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2 years ago

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Plot StartRoot 1.5 EndRoot and StartRoot 1.9 EndRoot on the number line to find which inequalities are true. Check all that appl

y. StartRoot 1.5 EndRoot < 1.5 StartRoot 1.9 EndRoot < 1.9 StartRoot 1.5 EndRoot < StartRoot 1.9 EndRoot 1.2 > StartRoot 1.5 EndRoot > 1.3 1.3 < StartRoot 1.9 EndRoot < 1.4

2 answers:

Vera_Pavlovna [14]2 years ago

4 0

Answer:

A, B, C, E

Explanation:

now gimme thanks please

Natali5045456 [20]2 years ago

4 0

Answer: 1 2 3 5 I got it right :D

Explanation:

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A charge is divided q1 and (q-q1)what will be the ratio of q/q1 so that force between the two parts placed at a given distance i

Arturiano [62]

Answer:

$q / q_{1} = 2$ , assuming that $q_{1}$ and $(q - q_{1})$ are point charges.

Explanation:

Let $k$ denote the coulomb constant. Let $r$ denote the distance between the two point charges. In this question, neither $k$ and $r$ depend on the value of $q_{1}$ .

By Coulomb's Law, the magnitude of electrostatic force between $q_{1}$ and $(q - q_{1})$ would be:

$\begin{aligned}F &= \frac{k\, q_{1}\, (q - q_{1})}{r^{2}} \\ &= \frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\end{aligned}$ .

Find the first and second derivative of $F$ with respect to $q_{1}$ . (Note that $0 < q_{1} < q$ .)

First derivative:

$\begin{aligned}\frac{d}{d q_{1}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\right] \\ &= \frac{k}{r^{2}}\, \left[\frac{d}{d q_{1}} [q\, q_{1}] - \frac{d}{d q_{1}}[{q_{1}}^{2}]\right]\\ &= \frac{k}{r^{2}}\, (q - 2\, q_{1})\end{aligned}$ .

Second derivative:

$\begin{aligned}\frac{d^{2}}{{d q_{1}}^{2}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q - 2\, q_{1})\right] \\ &= \frac{(-2)\, k}{r^{2}}\end{aligned}$ .

The value of the coulomb constant $k$ is greater than $0$ . Thus, the value of the second derivative of $F$ with respect to $q_{1}$ would be negative for all real $r$ . $F\!$ would be convex over all $q_{1}$ .

By the convexity of $\! F$ with respect to $\! q_{1} \!$ , there would be a unique $q_{1}$ that globally maximizes $F$ . The first derivative of $F\!$ with respect to $q_{1}\!$ should be $0$ for that particular $\! q_{1}$ . In other words:

$\displaystyle \frac{k}{r^{2}}\, (q - 2\, q_{1}) = 0$ <em>.</em>

$2\, q_{1} = q$ .

$q_{1} = q / 2$ .

In other words, the force between the two point charges would be maximized when the charge is evenly split:

$\begin{aligned} \frac{q}{q_{1}} &= \frac{q}{q / 2} = 2\end{aligned}$ .

3 0

2 years ago

Using monochromatic light of 410 nm wavelength, a single thin slit forms a diffraction pattern, with the first minimum at an ang

VLD [36.1K]

Answer:

c. 307 nm

Explanation:

angular position of first dark fringe = λ / d , λ is wavelength and d is width of slit .

(40 x π ) / 180 = 410 / d

angular position of second dark fringe = 2 x λ / d , λ is wavelength and d is width of slit .

(60 x π ) / 180 = 2 x λ / d

Dividing these equations

60 / 40 = 2 x λ / 410

λ = 307.5 nm.

5 0

3 years ago

A hollow cylinder of mass 2.00 kg, inner radius 0.100 m, and outer radius 0.200 m is free to rotate without friction around a ho

Aneli [31]

Answer:

h=2.86m

Explanation:

In order to give a quick response to this exercise we will use the equations of conservation of kinetic and potential energy, the equation is given by,

$\Delta PE_i + \Delta KE_i = \Delta PE_f +\Delta KE_f$

There is no kinetic energy in the initial state, nor potential energy in the end,

$mgh+0=0+KE_f$

In the final kinetic energy, the energy contributed by the Inertia must be considered, as well,

$mgh = (\frac{1}{2}mv^2+\frac{1}{2}I\omega^2)$

The inertia of the bodies is given by the equation,

$I=\frac{m(R_1^2+R^2_2)}{2}$

$I=\frac{2(0.2^2+0.1^2)}{2}$

$I=0.05Kgm^2$

On the other hand the angular velocity is given by

$\omega =\frac{v}{R_2}=\frac{4}{1/5} = 2rad/s$

Replacing these values in the equation,

$(0.5)(9.8)(h) =\frac{1}{2}*0.5*4^2+\frac{1}{2}*0.05*20^2$

Solving for h,

$h=2.86m$

5 0

2 years ago

Determine the binding energy per nucleon of an mg-24 nucleus. the mg-24 nucleus has a mass of 24.30506. a proton has a mass of 1

My name is Ann [436]

The mass of Mg-24 is 24.30506 amu, it contains 12 protons and 12 neutrons.

Theoretical mass of Mg-24:

The theoretical mass of Mg-24 is:

Hydrogen atom mass = 12 × 1.00728 amu = 12.0874 amu

Neutron mass = 12 x 1.008665 amu = 12.104 amu

Theoretical mass = Hydrogen atom mass + Neutron mass = 24.1913 amu

Note that the mass defect is:

Mass defect = Actual mass - Theoretical mass : 24.30506 amu- 24.1913 amu= 0.11376 amu

Calculating the binding energy per nucleon:

$\frac{B.E.}{nucleon}=\frac{(0.11376amu)(931Mev/amu}{24nucleons} = 4.41294 Mev/nucleon$

So approximately 4.41294 Mev/necleon

3 0

2 years ago

What is the average speed of a car that travels 60 meters in 2<br> seconds?

blondinia [14]

Answer:

30 m/s

Explanation:

Speed is distance over time. 60 meters / 2 seconds, = 30 m/s.

6 0

2 years ago