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2 years ago

is using electrolysis to separate water molecules in hydrogen and oxygen atoms , a physical or chemical change?

Physics

1 answer:

DedPeter [7]2 years ago

8 0

Technically it's a chemical change even though the process is reversible.

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Answer plz also give reason

user100 [1]

See this suggested solution.

1. Let a force F' is the vector sum of the forces P and Q, then it is shown on the attached picture and marked with red color.

2. according to the condition the force F holds the object, then F should have the same length as the force F' and the opposite direction.

3. using the conditions described in 2. the answer is C.

4 0

2 years ago

3. A concrete highway is built of slabs 12 m long (20o C). How wide should the expansion cracks between the slabs be (at 20o C)

lesantik [10]

Answer:

$\Delta x=2.304\times 10^{-2}\ m=2.304\ cm$ is the minimum gap between the slabs

Explanation:

Given:

length of the concrete highway, $l=12\ m$

coefficient of thermal expansion, $\alpha=12\times 10^{-6}\ ^{\circ}C^{-1}$

range of temperature variation, $(-30^{\circ}C\ to\ 50^{\circ}C) \Rightarrow \Delta T=80^{\circ}C$

<u>Now form the equation of thermal expansion:</u>

$\Delta l=l\times \alpha\times \Delta T$

$\Delta l=12\times (12\times 10^{-6})\times 80$

$\Delta l=1.152\times 10^{-2}\ m$

Since each slab of the highway expands by the above length so the minimum gap between the slabs to prevent buckling:

$\Delta x=2\times \Delta l$

$\Delta x=2\times( 1.152\times 10^{-2})$

$\Delta x=2.304\times 10^{-2}\ m=2.304\ cm$ is the minimum gap between the slabs

4 0

2 years ago

Consider the mass and velocity values of Objects A and B below.

777dan777 [17]

Answer:

the same

Explanation:

8 0

1 year ago

Where is the electron located in an atom?

stepan [7]

Around the nucleus or in the shell of the atom. Hope this makes sense.

8 0

2 years ago

One recently discovered extrasolar planet, or exoplanet, orbits a star whose mass is 0.70 times the mass of our sun. This planet

Stels [109]

0.078 times the orbital radius r of the earth around our sun is the exoplanet's orbital radius around its sun.

Answer: Option B

<u>Explanation:</u>

Given that planet is revolving around the earth so from the statement of centrifugal force, we know that any

$\frac{G M m}{r^{2}}=m \omega^{2} r$

The orbit’s period is given by,

$T=\sqrt{\frac{2 \pi}{\omega r^{2}}}=\sqrt{\frac{r^{3}}{G M}}$

Where,

$T_{e}$ = Earth’s period

$T_{p}$ = planet’s period

$M_{s}$ = sun’s mass

$r_{e}$ = earth’s radius

Now,

$T_{e}=\sqrt{\frac{r_{e}^{3}}{G M_{s}}}$

As, planet mass is equal to 0.7 times the sun mass, so

$T_{p}=\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}$

Taking the ratios of both equation, we get,

$\frac{T_{e}}{T_{p}}=\frac{\sqrt{\frac{r_{e}^{3}}{G M_{s}}}}{\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}}$

$\frac{T_{e}}{T_{p}}=\sqrt{\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}}$

$\left(\frac{T_{e}}{T_{p}}\right)^{2}=\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}$

$\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}=\frac{r_{e}^{3}}{r_{p}^{3}}$

$\frac{r_{e}}{r_{p}}=\left(\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}$

Given $T_{p}=9.5 \text { days }$ and $T_{e}=365 \text { days }$

$\frac{r_{e}}{r_{p}}=\left(\left(\frac{365}{9.5}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=\left(\frac{133225}{90.25} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=(2108.82)^{\frac{1}{3}}$

$r_{p}=\left(\frac{1}{(2108.82)^{\frac{1}{3}}}\right) r_{e}=\left(\frac{1}{12.82}\right) r_{e}=0.078 r_{e}$

7 0

2 years ago