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Nezavi [6.7K]
3 years ago
5

A small flashlight bulb provides a resistance of 3 Ω to the 300 mA current that runs through it. Determine the voltage of the ba

ttery in the flashlight.
Physics
1 answer:
Contact [7]3 years ago
3 0

V=IR

V= 0.3 x 3

V= 0.9V

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How many dB greater than 4 Watts is 64 Watts?
Novay_Z [31]

Answer:

64 Watts is 12dB greater than 4 Watts

6 0
3 years ago
You are part of a design team assigned the task of making an electronic oscillator that will be the timing mechanism of a micro-
Snowcat [4.5K]

Solution :

We assume that there is a ring having a charge +Q and radius r. Electric field due to the ring at a point P on the axis is given by :

E_P=\int dE \cos

E_P=\int \frac{KdQ}{(\sqrt{r^2+x^2})^2}\times \frac{x}{\sqrt{r^2+x^2}}

\vec{E_P}=\frac{Kx}{r^2+x^2} \int dQ

\vec{E_P}=\frac{KxQ}{(r^2+x^2)^{3/2}} \hat{i}

If we put an electron on point P, then force on point e is :

\vec{F}=-|e|\vec{E_P}

F= \frac{-eKQx}{(r^2+x^2)^{3/2}}= \frac{-eKQx}{r^3[1+\frac{x^2}{r^2}]^{3/2}}

If r >> x , then    $\frac{x^2}{r^2} \approx 0$

Then,  $\frac{-eKQ}{r^3}x$

$ma =\frac{-eKQ}{r^3}x$

$a =\frac{-eKQ}{mr^3}x$

Compare, a = -ω²x

We get,

$\omega^2 = \frac{eKQ}{R^3m}$

$\omega = \sqrt{\frac{eKQ}{r^3m}}$

$2 \pi f = \sqrt{\frac{eKQ}{r^3m}}$

$f = \frac{1}{2 \pi}\sqrt{\frac{eKQ}{mr^3}}$

6 0
3 years ago
The water flowing through a 2.0 cm (inside diameter) pipe flows out through three 1.3 cm pipes. (a) If the flow rates in the thr
Verdich [7]

Answer:

a)54L/min

b)0.845

Explanation:

a) A x V=A_1V_1+ A_2V_2+A_3V_3

where suffix 1,2,3 refers to the three pipes.

            =27L/min+16L/min+11 L/min

            =54L/min

b) A x V=54L/min => \frac{\pi }{4} d^2 x v

   d= 2 cm

\frac{\pi }{4} d^2 x v = 54

v= \frac{4}{\pi } x \frac{54}{2^2}

-> A_1 x V_1=27L/min => \frac{\pi }{4} d_1^2 x v_1

d_1= 1.3cm

\frac{\pi }{4} d^2 x v_1 = 27

v_1= \frac{4}{\pi } x \frac{27}{1.3^2}

Next is to find the ratio of speed i.e \frac{v}{v_1}

\frac{4}{\pi } x \frac{54}{2^2} / \frac{4}{\pi } x \frac{27}{1.3^2} => \frac{54}{27} \frac{1.3^2}{2^2}

\frac{v}{v_1}= 0.845

8 0
2 years ago
0.002 written in scientific notation
Oliga [24]

Answer:0,002 = 2 x 10⁻³

Explanation:

0,002 = 2 / 1000 = 2 / 10³ = 2 x 10⁻³

3 0
3 years ago
Which of these components is present in this circuit schematic?
Svetlanka [38]
Its resistor :}...............

4 0
3 years ago
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