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3 years ago

Thermal energy is added to four identical samples of water what increases in each sample

2 answers:

8 0

Thermal energy is added to four identical 1.0 kg samples of water at room temperature. Which of the following increases in each sample? average charge of an electron; average density of a nucleus; average mass of a proton; average speed of a molecule. Your answer: -. Answer: D - average speed of amolecule.

belka [17]3 years ago

8 0

Answer:

Average speed of a molecule

Explanation:

If thermal energy (heat) is added to a substance or sample, according to the first law of thermodynamics this will cause the internal energy in that sample to increase.

The increase in internal energy will be manifested in that the speed of the molecules that make up the substance will increase, and therefore the average speed of a molecule increases.

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A textbook of mass 2.02 kg rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose d

Anika [276]

Answer:

a. 7.38 N b. 40.87 N c. 0.113 kg-m²

Explanation:

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3 years ago

A bicycle tire has a radius of 0.23 m. the tire spins at a constant speed with a centripetal acceleration of 42 m/s2. what is th

vredina [299]

Answer:

a = v^2/r centripetal acceleration

v = (a * r)^1/2 = (42 * .23)^1/2 = 3.1 m/s

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2 years ago

Read 2 more answers

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

Explanation:

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

If an unbalanced force of 25 newtons acts on an object that accelerates at 2.5 meters

Alinara [238K]

Answer:

Explanation:

B. Must be the answer I. E 10kg

7 0

2 years ago

40 g of magnesium and 40 g of oxygen react in a chemical reaction in a close system which of the following must result

swat32

Answer:

Option "D" is the correct answer to the following question.

Explanation:

According to The law of conservation of mass, in a chemical reaction mass neither created nor destroyed.

So,

Total product = Mass of magnesium + Mass of oxygen

Total product = 40 gram + 40 gram

Total product = 80 gram

Therefore, option "D" is the correct answer

6 0

3 years ago