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3 years ago

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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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A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.62 m/s2 for t1 = 20 s. At that point the

arsen [322]

Total displacement of the car: 405 m

Explanation:

The first part of the motion of the car is a uniformly accelerated motion, so we can use the suvat equation

$s_1=ut_1+\frac{1}{2}a_1 t_1^2$

where:

u = 0 is the initial velocity (the car starts from rest)

$t_1 = 20 s$ is the time elapsed in the 1st part

$a_1=1.62 m/s^2$ is the acceleration of the car in the 1st part

$s_1$ is the displacement of the car in the 1st part

Solving for $s_1$ ,

$s_1=0+\frac{1}{2}(1.62)(20)^2=324 m$

We can also find the velocity of the car after these 20 seconds using the equation:

$v_1 = u +a_1 t_1 = 0 + (1.62)(20)=32.4 m/s$

Now we can find the distance covered by the car in the 2nd part, where it decelerates after having seen the tree limb on the road. We can do it by using the suvat equation:

$s_2 = (\frac{v_1 + v_2}{2})t_2$

where:

$v_1=32.4 m/s$ is the initial velocity at the beginning of the 2nd phase

$v_2=0$ is the final velocity (the car comes to a stop)

$t_2=5 s$ is the time elapsed in the 2nd phase

Substituting,

$s_2=\frac{32.4+0}{2}(5)=81 m$

So, the total displacement of the car is

$s=s_1+s_2=324+81=405 m$

Learn more about accelerated motion:

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3 years ago

When is an object magnetic?

Sliva [168]

When it has a magnetic field

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3 years ago

which statement is a Hypothesis? A: Most of the earthworms moved to the shaded area during experiment. B: If an earthworm is giv

True [87]

Answer:

B

Explanation:

It's saying what you think

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3 years ago

An 800 N man climbs 5 m up a ladder. How much gravitational potential energy does he gain?

Artemon [7]

Answer:

4000J

Explanation:

Given parameters:

Weight of the man = 800N

Height of ladder = 5m

Unknown:

Gravitational potential energy gained = ?

Solution:

The gravitational potential energy is due to the position of a body.

Gravitational potential energy = weight x height

Now insert the parameters;

Gravitational potential energy = 800 x 5 = 4000J

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3 years ago

Through what potential difference would you need to accelerate an alpha particle, starting from rest, so that it will just reach

Svetlanka [38]

Answer:

$\Delta V = 1.8 \times 10^7 V$

Explanation:

GIVEN

diameter = 15 fm = $15 \times 10^{-15}$ m

we use here energy conservation

$K_{i}+U_{i} =K_{f}+U_{f}$

there will be some initial kinetic energy but after collision kinetic energy will zero

$K_{i} + 0 = 0 + \frac{1}{4 \pi \epsilon _{0}} \frac{(2e)(92e)}{7.5 \times 10^{-15}}$

on solving these equations we get kinetic energy initial

$KE_{i} = 5.65\times 10 ^{-12} \times \frac {1 eV}{1.6 \times 10^{-19}}$

$KE_{i} = 35.33$ J ..............(i)

That is, the alpha particle must be fired with 35.33 MeV of kinetic energy. An alpha particle with charge q = 2 e

and gains kinetic energy K =e∆V ..........(ii)

by accelerating through a potential difference ∆V

Thus the alpha particle will

just reach the ${238}_U$ nucleus after being accelerated through a potential difference ∆V

equating (i) and second equation we get

e∆V = 35.33 Me V

$\Delta V = \frac{35.33}{2} MV\\\Delta V = 1.8 \times 10^7 V$

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3 years ago