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2 years ago

If the trend changed toward traditional (pre-World War II) families, how would that affect women’s rights?

1 answer:

7 0

If the trend changed toward traditional (pre-World War II) families, the women’s rights are employment in manufacturing sector.

<h3>What is World war?</h3>

The war between two countries to take over each other's kingdom using weapons to kill each other.

Before the world war, the army needs armor, weapons, guns and tanks. Their manufacturing is only possible with many workers to work for long hours. If the men are not enough, then women are given opportunities to work with them.

Thus, the women’s rights are employment in manufacturing sector when trend changed traditional families.

Learn more about world war.

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In order to create a charged object, you need to ________.

Brums [2.3K]

In order to create a charged object you need to transfer electrons either away or to the object by induction, conduction, or friction

Induction is without contact(like bringing a charged object to a electroscope charges the leaves at the bottom)
Conduction is with contact(like the previous answer, wore touching transfers the charge from a source to the object)
Friction(rubbing a balloon on wool)

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2 years ago

Complete the equation???

soldi70 [24.7K]

Here in nuclear reaction we can say that sum of neutrons and protons in reactant side and product side will be same always

Here mass number on the product side is given to us

so sum of mass number is given as

$A_1 + A_2 = 265 + 1 = 266$

now on the reactant side also the number must be same

$A_1' + A_2' = 58 + x$

now we will have

$58 + x = 266$

$x = 208$

Now number of protons on product side is given as

$P_1 + P_2 = 108 + 0$

now we also know that atomic number of Fe is 26

so now we will have

$P_1' + P_2' = 108$

$26 + P_2' = 108$

$P_2' = 82$

now the equation is given as

$_{26}^{58}Fe + _{82}^{208}Pb = _{108}^{266}Hs + _0^1X$

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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

3 years ago

A camcorder has a power rating of 17 watts. If the output voltage from its battery is 5 volts, what current does it use?

shtirl [24]

power = (voltage) x (current)

17 w = (5 v) x (current)

current = (17 w) / (5 v) = 3.4 Amperes

6 0

3 years ago

If the partial pressures of the three gases in the tank are 38.39 atm of O2, 3.38 atm of He, and 25.33 atm of N2, what is the to

Tasya [4]

Answer:

67.1 atm

Explanation:

From Dalton's law of Partial Pressures, the total partial pressure equals the sum of the individual partial pressures.

P = P₁ + P₂ + P₃

P₁ = 38.39 atm of O₂

P₂ = 3.38 atm of He

P₃ = 25.33 atm of N₂

So, P = P₁ + P₂ + P₃ = 38.39 atm + 3.38 atm + 25.33 atm = 67.1 atm

4 0

3 years ago