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pickupchik [31]

2 years ago

13

you place a one-liter substance in a 2-liter bottle and tightly cover the bottle. the substance expands until it completely fill

s the bottle. what state is the substance in?

1 answer:

9966 [12]2 years ago

4 0

Answer:

The substance in is at state of gas.

Explanation:

Please see attavhed document

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A particle initially located at the origin has an acceleration of vector a = 2.00ĵ m/s2 and an initial velocity of vector v i =

natali 33 [55]

With acceleration

$\mathbf a=\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)\,\mathbf j$

and initial velocity

$\mathbf v(0)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i$

the velocity at time <em>t</em> (b) is given by

$\mathbf v(t)=\mathbf v(0)+\displaystyle\int_0^t\mathbf a\,\mathrm du$

$\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\displaystyle\int_0^t\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)\,\mathbf j\,\mathrm du$

$\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\bigg|_{u=0}^{u=t}$

$\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf j$

We can get the position at time <em>t</em> (a) by integrating the velocity:

$\mathbf x(t)=\mathbf x(0)+\displaystyle\int_0^t\mathbf v(u)\,\mathrm du$

The particle starts at the origin, so $\mathbf x(0)=\mathbf0$ .

$\mathbf x(t)=\displaystyle\int_0^t\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\,\mathrm du$

$\mathbf x(t)=\left(\left(8.00\dfrac{\rm m}{\rm s}\right)u\,\mathbf i+\dfrac12\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf j\right)\bigg|_{u=0}^{u=t}$

$\mathbf x(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)t\,\mathbf i+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)t^2\,\mathbf j$

Get the coordinates at <em>t</em> = 8.00 s by evaluating $\mathbf x(t)$ at this time:

$\mathbf x(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)(8.00\,\mathrm s)\,\mathbf i+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)(8.00\,\mathrm s)^2\,\mathbf j$

$\mathbf x(8.00\,\mathrm s)=(64.0\,\mathrm m)\,\mathbf i+(64.0\,\mathrm m)\,\mathbf j$

so the particle is located at (<em>x</em>, <em>y</em>) = (64.0, 64.0).

Get the speed at <em>t</em> = 8.00 s by evaluating $\mathbf v(t)$ at the same time:

$\mathbf v(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)(8.00\,\mathrm s)\,\mathbf j$

$\mathbf v(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(16.0\dfrac{\rm m}{\rm s}\right)\,\mathbf j$

This is the <em>velocity</em> at <em>t</em> = 8.00 s. Get the <em>speed</em> by computing the magnitude of this vector:

$\|\mathbf v(8.00\,\mathrm s)\|=\sqrt{\left(8.00\dfrac{\rm m}{\rm s}\right)^2+\left(16.0\dfrac{\rm m}{\rm s}\right)^2}=8\sqrt5\dfrac{\rm m}{\rm s}\approx17.9\dfrac{\rm m}{\rm s}$

5 0

3 years ago

How can you determine the number of valence electrons in atom without drawing a Bohr model?

grigory [225]

Using your periodic table if you look at it 3-11 are tansition metals so the horizontal Group Number will help if the group number has to digits just remove the one so if it were to be 13, the valence would be 3, if it were 14 the valence would be ,4 if it were 15, the valence would be 5, if it were 16 the valence would be 6, if it were 17 the valence would be 7 if it were group 18 the valence would be 8 so if anymore help needed to explain hit me up

5 0

3 years ago

Read 2 more answers

Which process is involved in the formation of a galaxy

zubka84 [21]

Answer:

SUN

Explanation:

EARTH

8 0

4 years ago

Read 2 more answers

A 2200-kg railway freight car coasts at 4.1 m/s underneath a grain terminal, which dumps grain directly down into the freight ca

777dan777 [17]

Answer:

The answer is " $2.41 \times 10^3$ "

Explanation:

Given:

$m_i = 2000 \ kg \\\\v_i= 4.1 \ \frac{m}{s} \\\\v_f = 3.4 \ \frac{m}{s} \\$

Using formula:

$\to m_iv_i = m_fv_f \\\\\to m_f= \frac{m_iv_i}{v_f}$

$p_i, p_f$ = system initial and final linear momentum.

$V_i, v_f$ = system original and final linear pace.

$m_i$ = original weight of the car freight.

$m_f$ = car's maximum weight

$= \frac{ 2000 \times 4.1}{3.4}\\\\= \frac{ 8.2\times 10^3}{3.4}\\\\= 2.41 \times 10^3$

$\boxed{m_f = 2.41 \times 10^3}$

8 0

3 years ago

A pendulum is made of a small sphere of mass 0.250 kg attached to a lightweight string 1.20 m in length. As the pendulum swings

olchik [2.2K]

Answer:

0.833 N

Explanation:

Formula for Kinetic Energy $E_k = \frac{mv^2}{2}$

Formula for Potential Energy $E_p = mgy$

First we need to find the vertical distance between the maximum-angle position and the pendulum lowest point:

Using the swinging point as the reference, the vertical distance from the maximum-angle (34 degree) position to the swinging point is:

$L * cos(34^o) = 1.2cos(34^o) = 1.2*0.83 = 0.995 \approx 1 m$

At the lowest position, pendulum is at string length to the swinging point, which is 1.2 m. Therefore, the vertical distance between the maximum-angle position and the pendulum lowest point would be

y = 1.2 - 1 = 0.2 m.

As the pendulum is traveling from the maximum-angle position to the lowest point position, its potential energy would be converted to the kinetic energy.

By law of energy conservation:

$E_k = E_p$

$\frac{mv^2}{2} = mgy$

$v^2 = 2gy$

$v = \sqrt{2gy}$

Substitute $g = 10 m/s^2$ and y = 0.2 m:

$v = \sqrt{2 * 10 * 0.2} = \sqrt{4} = 2 m/s$

At lowest point, pendulum would generate centripetal tension force on the string:

$F = m\frac{v^2}{L}$

We can substitute mass m = 0.25, rotation radius L = 1.2 m and v = 2 m/s:

$F = 0.25\frac{2^2}{1.2} = 0.833 N$

5 0

3 years ago