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2 years ago

When baking soda is mixed with lemon juice, bubbles are formed with the evolution of a gas. What type of change is it? Explain.

Physics

1 answer:

lorasvet [3.4K]2 years ago

5 0

$\large{\underline{\underline{\pmb{\frak {\color {red}{Question:}}}}}}$

$\sf \red{When \: baking \: soda \: is \: mixed \: with \: lemon \: juice, \: bubbles \: are \: formed \: with \: the \: evolution \: of \: a \: gas.}$

$\large{\underline{\underline{\pmb{\frak {\color {blue}{Answer:}}}}}}$

When baking soda is mixed with lemon juice, bubbles are formed with the evolution of a gas. The gas is formed in the reaction is Carbon dioxide. $CO_{2}$ is formed.

The change which happened in this reaction is a chemical change.

$\boxed{ \frak \green{Explanation}}$

Since, in chemical change we can't bring a substance to it's actual form how it was in earlier.

Examples: burning of paper is chemical, since we can't get the fine paper again after it is burnt.

Thus, the above reaction is also a chemical change, since we can't get back the lemon juice how it was earlier.

$\boxed{ \frak \red{Brainlysamurai}}$

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3 years ago

Sandy is whirling a ball attached to a string in a horizontal circle over his head. If Sandy doubles the speed of the ball, what

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The tension in the string B) It quadruples.

Explanation:

The ball is in uniform circular motion in a horizontal circle, so the tension in the string is providing the centripetal force that keeps the ball in circular motion. So we can write:

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T is the tension in the string

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r is the radius of the circle (the lenght of the string)

In this problem, we are told that the speed of the ball is doubled, so

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Substituting into the previous equation, we find the new tension in the string:

$T' = m \frac{(2v)^2}{r}=4(m\frac{v^2}{r})=4T$

Therefore, the tension in the string will quadruple.

Learn more about circular motion:

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3 years ago

Army is standing still on the ground; Bill is riding his bicycle at 5 m/s eastward: and Carlos is driving his car at 15 m/s west

Aleks04 [339]

Explanation:

Given that,

Bill is riding his bicycle at 5 m/s eastward: and Carlos is driving his car at 15 m/s westward.

Taking eastward as positive direction, we have:

$v_B=+5\ m/s$ is the velocity of Bill with respect to Amy (which is stationary)

$v_c=15\ m/s$ is the velocity of Carlos with respect to Amy.

Bill is moving 5 m/s eastward compared to Amy at rest, so the velocity of Bill's reference frame is

$v_B=+5\ m/s$

Therefore, Carlos velocity in Bill's reference frame will be

$v_c'=-15\ m/s-(+5\ m/s)\\\\=-20\ m/s$

So, the magnitude is 20 m/s and the direction is westward (negative sign).

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3 years ago

SCALCET8 3.9.018.MI. A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the

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Answer:

The length of his shadow is decreasing at a rate of 1.13 m/s

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Also, the height of the man, h which is parallel to H forms a right-angled triangle of width, L which is the length of the shadow.

By similar triangles,

H/D = h/L

L = hD/H

Also, when the man is 4 m from the building, the length of his shadow is L = D - 4

So, D - 4 = hD/H

H(D - 4) = hD

H = hD/(D - 4)

Since h = 2 m and D = 12 m,

H = 2 m × 12 m/(12 m - 4 m)

H = 24 m²/8 m

H = 3 m

Since L = hD/H

and h and H are constant, differentiating L with respect to time, we have

dL/dt = d(hD/H)/dt

dL/dt = h(dD/dt)/H

Now dD/dt = velocity(speed) of man = -1.7 m/s ( negative since he is moving towards the building in the negative x - direction)

Since h = 2 m and H = 3 m,

dL/dt = h(dD/dt)/H

dL/dt = 2 m(-1.7 m/s)/3 m

dL/dt = -3.4/3 m/s

dL/dt = -1.13 m/s

So, the length of his shadow is decreasing at a rate of 1.13 m/s

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3 years ago