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ASHA 777 [7]
3 years ago
10

Litmus paper is made from water-soluble dyes which are extracted from lichens. This paper is used as an acid-base indicator. Whi

ch of these common household substances would turn blue litmus paper red? A) ammonia B) vinegar C) baking soda D) distilled water
Physics
2 answers:
shtirl [24]3 years ago
8 0
The answer is B- Vinegar.
Alex777 [14]3 years ago
7 0

Answer: B) vinegar

Explanation: Acids are those substances which when dissolved in water give H^+ ions. They turn blue litmus red.

Bases are those substances which when dissolved in water give OH^- ions. They turn red litmus blue.

Ammonia is a base which when dissolved in water forms ammonium hydroxide and gives OH^- ions and turns red litmus blue.

NH_4OH\rightarrow NH_4^++OH^-

Vinegar is acetic acid and thus gives H^+ ions and turn blue litmus red.

CH_3COOH\rightarrow CH_3COO^-+H^+

Baking soda is a basic salt NaHCO_3 which turns red litmus blue.

Distilled water H_2O is neutral which shows no color change with red or blue litmus.

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A satellite in a circular orbit of radius R around planet X has an orbital period T. If Planet X had one-fourth as much mass, th
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<h2>Answer: 2T</h2>

According to the Third Kepler’s Law of Planetary motion <em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.</em>

In other words, this law states a relation between the orbital period T of a body (moon, planet, satellite) orbiting a greater body in space with the size R of its orbit.

This Law is originally expressed as follows (in the case of planet X and assuming we have a circular orbit):

T^{2}=\frac{4\pi^{2}}{GM}R^{3}    (1)

Where:

G is the Gravitational Constant

M=1.9(10)^{27}kg is the mass of planet X

R  is the radius of the orbit of the satellite around planet X

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

T=2\pi\sqrt{\frac{R^{3}}{GM}}   (2)

Now, we are asked to find the period when tha mass of the planet is \frac{1}{4}M. In order to do this, we have to rewrite equation (2) with this new value:

T=2\pi\sqrt{\frac{R^{3}}{G(\frac{1}{4}M)}}  (3)

Solving:

T=4\pi\sqrt{\frac{R^{3}}{G(\frac{1}{4}M)}}   (4)

On the other hand, if we multiply both sides of equation (2) by 2, we have:

2T=4\pi\sqrt{\frac{R^{3}}{GM}}    (5)

As we can see, (5) is equal to (4). This means the orbital period is twice the orignal period.

Hence, the answer is:

If Planet X had <u>one-fourth </u>as much mass, the <u>orbital period</u> of this satellite in an orbit of the same radius would be <u>2T.</u>

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