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3 years ago

Is language an innate or an acquired one

1 answer:

3 0

It's acquired. Innate means you have it at birth, acquired means you picked it up. Obviously babies can't speak ;) so it's acquired.

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Max hears on the news that a total solar eclipse will be visible from his town in the coming week. Which of the following explai

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What's the energy transfer for a catapult

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Kinetic energy is the energy for a catapult.

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3 years ago

The velocity of a ball changes from ‹ 9, −6, 0 › m/s to ‹ 8.96, −6.12, 0 › m/s in 0.02 s, due to the gravitational attraction of

Alenkasestr [34]

Answer:

a) $a=(-2,-6,0)m/s^2$ , with a magnitude of $6.3m/s^2$

b) $\frac{\Delta p}{\Delta t}=(-0.24,-0.72,0)Kgm/s^2$ , with a magnitude of $0.76Kgm/s^2$

c) $F=(-0.24,-0.72,0)N$ , with a magnitude of $0.76N$

Explanation:

We have:

$v_{ix}=9m/s, v_{iy}=-6m/s, v_{iz}=0m/s\\v_{fx}=8.96m/s, v_{fy}=-6.12m/s, v_{fz}=0m/s\\t=0.02s, m=0.12Kg$

We can calculate each component of the acceleration using its definition $a=\frac{\Delta v}{\Delta t}$

$a_x=\frac{v_{fx}-v_{ix}}{t} = \frac{(8.96m/s)-(9m/s)}{0.02s} =-2m/s^2\\a_y=\frac{v_{fy}-v_{iy}}{t} = \frac{(-6.12m/s)-(-6m/s)}{0.02s} =-6m/s^2\\a_y=\frac{v_{fz}-v_{iz}}{t} = \frac{(0m/s)-(0m/s)}{0.02s} =0m/s^2\\$

The rate of change of momentum of the ball is $\frac{\Delta p}{\Delta t} = \frac{\Delta mv}{\Delta t} = \frac{m\Delta v}{\Delta t} = ma$

So for each coordinate:

$\frac{\Delta p_x}{\Delta t}=-0.24Kgm/s^2\\\frac{\Delta p_y}{\Delta t}=-0.72Kgm/s^2\\\frac{\Delta p_z}{\Delta t}=0Kgm/s^2$

And these are equal to the components of the net force since F=ma.

If magnitudes is what is asked:

$a=\sqrt{a_x+a_y+a_z} =6.3m/s^2\\F=ma=\frac{\Delta p}{\Delta t}=0.76N$

3 0

3 years ago

A person of mass 55 kg swings on a rope length 4 m from rest (when the rope makes an angle of 30 degrees with the vertical) and

vovangra [49]

Answer:

θ = 19.66°

Explanation:

To determine the angle that the rope makes with the vertical for the two people, you first take into account the potential energy of the first person before he swings on the rope:

$U=mgh$

h: distance to the ground

g: gravitational acceleration = 9.8m/s^2

m: mass of the first person = 55 kg

In the image attache below you can notice that the height h is:

$h=4-4cos(30\°)=0.53m$

Then, the potential energy is:

$U=(55kg)(9.8m/s^2)(0.53m)=285.67J$

When the first person picks up the second person (when the rope is exactly vertical), all the potential energy becomes kinetic energy. Next, when both people reaches the maximum height h' the energy must be equal to the initial potential energy of the first person:

$U'=(m_1+m_2)gh'=285.67\ J$

From the previous equation you can get h':

$h'=\frac{285.67J}{(55kg+70kg)(9.8m/s^2)}=0.2332m$

Finally, you obtain the angle between the rope at the height h,' and the vertical, by calculating the following:

$h'=4-4cos(\theta)\\\\\theta=cos^{-1}(\frac{4-h'}{4})=cos^{-1}(\frac{4-0.2332}{4})=19.66\°$

hence, the angle between the rope and the vertical, when the two people are in the rope is 19.66°

8 0

3 years ago