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3 years ago

Which phase of cell division is shown?

Physics

1 answer:

timurjin [86]3 years ago

5 0

Answer:

This is prophase as the nucleus disappears. Anaphase is when the chromosomes separate. Telophase is when the cell is about the divide.

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A 40.0-mH inductor is connected to a North American electrical outlet (ΔVrms = 120 V, f = 60.0 Hz). Assuming the energy stored i

trasher [3.6K]

Explanation:

It is given that,

Inductance, $L=40\ mH=40\times 10^{-3}\ H$

RMS value of voltage, $v_{rms}=120\ V$

Frequency, f = 60 Hz

We need to find the energy stored at t = (1 /185) s. It is assumed that energy stored in the inductor is zero at t = 0. So,

The current flowing through the inductor is given by :

$I_t=\dfrac{V_o}{X_L}\ (sin\ \omega t-\dfrac{\pi}{2})$

$I_t=\dfrac{\sqrt{2} V_{rms}}{X_L}\ (sin\ \omega t-\dfrac{\pi}{2})$

$I_t=\dfrac{120\sqrt{2}}{2\pi f L}\ sin(2\pi f t-\dfrac{\pi}{2})$

$I_t=\dfrac{120\sqrt{2}}{2\pi\times 60\times 40\times 10^{-3}}\ sin(2\pi \times 60\times \dfrac{1}{185})-\dfrac{\pi}{2})$

$I_t=\dfrac{120\sqrt2}{15.07}\ sin(2\pi \times 60\times \dfrac{1}{185}-\dfrac{\pi}{2})$

I = 0.091 A

Energy stored in the inductor is, $U=\dfrac{1}{2}LI^2$

$U=\dfrac{1}{2}\times 40\times 10^{-3}\times (0.091)^2$

U = 0.000165 Joules

Hence, this is the required solution.

6 0

3 years ago

The tape in a videotape cassette has a total

Softa [21]

Answer:

1.37 rad/s

Explanation:

Given:

Total length of the tape is, $d= 297$ m

Total time of run is, $t = 2.1$ hours

We know, 1 hour = 3600 s

So, 2.1 hours = 2.1 × 3600 = 7560 s

So, total time of run is, $t= 7560$ s

Inner radius is, $r = 10\ mm = 0.01\ m
$

Outer radius is, $R = 47\ mm = 0.047\ m
$

Now, linear speed of the tape is, $v =\frac{d}{t}=\frac{297}{7560}=0.039\ m/s
$

Let the same angular speed be $\omega$ .

Now, average radius of the reel is given as the sum of the two radii divided by 2.

So, average radius is, $R_{avg}=\frac{R+r}{2}=\frac{0.047+0.01}{2}=\frac{0.057}{2}=0.0285\ m
$

Now, common angular speed is given as the ratio of linear speed and average radius of the tape. So,

$\omega=\dfrac{v}{R_{avg}}\\\\\\\omega=\dfrac{0.039}{0.0285}\\\\\\\omega=1.37\ rad/s
$

Therefore, the common angular speed of the reels is 1.37 rad/s.

5 0

3 years ago

Two small objects each with a net charge of +Q exert a force of magnitude F on each other. We replace one of the objects with an

Alona [7]

Answer:

F'= 4F/9

Explanation:

Two small objects each with a net charge of +Q exert a force of magnitude F on each other. If r is the distance between them, then the force is given by :

$F=\dfrac{kQ^2}{r^2}$ ...(1)

Now, if one of the objects with another whose net charge is + 4Q is replaced and also the distance between +Q and +4Q charges is increased 3 times as far apart as they were. New force is given by :

$F'=\dfrac{kQ\times 4Q}{(3r)^2}\\\\F'=\dfrac{4kQ^2}{9r^2}$ .....(2)

Dividing equation (1) and (2), we get :

$\dfrac{F}{F'}=\dfrac{\dfrac{kQ^2}{r^2}}{\dfrac{4kQ^2}{9r^2}}\\\\\dfrac{F}{F'}=\dfrac{kQ^2}{r^2}\times \dfrac{9r^2}{4kQ^2}\\\\\dfrac{F}{F'}=\dfrac{9}{4}\\\\F'=\dfrac{4F}{9}$