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3 years ago

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Andy is entering early adolescence. although he used to be a big help to his mother by doing household chores, he now refuses, s

aying that the chores are "women's jobs." he also has a new interest in competitive sports and is being assertive in his interactions with friends. andy is most likely experiencing
a. gender intensification.
b. gender segregation.
c. gender typing.
d. the timing of puberty effect

1 answer:

o-na [289]3 years ago

8 0

Answer:

a. gender intensification

Explanation:

This theory states that in the adolescence both girls and boys are pressured to follow gender roles established culturally.

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Two positive point charges that are equal in magnitude are fixed in place, one at x = 0.00 m and the other at x = 1.00 m, on the

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Answer:

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Explanation:

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Let the third charge is Q is placed at a distance x from the origin so that the charge is in equilibrium.

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So,

$\frac{KQq}{x^{2}}=\frac{KQq}{\left ( 1-x \right )^{2}}$

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4 years ago

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An airplane with a speed of 92.3 m/s is climbing upward at an angle of 51.1 ° with respect to the horizontal. When the plane's a

devlian [24]

Answer:

a

$D = 1162.7 \ m$

b

$\beta =- 65.55^o$

Explanation:

From the question we are told that

The speed of the airplane is $u = 92.3 \ m/s$

The angle is $\theta = 51.1^o$

The altitude of the plane is $d = 532 \ m$

Generally the y-component of the airplanes velocity is

$u_y = v * sin (\theta )$

=> $u_y = 92.3 * sin ( 51.1 )$

=> $u_y = 71.83 \ m/s$

Generally the displacement traveled by the package in the vertical direction is

$d = (u_y)t + \frac{1}{2}(-g)t^2$

=> $-532 = 71.83 t + \frac{1}{2}(-9.8)t^2$

Here the negative sign for the distance show that the direction is along the negative y-axis

=> $4.9t^2 - 71.83t - 532 = 0$

Solving this using quadratic formula we obtain that

$t = 20.06 \ s$

Generally the x-component of the velocity is

$u_x = u * cos (\theta)$

=> $u_x = 92.3 * cos (51.1)$

=> $u_x = 57.96 \ m/s$

Generally the distance travel in the horizontal direction is

$D = u_x * t$

=> $D = 57.96 * 20.06$

=> $D = 1162.7 \ m$

Generally the angle of the velocity vector relative to the ground is mathematically represented as

$\beta = tan ^{-1}[\frac{v_y}{v_x } ]$

Here $v_y$ is the final velocity of the package along the vertical axis and this is mathematically represented as

$v_y = u_y - gt$

=> $v_y = 71.83 - 9.8 * 20.06$

=> $v_y = -130.05 \ m/s$

and v_x is the final velocity of the package which is equivalent to the initial velocity $u_x$

So

$\beta = tan ^{-1}[-130.05}{57.96 } ]$

$\beta =- 65.55^o$

The negative direction show that it is moving towards the south east direction

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Answer:

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Explanation:

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