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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.

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I'm very confused. Thanks for whoever helps me :)

sergij07 [2.7K]

(C). Remember gravity provides an acceleration of 9.81m/s^2, so the y component of velocity initial is zero because it isn’t already falling, and we have the height, so basically we use the kinematic equation vf^2=vi^2+2ad, substitute given values and you get vf^2=2(9.81)(65) which is 1275, when you take the square root you get 35.7m/s for final velocity
(B). Then you use vf=vi+at to get the equation 35.7=(9.81)t, when you divide out you get 3.64s for time t
(A). Finally, since we assume that there is no acceleration or deceleration horizonatally, we just multiply the time taken for it to hit the ground and the initial speed ((3.64)(35.7)) to get 129.96, with significant figures I would round that to 130 metres.
**this is in the order that I felt was easiest to answer**

6 0

3 years ago

Rank the wavelengths of the following quantum particles from the largest to the smallest. If any have equal wavelengths, display

goldfiish [28.3K]

The wavelengths of the following quantum particles from the largest to the smallest is (d) > (a) = (e) > (b) > (c)

De Broglie proposed that because light has both wave and particle properties, matter exhibits both wave and particle properties. This property has been explained as the dual behavior of matter.
From his observations, de Broglie derived the relationship between the wavelength and momentum of matter. This relationship is known as de Broglie's relationship.

De Broglie's relationship is given by $\lambda=\frac{h}{mv}$ .

This can be written as $\lambda=\frac{h}{p}$ .....(1) where λ is known as de Broglie wavelength and p is momentum , h = Plank’s constant .

As we know that mass of proton is greater than electron and photon .

(a)For photon , the momentum is given by $p=\frac{E}{c}$ ...(2) where c is the speed is the speed of light .

Putting E = 3eV in equation (2) , we get

$p=\frac{3\times 1.6\times10^-^1^9}{3\times 10^8} \\p=1.6\times10^-^2^7Js/m$

Putting this value of p in equation (1) , we get

$\lambda=\frac{6.62\times10^{-34}}{1.6\times10^{-27}}\\\lambda=4.13\times10^{-7}$

(b) As we know that kinetic energy is given by

$K.E=\frac{1}{2} mv^2\\\\2K.E = mv^2\\2K.E\times m = m^2v^2\\2K.E\times m = (mv)^2\\2K.E\times m = p^2\\\\\sqrt{2K.E\times m }=p$ ...(3)

Where mass of electron is $9.1\times10^-^3^1 kg$ .

Putting K.E = 3eV in equation (3) , we get

$\sqrt{2K.E\times m }=p\\p=\sqrt{2\times3\times1.6\times10^{-19} \times 9.31\times10^{-31} }\\p=\sqrt{89.376\times10^{-50}} \\p=9.45\times10^{-25}Js/m$

Putting this value of p in equation (1) , we get

$\lambda=\frac{6.62\times10^{-34}}{9.45\times10^{-25}}\\\lambda=0.7005\times10^{-9}$

$\sqrt{2K.E\times m }=p\\p=\sqrt{2\times3\times1.6\times10^{-19} \times 1.67\times10^{-27} }\\p=\sqrt{16.032\times10^{-46}} \\p=4.003\times10^{-23}Js/m$

Putting this value of p in equation (1) , we get

$\lambda=\frac{6.62\times10^{-34}}{4.003\times10^{-23}}\\\lambda=1.65\times10^{-11}$

(d) Putting E = 0.3eV in equation (2) , we get

$p=\frac{0.3\times 1.6\times10^-^1^9}{3\times 10^8} \\p=1.6\times10^-^2^8Js/m$

Putting this value of p in equation (1) , we get

$\lambda=\frac{6.62\times10^{-34}}{1.6\times10^{-28}}\\\lambda=4.13\times10^{-6}$

(e) Putting $p=3eV/c$ in equation (1) , we get

$\lambda=\frac{6.62\times10^{-34}\\\times3\times10^8}{3\times1.6\times10^{-19}}\\\lambda=4.13\times10^{-7}$

On comparing the wavelength order should be (d) > (a) = (e) > (b) > (c) .

Learn more about de brogile here :

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6 0

2 years ago

A 200.0 g block rests on a frictionless, horizontal surface. It is pressed against a horizontal spring with spring constant 4500

kenny6666 [7]

Answer:

6 m/s

Explanation:

Given that :

mass of the block m = 200.0 g = 200 × 10⁻³ kg

the horizontal spring constant k = 4500.0 N/m

position of the block (distance x) = 4.00 cm = 0.04 m

To determine the speed the block will be traveling when it leaves the spring; we applying the work done on the spring as it is stretched (or compressed) with the kinetic energy.

i.e $\frac{1}{2} kx^2 = \frac{1}{2} mv^2$

$kx^2 = mv^2$

$4500* 0.04^2 = 200*10^{-3} *v^2$

$7.2 =200*10^{-3}*v^{2}$

$v^{2} =\frac{7.2}{200*10^{-3}}$

$v =\sqrt{\frac{7.2}{200*10^{-3}}}$

v = 6 m/s

Hence,the speed the block will be traveling when it leaves the spring is 6 m/s

5 0

4 years ago

Which picture correctly shows the path of refracted light rays given an object outside the focal point? Select one: a. A b. B c.

Lina20 [59]

Answer:

Answer is C because light travels in a sight line but when light pass through a refractor the light from the source changes direction when passes through a refractor

3 0

2 years ago

A squirrel runs along an overhead telephone wire that stretches from the top of one pole to the next. The creature is initially

xxTIMURxx [149]

Answer:

xf = - 3.16 m

Explanation:

the squirrel was initially in the position xi = 3.65 m, then it had a displacement of Δx = -6.81 m.

The negative sign indicates that it moved in the opposite direction, so we must subtract this displacement Δx = -6.81 m to the initial position xi = 3.65 m, to find its final position.

3.65 m - 6.81 m = - 3.16 m

4 0

3 years ago