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3 years ago

Resolve the weight of the box to find the component of the weight acting parallel to the slope.

Physics

1 answer:

tekilochka [14]3 years ago

4 0

Answer:

here's your answer below

Explanation:

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A 12 V battery is connected to a 1200 Ω resistor. How much current is flowing through the resistor?

Oduvanchick [21]

Explanation:

Use Ohm's law.

V = IR

12 V = I (1200 Ω)

I = 0.01 A

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3 years ago

Need help with question 1) and 2) please ASAP ??? 15 points ?

Vaselesa [24]

1- interaction between 2 objects
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3 years ago

A police siren of frequency fsiren is attached to a vibrating platform. The platform and siren oscillate up and down in simple h

babunello [35]

Answer:

he maximum frequency occurs when the denominator is minimum

f’= f₀ $\frac{343}{343 + v_s}$

Explanation:

This is a doppler effect exercise, where the sound source is moving

f = fo $\frac{v}{v-v)s}$ when the source moves towards the observer

f ’=f_o $\frac{v}{v+v_{sy}}$ Alexandrian source of the observer

the maximum frequency occurs when the denominator is minimum, for both it is the point of maximum approach of the two objects

f’= f₀ $\frac{343}{343 + v_s}$

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3 years ago

Need help with this science!!

noname [10]

Lol, I remember I helped my sister with this question but anyways, the card flew because of inertia and the quarter fell in the glass because of gravity :).

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3 years ago

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: