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4 years ago

I need help for a review sheet

Physics

1 answer:

siniylev [52]4 years ago

3 0

Tin is Sn, atomic number 47 is Silver, Mass of sodium is 22.9 u

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A particle travels clockwise on a circular path of diameter R, monitored by a sensor on the circle at point P; the other endpo

kotykmax [81]

We make a graphic of this problem to define the angle.

The angle we can calculate through triangle relation, that is,

$sin\theta = \frac{c}{QP}\\sin\theta = \frac{c}{R}\\\theta=sin^{-1}\frac{c}{R}$

With this function we should only calculate the derivate in function of c

$\frac{d\theta}{dc} = \frac{1}{\sqrt{1-\frac{c^2}{R^2}}}(\frac{c}{R})'\\\frac{d\theta}{dc} = \frac{1}{\sqrt{R^2-c^2}}$

That is the rate of change of $\theta$ .

b) At this point we need only make a substitution of 0 for c in the equation previously found.

$\frac{d\theta}{dc}\big|_{c=0} = \frac{1}{\sqrt{R^2-0}}\\\frac{d\theta}{dc}\big|_{c=0} = \frac{1}{R}$

Hence we have finally the rate of change when c=0.

6 0

3 years ago

A flutist assembles her flute in a room where the speed of sound is 342 m/s. When she plays the note A, it is in perfect tune wi

sertanlavr [38]

Answer:

5.15348 Beats/s

4.55 mm

Explanation:

$v_1$ = Velocity of sound = 342 m/s

$v_2$ = Velocity of sound = 346 m/s

$f_1$ = First frequency = 440 Hz

Frequency is given by

$f_2=\frac{v_2}{2L_1}\\\Rightarrow f_2=\frac{346}{2\times 0.38863}\\\Rightarrow f_2=445.15348\ Hz$

Beat frequency is given by

$|f_1-f_2|=|440-445.15348|=5.15348\ Beats/s$

Beat frequency is 5.15348 Hz

Wavelength is given by

$\lambda_1=\frac{v_1}{f}\\\Rightarrow \lambda_1=\frac{342}{440}\\\Rightarrow \lambda_1=0.77727\ m$

Relation between length of the flute and wavelength is

$\lambda_1=2L_1\\\Rightarrow L_1=\frac{\lambda_1}{2}\\\Rightarrow L_1=\frac{0.77727}{2}\\\Rightarrow L_1=0.38863\ m$

At v = 346 m/s

$\lambda_2=\frac{v_2}{f}\\\Rightarrow \lambda_2=\frac{346}{440}\\\Rightarrow \lambda_1=0.78636\ m$

$L_2=\frac{\lambda_2}{2}\\\Rightarrow L_2=\frac{0.78636}{2}\\\Rightarrow L_2=0.39318\ m$

Difference in length is

$\Delta L=L_2-L_1\\\Rightarrow \Delta L=0.39318-0.38863\\\Rightarrow \Delta L=0.00455\ m=4.55\ mm$

It extends to 4.55 mm

7 0

3 years ago

Several light bulbs, each of resistance 1.5 Ω, are connected in a series across a 120 V source of emf. If the current through th

Leni [432]

<h3><u>Answer;</u></h3>

40 light bulbs

<h3><u>Explanation</u>;</h3>

The total resistance of components or bulbs in series is given as the sum of resistance of all the components.

Thus; if there are bulbs in series each with a resistance of 1.5 Ω, the the total resistance will be; 1.5nΩ

From the ohms law;

V = IR , where V is the voltage, I is the current and R is the resistor.

Thus; R = V/i

R = 120/2

= 60 Ω

But, there are n bulbs each with 1.5 Ω; thus there are;

n = 60/1.5

<u> = 40 Bulbs </u>

7 0

3 years ago

A capacitor with an initial potential difference of 185 V is discharged through a resistor when a switch between them is closed

GrogVix [38]

Answer:

a. $\tau = 2.1161 s$

b. $V(18.8 \ s) = 0.0256 \ V$

Explanation:

The equation for the voltage V of discharging capacitor in an RC circuit at time t is:

$V(t) = V_0 e^{(- \frac{t}{\tau}) }$

where $V_0$ is the initial voltage, and $\tau$ is the time constant.

For our problem, we know

$V_0 = 185 \ V$

and

$V(10 \ s) = V_0 e^{(- \frac{10 \ s}{\tau}) } = 1.64 \ V$

$185 \ V \ e^{(- \frac{10 \ s}{\tau}) } = 1.64 \ V$

$e^{(- \frac{10 \ s}{\tau}) } = \frac{1.64 \ V}{ 185 \ V }$

$ln (e^{(- \frac{10 \ s}{\tau}) } ) = ln (\frac{1.64 \ V}{ 185 \ V })$

$- \frac{10 \ s}{\tau} = ln (\frac{1.64 \ V}{ 185 \ V })$

$\tau = \frac{-10 \s}{ln (\frac{1.64 \ V}{ 185 \ V }) }$

This gives us

$\tau = 2.1161 s$

and this is the time constant.

At t = 18.8 s we got:

$V(18.8 \ s) = 185 \ V \ e^{(- \frac{18.8 \ s}{2.1161 s}) }$

$V(18.8 \ s) = 0.0256 \ V$

4 0

3 years ago

In a pith ball experiment, the two pith balls are at rest. The magnitude of the tension in each string is |T|=0.55N, and the ang

Len [333]

Answer:

Explanation: Two pith ball will repel each other . they will remain balaced due to tension in the spring whose one component balances the weight and the other balances the repulsive force on each.

The gravitational force will be balanced by T cos 27.33 and the electrostatic repulsive force will be balanced by T sin27.33

Fg =T cos 27.33

= .55 X .888

= .49 N

Fq = T sin27.33

=.55 x .459

= .25 N.

5 0

3 years ago