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Vikentia [17]
3 years ago
11

Red cabbage juice can be used as an acid-base indicator. When a solution is highly acidic, the juice turns red. As the acidity d

ecreases and the solution becomes basic, the color changes from red to violet to green to yellow. Red cabbage juice is used as an indicator of some common household products. Which product will turn yellow when cabbage juice is added? A) coffee B) ammonia C) vinegar D) bottled water
Physics
2 answers:
Viktor [21]3 years ago
5 0

Answer:

B) Ammonia

Explanation:

Red cabbage juice is an acid-base indicator. It turns red for acidic medium and as the pH increase it turns to violet, green and then yellow.

It means for a basic solution, it would turn yellow. From the given options, coffee and vinegar are acidic. Therefore, they would turn red. Bottled water is neutral. Ammonia is a weak base. It would the indicator yellow.

Vesna [10]3 years ago
4 0
Answer is B ammonia cuz its basic
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Explanation:

Given that,

Velocity of the proton in lab frame u_{x} = 0.6c

Velocity of the observer v= 0.8c

We need to calculate the velocity of the proton with respect to the observer

Using formula of velocity

u'=\dfrac{v-u_{x}}{1-\dfrac{u_{x}v}{c^2}}

u'=\dfrac{0.8c-0.6c}{1-\dfrac{0.6c\times0.8c}{c^2}}

u'=c(\dfrac{0.8-0.6}{1-(0.8\times0.6)})

u'=0.385c

(a). We need to calculate the total energy of the proton in the lab frame

Using formula of kinetic energy

K.E=m_{0}c^2(\dfrac{1}{\sqrt{1-\dfrac{u_{x}^2}{c^2}}}-1)

Where, Proton mass energy = m₀c²

Put the value into the formula

K.E=938.28\times(\dfrac{1}{\sqrt{1-\dfrac{(0.6c)^2}{c^2}}}-1)

K.E=938.28\times(\dfrac{1}{\sqrt{1-(0.6)^2}}-1)

K.E=234.57\ MeV

(b). We need to calculate the kinetic energy of the proton in the observer

Using formula of kinetic energy

K.E=m_{0}c^2(\dfrac{1}{\sqrt{1-\dfrac{u'^2}{c^2}}}-1)

K.E=938.28\times(\dfrac{1}{\sqrt{1-\dfrac{(0.385)^2}{c^2}}}-1)

K.E=938.28\times(\dfrac{1}{\sqrt{1-(0.385)^2}}-1)

K.E=78.366\ MeV

(c). We need to calculate the momentum of the proton with respect to observer

Using formula of momentum

P_{obs}=\dfrac{m_{0}u'^2}{\sqrt{1-\dfrac{u'^2}{c^2}}}

We know that,

Proton mass energy = m₀c²

m_{0}=\dfrac{938.28}{c^2}

P_{obs}=\dfrac{\dfrac{938.28}{c^2}\times(0.385c)^2}{\sqrt{1-\dfrac{(0.385c)^2}{c^2}}}

P_{obs}=\dfrac{938.28\times(0.385)^2}{\sqrt{1-0.385^2}}

P_{obs}=150.69\ MeV/c

Hence, This is required solution.

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3 years ago
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