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3 years ago

Kim was adopted as a baby and raised by loving parents in an enriched environment. Studies have shown that ________

Physics

1 answer:

erik [133]3 years ago

8 0

Answer:

Kim was adopted as a baby and raised by loving parents in an enriched environment. Studies have shown that the correlation between IQ

( intelligence quotient) scores of adopted children and those of their biologically unrelated family members reduce to zero as they grow into adulthood.

Explanation:

Kim was adopted as a baby and raised by loving parents in an enriched environment. Studies have shown that the correlation between IQ

( intelligence quotient) scores of adopted children and those of their biologically unrelated family members reduce to zero as they grow into adulthood.

Also, a loving environment is equally important for a child's upbringing.

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A rocket is launched from rest and moves in a straight line at 30.0 degrees above the horizontal with an acceleration of 35.0 m/

klemol [59]

Answer:

t = 123.59s

Explanation:

For the launch pad section:

Vf = Vo + a*t where Vo=0.

Vf = 35*25 = 875m/s

The distance traveled during the launch:

$d = Vo*t+\frac{a*t^2}{2} = 0+\frac{35*25^2}{2} = 10937.5m$

Now the projectile motion, we know that its initial speed is the speed calculated previously and the initial height is the y-component of the previously calculated distance.

$\Delta Y = Vo*sin(30)*t - \frac{g*t^2}{2}$

$-d*sin(30) = Vo*sin(30)*t - \frac{g*t^2}{2}$ where d= 10937.5m; Vo=875m/s.

Solving for t:

t1 = -11.093s t2 = 98.59s

So, the total time of flight will be:

$t_{total} = t_{launch}+t_{projectile}=25+98.59 = 123.59s$

6 0

3 years ago

Convert 15 litre into cubic metre

atroni [7]

Answer:

0.015m^3

Explanation:

1 m^3 = 1000 liters

x m^3 = 15 liters

Cross multiply

xm^3 x 1000 l = 15 l

Divide both sides by 1000

xm^3 x1000/1000 = 15/1000

xm^3 = 0.015m^3

Therefore 15 liter = 0.015m^3

5 0

3 years ago

A 20.0-kg block is initially at rest on a horizontal surface. A horizontal force of 77.0 N is required to set the block in motio

saveliy_v [14]

Answer: The coefficient of static friction is 3.85 and The coefficient of kinetic friction is 2.8

Explanation:

in the attachment

6 0

3 years ago

Hanna tosses a ball straight up with enough speed to rermain in the air for several seconds

Ipatiy [6.2K]

What a relief ! That gives her time to step out of the way, before the ball
comes crashing down in the same place where she was standing.

7 0

3 years ago

A small logo is embedded in a thick block of crown glass (n = 1.52), 4.70 cm beneath the top surface of the glass. The block is

harkovskaia [24]

The concept required to solve this problem is the optical relationship that exists between the apparent depth and actual or actual depth. This is mathematically expressed under the equations.

$d'w = d_w (\frac{n_{air}}{n_w})+d_g (\frac{n_{air}}{n_g})$

Where,

$d_g =$ Depth of glass

$n_w =$ Refraction index of water

$n_g =$ Refraction index of glass

$n_{air} =$ Refraction index of air

$d_w =$ Depth of water

I enclose a diagram for a better understanding of the problem, in this way we can determine that the apparent depth in the water of the logo would be subject to

$d'w = d_w (\frac{n_{air}}{n_w})+d_g (\frac{n_{air}}{n_g})$

$d'w = (1.7cm) (\frac{1}{1.33})+(4.2cm)(\frac{1}{1.52})$

$d'w = 4.041cm$

Therefore the distance below the upper surface of the water that appears to be the logo is 4.041cm

3 0

3 years ago