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4 years ago

The flame in an operational Bunsen burner undergoes a

2 answers:

6 0

<span> The flame in an operational Bunsen burner undergoes a
A. phase change
B. nuclear change
C. chemical change
D. physical change

Flames are a chemical change since you cannot separate the components physically once you've combined them so the answer is C. chemical change</span>

pshichka [43]4 years ago

3 0

Can you explain the question more?

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An object that is moving in a linear path with an acceleration in the direction opposite to the motion has a(n) ______________ v

Bas_tet [7]

An object that is moving in a linear path with an acceleration in the direction opposite to the motion has a(n) _______changing_______ velocity.

5 0

4 years ago

A stereo speaker is rated at P1000 = 52 W of output at 1000 Hz. At 20 Hz, the sound intensity level LaTeX: \betaβ decreases by 1

baherus [9]

Answer:

The value of the power is $P_c = 38.55 \ W$

Explanation:

From the question we are told that

The power rating $P_{1000} =P_b= 52 \ W$

The frequency is $f = 1000 \ Hz$

The frequency at which the sound intensity decreases $f_k = 20 \ Hz$

The decrease in intensity is by $\beta = 1.3 dB$

Generally the initial intensity of the speaker is mathematically represented as

$\beta_1 = 10 log_{10} [\frac{P_b}{P_a} ]$

Generally the intensity of the speaker after it has been decreased is

$\beta_2 = 10 log_{10} [\frac{P_c}{P_a} ]$

$\beta_1-\beta_2 = 10 log_{10} [\frac{P_c}{P_a} ]- 10 log_{10} [\frac{P_b}{P_a} ]$

=> $\beta = 10 log_{10} [\frac{P_c}{P_a} ]- 10 log_{10} [\frac{P_b}{P_a} ]= 1.3$

=> $\beta =10log_{10} [\frac{\frac{P_b}{P_a}}{\frac{P_c}{P_a}} ] = 1.3$

=> $\beta =10log_{10} [\frac{P_b}{P_c} ] = 1.3$

=> $10log_{10} [\frac{P_b}{P_c} ] = 1.3$

=> $log_{10} [\frac{P_b}{P_c} ] = 0.13$

taking atilog of both sides

$[\frac{P_b}{P_c} ] = 10^{0.13}$

=> $[\frac{52}{P_c} ] = 10^{0.13}$

=> $P_c = \frac{52}{1.34896}$

=> $P_c = 38.55 \ W$

3 0

3 years ago

Identify the energy transformation that takes when place when you apply the brakes on a bicycle.

valina [46]

Um Kinetic, mechanical, potential idk

The rider is riding the nick so kinetic then mechanical making the bike move then potential stopping the bike

4 0

3 years ago

A body is traveling at 5.0 m/s along the positive direction of an x axis; no net force acts on the body. An internal explosion s

loris [4]

To solve this problem it is necessary to apply the concepts related to the conservation of momentum and conservation of kinetic energy.

By definition kinetic energy is defined as

$KE = \frac{1}{2} mv^2$

Where,

m = Mass

v = Velocity

On the other hand we have the conservation of the moment, which for this case would be defined as

$m*V_i = m_1V_1+m_2V_2$

Here,

m = Total mass (8Kg at this case)

$m_1=m_2 =$ Mass each part

$V_i =$ Initial velocity

$V_2 =$ Final velocity particle 2

$V_1 =$ Final velocity particle 1

The initial kinetic energy would be given by,

$KE_i=\frac{1}{2}mv^2$

$KE_i = \frac{1}{2}8*5^2$

$KE_i = 100J$

In the end the energy increased 100J, that is,

$KE_f = KE_i KE_{increased}$

$KE_f = 100+100 = 200J$

By conservation of the moment then,

$m*V_i = m_1V_1+m_2V_2$

Replacing we have,

$(8)*5 = 4*V_1+4*V_2$

$40 = 4(V_1+V_2)$

$V_1+V_2 = 10$

$V_2 = 10-V_1$ (1)

In the final point the cinematic energy of EACH particle would be given by

$KE_f = \frac{1}{2}mv^2$

$KE_f = \frac{1}{2}4*(V_1^2+V_2^2)$

$200J=\frac{1}{2}4*(V_1^2+V_2^2)$ (2)

So we have a system of 2x2 equations

$V_2 = 10-V_1$

$200J=\frac{1}{2}4*(V_1^2+V_2^2)$

Replacing (1) in (2) and solving we have to,

200J=\frac{1}{2}4*(V_1^2+(10-V_1)^2)

PART A: $V_1 = 10m/s$

Then replacing in (1) we have that

PART B: $V_2 = 0m/s$

8 0

3 years ago

A student runs 120 meters in 10 seconds. What is their speed? Question 6 options: 1.2 m/s 1200 m/s 120 m/s 12 m/s

aleksklad [387]

12m/s

Explanation:

Given parameters:

Distance run by student = 120 meters

Time = 10 seconds

Unknown:

Speed = ?

Solution:

Speed is the rate of change of distance with time:

Speed = $\frac{distance}{time}$

Now input the parameters:

Speed = $\frac{120}{10}$ = 12m/s

Learn more:

Speed brainly.com/question/4581270

#learnwithBrainly

5 0

4 years ago