error is observation #1

\begin{array}{|c|c|c|}\mathbf{Observation}  &\mathbf{Before \ Reaction}&\mathbf{During /After \ Reaction}\\&&\\1&Mass = 30 \ g&Mass = 42 \ g\end{array}

From the given data, the mass of the items in the system change from <u>30 grams</u> to <u>42 grams</u>, which is an error based on Law of Conservation of Mass

The option that gives the observation that contains an error is option D.)

D.) <u>1</u>

Learn more about the Law of Conservation of Mass, chemical and physical changes here:

brainly.com/question/14466325

brainly.com/question/18515296

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A runner starts at point A, runs around a 1-mile track and finishes the run back at point A. Which of the following statements i
Lyrx [107]

This question is incomplete because the options are missing; here is the complete question:

A runner starts at point A, runs around a 1-mile track, and finishes the run back at point A. Which of the following statements is true?

A. The runner's displacement is 1 mile.

B. The runner's displacement is zero.

C. The distance the runner covered is zero.

D. The runner's speed was zero.

The answer to this question is B. The runner's displacement is zero

Explanation:

Displacement always implies a change of position; this means an object or individual moves from point A to point B, and therefore the original position is different from the final position. Additionally, in displacement, other related factors such as the total distance the body moved and the direction of movement. In the case presented, it can be concluded there was no displacement or the displacement is zero because even when the runner moved and ran two miles, he returned to the initial position, and without a change in the position, there is no displacement.

4 0
3 years ago
I need both parts please (a) Given a material with an attenuation coefficient (a) of 0.6/cm, what is the intensity of a beam (wi
Masteriza [31]

Answer:

<h3>a.</h3>
  • After it has traveled through 1 cm : I(1 \ cm) = 0.5488 I_0
  • After it has traveled through 2 cm : I(2 \ cm) = 0.3012 I_0
<h3>b.</h3>
  • After it has traveled through 1 cm : od( 1\ cm) =  0.2606
  • After it has traveled through 2 cm :  od( 2\ cm) =  0.5211

Explanation:

<h2>a.</h2>

For this problem, we can use the Beer-Lambert law. For constant attenuation coefficient \mu the formula is:

I(x) = I_0 e^{-\mu x}

where I is the intensity of the beam, I_0 is the incident intensity and x is the length of the material traveled.

For our problem, after travelling 1 cm:

I(1 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 1 cm}

I(1 \ cm) = I_0 e^{- 0.6}

I(1 \ cm) = I_0 e^{- 0.6}

I(1 \ cm) = 0.5488 \ I_0

After travelling 2 cm:

I(2 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 2 cm}

I(2 \ cm) = I_0 e^{- 1.2}

I(2 \ cm) = I_0 e^{- 1.2}

I(2 \ cm) = 0.3012 \ I_0

<h2>b</h2>

The optical density od is given by:

od(x) = - log_{10} ( \frac{I(x)}{I_0} ).

So, after travelling 1 cm:

od( 1\ cm) = - log_{10} ( \frac{0.5488 \ I_0}{I_0} )

od( 1\ cm) = - log_{10} ( 0.5488 )

od( 1\ cm) = - (  - 0.2606)

od( 1\ cm) =  0.2606

After travelling 2 cm:

od( 2\ cm) = - log_{10} ( \frac{0.3012 \ I_0}{I_0} )

od( 2\ cm) = - log_{10} ( 0.3012 )

od( 2\ cm) = - (  - 0.5211)

od( 2\ cm) =  0.5211

3 0
3 years ago
You wish to watch TV at exactly 85 dB and no louder to avoid long term damage to your hearing. You record the sound intensity le
BigorU [14]

Answer:

1) the new power coming from the amplifier is 19.02 W

2) The distance away from the amplifier now is 5.50 m

3) u₁ = 69.24 m

Therefore have to move u₁ - u ( 69.24 - 5.50) = 63.74 farther

Explanation:

Lets say that I am at a distance "u" from the TV,

Let I₁ be the corresponding intensity of the sound at my location when sound level is 125dB

SO

S(indB) = 10log (I₁/1₀)

we substitute

125 = 10(I₁/10⁻¹²)

12.5 = log (I₁/10⁻¹²)

10^12.5 = I₁/10^-12

I₁ = 10^12.5 × 10^-12

I₁ = 10^0.5 W/m²

Now I₂ will be intensity of sound when corresponding sound level is 107 dB

107 = 10log(I₂/10⁻²)

10.7 = log(I₂/10⁻¹²)

10^10.7 = I₂ / 10^-12

I₂ = 10^10.7  ×  10^-12

I₂ = 10^-1.3 W/m²

Now since we know that

I = P/4πu² ⇒ p = 4πu²I

THEN P₁ = 4πu²I₁ and P₂ =4πu²I₂

Therefore

P₁/P₂ = I₁/I₂

WE substitute

P₂ = P₁(I₂/I₁) = 1200 × ( 10^-1.3 / 10^0.5)

P₂ = 19.02 W

the new power coming from the amplifier is 19.02 W

2)

P₁ = 4πu²I₁

u =√(p₁/4πI₁)

u = √(1200/4π × 10^0.5)

u = 5.50 m

The distance away from the amplifier now is 5.50 m

3)

Let I₃ be the intensity corresponding to required sound level 85 dB

85 = 10log(I₃/10⁻¹²)

8.5 = log (I₃/10⁻¹²)

10^8.5 = I₃ / 10^-12

I₃ = 10^8.5  × 10^-12

I₃ = 10^-3.5 w/m²

Now, I ∝ 1/u²

so I₂/I₃ = u₁²/u²

u₁ = √(I₂/I₃) × u

u₁ = √(10^-1.3 / 10^-3.5) ×  5.50

u₁ = 69.24 m

Therefore have to move u₁ - u ( 69.24 - 5.50) = 63.74 farther

8 0
3 years ago
A water tank is filled with water. What will be the pressure of the water when the level of the water is 6m?
Luden [163]

Answer:

pressure = density x g x height

= 1000 x 10 x 6 Pascal

=60000 Pascal

OR 60 kP

3 0
3 years ago
Read 2 more answers
A simple pendulum takes 128s to complete 40 oscillations what is the time period of the pendulum
Serhud [2]

Explanation:

Parameters

Number of oscillation = 40 oscillations

Time of oscillation = 128s

Formular

Period (T) = <u>No of oscillation</u>

Time taken

T = <u>128</u>

40

T=3.2s

3 0
3 years ago
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