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3 years ago

8

La _______________ abarca de los dos a los doce años. Se caracteriza por un continuo proceso de adaptación motora, cognoscitiva,

emotiva y social del niño en el ambiente en el que vive.

1 answer:

Anna11 [10]3 years ago

5 0

Answer:

please do well to ask questions in English. This will help people provide you answers ASAP. Thank you

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A student drops a rock from a bridge to the water 12.7 m below. with what speed does the rock strike the water? the acceleration

kolezko [41]

The rock strike the water with the speed of 15.78 m/sec.

The speed by which rock hit the water is calculated by the formula

v= $\sqrt{2gh}$

v= $\sqrt{2*9.8*12.7}$

v=15.78 m/sec

Hence, the rock strike the water with the speed of 15.78 m/sec.

7 0

3 years ago

The temperature of a black body is 500 and its radiation is of wavelength 600 . If the number of oscillators with energy is 100

stiks02 [169]

Answer: An equation is missing in your question below is the missing equation

a) ≈ 8396

b) 150 nm/k

Explanation:

<u>A) Determine the number of Oscillators in the black body</u>

number of oscillators = 8395

attached below is the detailed solution

<u>b) determine the peak wavelength of the black body </u>

Black body temperature = 20,000 K

applying Wien's law / formula

λmax = b / T ------ ( 1 )

T = 20,000 K

b = 3 * 10^6 nm

∴ λmax = 150 nm/k

4 0

3 years ago

The force exerted by the wind on the sails of a sailboat is Fsail = 330 N north. The water exerts a force of Fkeel = 210 N east.

Elena L [17]

Answer:

The magnitude of the acceleration is $a_r = 1.50 \ m/s^2$

The direction is $\theta = 32.5 6^o$ north of east

Explanation:

From the question we are told that

The force exerted by the wind is $F_{sail} = (330 ) \ N \ north$

The force exerted by water is $F_{keel} = (210 ) \ N \ east$

The mass of the boat(+ crew) is $m_b = 260 \ kg$

Now Force is mathematically represented as

$F = ma$

Now the acceleration towards the north is mathematically represented as

$a_n = \frac{F_{sail}}{m_b}$

substituting values

$a_n = \frac{330 }{260}$

$a_n = 1.269 \ m/s^2$

Now the acceleration towards the east is mathematically represented as

$a_e = \frac{F_{keel}}{m_b }$

substituting values

$a_e = \frac{210}{260}$

$a_e =0.808 \ m/s^2$

The resultant acceleration is

$a_r = \sqrt{a_e^2 + a_n^2}$

substituting values

$a_r = \sqrt{(0.808)^2 + (1.269)^2}$

$a_r = 1.50 \ m/s^2$

The direction with reference from the north is evaluated as

Apply SOHCAHTOA

$tan \theta = \frac{a_e}{a_n}$

$\theta = tan ^{-1} [\frac{a_e}{a_n } ]$

substituting values

$\theta = tan ^{-1} [\frac{0.808}{1.269 } ]$

$\theta = tan ^{-1} [0.636 ]$

$\theta = 32.5 6^o$

5 0

3 years ago

Do you think there might be a point in space between Earth and the Moon where the gravity of each would pull on an object equall

Firlakuza [10]

Answer: Yes.

Explanation:

Assuming Earth and Moon are isolated is space, it is possible to have a point where Earth and Moon will pull at an object with equal force.

That point will be closer to the Moon than the Earth because Moon's gravitational field strength is weaker than Earth's gravitational field strength.

7 0

3 years ago

11. The primary job of the cell wall is to

hodyreva [135]

11. protect the cell and keep its shape.

12. chloroplast

3 0

3 years ago

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