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3 years ago

12

What happens to the chemical structure of water when it changes state?

1 answer:

Marta_Voda [28]3 years ago

5 0

Answer:

c) Water molecules melt into gas molecules.

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Erica is working in the lab. She wants to remove the fine dust particles suspended in a sample of oil. Which method is she most

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Option a is right answer that is reverse osmosis.

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This type of radiation is smallest and highest in energy: ____ ___.

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Gamma rays have the highest energies and the shortest wavelengths.

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The total magnification of an image is determined by.

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Magnification of the objective lens used and the magnification of the ocular lens.

Explanation: I hope you have/had an amazing day today<3

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2 years ago

As mentioned in the text, the tangent line to a smooth curve r(t) = ƒ(t)i + g(t)j + h(t)k at t = t0 is the line that passes thro

LiRa [457]

Answer:

$x = t$

$y = \frac{1}{3}t$

$z =t$

Explanation:

Given

$r(t) = f(t)i + g(t)j + h(t)k$ at $t = 0$

Point: $(f(t0), g(t0), h(t0))$

$r(t) = ln\ t_i + \frac{t-1}{t+2}j + t\ ln\ tk$ , $t0 = 1$ -- Missing Information

Required

Determine the parametric equations

$r(t) = ln\ ti + \frac{t-1}{t+2}j + t\ ln\ tk$

Differentiate with respect to t

$r'(t) = \frac{1}{t}i +\frac{3}{(t+2)^2}j + (ln\ t + 1)k$

Let t = 1 (i.e $t0 = 1$ )

$r'(1) = \frac{1}{1}i +\frac{3}{(1+2)^2}j + (ln\ 1 + 1)k$

$r'(1) = i +\frac{3}{3^2}j + (0 + 1)k$

$r'(1) = i +\frac{3}{9}j + (1)k$

$r'(1) = i +\frac{1}{3}j + (1)k$

$r'(1) = i +\frac{1}{3}j + k$

To solve for x, y and z, we make use of:

$r(t) = f(t)i + g(t)j + h(t)k$

This implies that:

$r'(1)t = xi + yj + zk$

So, we have:

$xi + yj + zk = (i +\frac{1}{3}j + k)t$

$xi + yj + zk = it +\frac{1}{3}jt + kt$

By comparison:

$xi = it$

Divide by i

$x = t$

$yj = \frac{1}{3}jt$

Divide by j

$y = \frac{1}{3}t$

$zk = kt$

Divide by k

$z = t$

Hence, the parametric equations are:

$x = t$

$y = \frac{1}{3}t$

$z =t$

3 0

3 years ago

A flat loop of wire consisting of a single turn ofcross-sectional area 7.90 cm2is perpendicular to a magnetic field that increas

Westkost [7]

To develop this problem, it is necessary to apply the concepts related to Faraday's law and Magnetic Flow, which is defined as the change that the magnetic field has in a given area. In other words

$\Phi = BA Cos\theta$

Where

B= Magnetic Field

A = Area

$\theta =$ Angle between magnetic field lines and normal to the area

The differentiation of this value allows us to obtain in turn the induced emf or electromotive force.

In this case we have that the flat loop of wire is perpendicular to the magnetic field, therefore the angle is 0 degrees, since its magnetic field acts parallel to the area:

$\theta =$ 0 then our expression can be written as

$\Phi = BA$

From the same value of the electromotive force we have to

$\epsilon = -\frac{d\Phi}{dt}$

Replacing we have

$\epsilon = -A\frac{B}{dt}$

Replacing with our values we have that

$\epsilon = -(7.9*10^{-4}m)\frac{(3.5-0.5)}{0.1}$

$\epsilon = -0.0237V$

Therefore the magnitude of the induced emf in the loop is 0.0237V

On the other hand we have that the current by Ohm's Law can be defined as

$I = \frac{\epsilon}{R}$

For the given value of the resistance and the previously found potential we have to

$I = \frac{0.237}{1.3}$

$I= 0.0182A$

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3 years ago