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3 years ago

11

Which of the following developmental milestones is likely achieved before a child's first birthday?

2 answers:

olga_2 [115]3 years ago

3 0

Answer by YourHope:

Hi! ^-^

Which of the following developmental milestones is likely achieved before a child's first birthday?

Reach for nearby objects!

I have a baby sister that did this so I know this is true!

Have a BEAUTIFUL day~

Lostsunrise [7]3 years ago

3 0

Answer:

Reach for nearby objects

Explanation:

Which of the following developmental milestones is likely achieved before a child's first birthday?

Kick a ball

Feel embarrassed

Stand on one foot

Reach for nearby objects

For a 12 month old child, he /she is suppose to recognise few words. recognise wants without tears. hold hands and reach out for nearby objects. Can walk few distances alone.

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1. What is the new kinetic energy of the 1900kg ship on the right moving at 4 m/s?

neonofarm [45]

Explanation:

That`s is the answer, just check

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3 years ago

To push a 26.0 kg crate up a frictionless incline, angled at 25.0° to the horizontal, a worker exerts a force of 209 N parallel

alukav5142 [94]

Answer:

(a) W = +397.1 J

(b) W = -204.6 J

(c) W = 0

(d) W= + 192.5 J

Explanation:

Work (W) is defined as the product of force (F) by the distance (d)the body travels due to this force. :

W= F*d Formula ( 1)

The forces that perform work on an object must be parallel to its displacement.

The forces perpendicular to the displacement of an object do not perform work on it.

The work is positive (W+) if the force has the same direction of movement of the object.

The work is negative (W-) if the force has the opposite direction of the movement of the object.

Problem development

(a) Work performed by the worker's applied force on the box .

W= 209 N * 1.9 m = +397.1 J

(b) Work performed by the gravitational force on the crate

We calculate the weight component parallel to the displacement of the box:

We define the x-axis in the direction of the inclined plane ,25.0° to the horizontal.

We define the y-axis and in the direction of the plane perpendicular to the inclined plane.

W= m*g=26*9.8= 254.8N : total box weight

Wx= W*sen25.0°= 254.8*sen25.0°= 107.68 N

W = -Wx *d =107.68 N *1.9 m= -204.6 J

(c) Work performed by normal force (N) exerted by the incline on the crate

The force N is perpendicular to the displacement, then:

W=0

(d) Total work done on the crate

W = 397.1 J -204.6 J

W = 192.5 J

4 0

3 years ago

Your mom is making lunch which forces are acting on your mom

N76 [4]

The forces acting on your mom while cooking is Air resistance and the force of friction

<u>Explanation:</u>

<u>1. Air resistance:</u>

In simple words, Air resistance can be stated as the type of friction between the air and the other materials.
In this scenario, there will be an air resistance and the air hits the mom while cooking via the doors or windows

<u>2. The force of friction:</u>

In simple words, friction can be stated as, the resistance that one surface or object encounters when moving over another.
While cooking the food mom would experience the friction since friction is the transfer of heat, and cooking is the process of receiving that heat.

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3 years ago

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If a glass rod rubbed with Silk piece is taken to a ball pen rubeed with wool the ball pen_____

WARRIOR [948]

Answer:

The glass rod remains charged due to electrification by the silk

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3 years ago

The resistance of a very fine aluminum wire with a 20 μm × 20 μm square cross section is 1200 Ω . A 1200 Ω resistor is made by w

notsponge [240]

Explanation:

The given data is as follows.

Resistance (R) = 1200 ohm, Area (A) = $20 \times 10^{-6}$ m (as $1 \mu m = 10^{-6} m$ )

Diameter (d) = 2.3 mm = $2.3 \times 10^{-3}$ m

First, we will calculate the length as follows.

R = $\rho \frac{L}{A}$

Here, $\rho$ = resistivity of aluminium = $2.65 \times 10^{-8}$

Putting the given values above and we will calculate the value of length as follows.

R = $\rho \frac{L}{A}$

1200 = $2.65 \times 10^{-8} \times \frac{L}{20 \times 10^{-6}}$

L = $9.056 \times 10^{5}$

As the circumference of circular wire = $2 \pi r$

or, = $2 \times \pi \times \frac{d}{2}$

= $\pi \times d$

And, number of turns will be calculated as follows.

No. of turns × Circumference = Length of wire

No. of turns × $3.14 \times 2.3 \times 10^{-3} = 9.056 \times 10^{5}$

= $1.25 \times 10^{8}$

Thus, we can conclude that $1.25 \times 10^{8}$ turns of wire are needed.

6 0

3 years ago