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4 years ago

6

What would be the result of voltage while joining 2 or more different voltages of batteries with parallel ? and explain me what

will happen on the circuit ?

1 answer:

cupoosta [38]4 years ago

3 0

That's not a good idea. You don't want to connect batteries with different voltages in parallel. What happens is large currents among the batteries themselves, and some of the batteries quickly die.

When the batteries are connected in SERIES, they behave as a single battery whose voltage is the sum of their individual voltages.

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Arrange an 8-, 12-, and 16-Ω resistor in a combination that has a total resistance of 8.89 Ω pls with de work

kolezko [41]

20 ohms in parallel with 16 ohm= 8.89

20x16/20+16. Product over sum

8 0

3 years ago

An airplane is flying in a horizontal circle at a speed of 480 km/h (). If its wings are tilted at angle =40° to the horizontal

german

Answer:

$R = 2162 m$

Explanation:

When wings of the airplane makes an angle of 40 degree with the horizontal so here we can say that force due to air is having two components

$F_y = mg$

$F_x = \frac{mv^2}{R}$

now we know that

$F_y = F cos40$

$F_x = F sin40$

also we know that

$v = 480 km/h$

$v = 133.3 m/s$

now plug in all data in above equations

$tan 40 = \frac{v^2}{Rg}$

$R = \frac{v^2}{g tan40}$

$R = \frac{133.3^2}{9.8 tan40}$

$R = 2162 m$

6 0

3 years ago

An earthquake emits both S-waves and P-waves which travel at different speeds through the Earth. A P-wave travels at 9000 m/s an

Sergio [31]

Answer:

The distance between earthquake center and the measuring station is 1350 kilometers.

Explanation:

Let the earthquake center be at a distance of 'S' meters from the recording station.

Now from the basic relation of distance, speed and time we know that

$Distance=Speed\times Time$

For a Primary wave (P wave) let us assume that it appraoches the measuring station after $t_{1}$ minutes

Thus making use of the above relation we have

$Distance=V_{p}\times Time\\\\\therefore D=9000\times t_{1}.......(i)$

Now since it is given that the secondary wave (S wave) reaches the measuring spot after 2 minutes or 120 seconds thus the time taken by secondary waves to reach recorder equals $t_{1}+120$ making use of the same relation we get

$Distance=V_{s}\times Time\\\\\therefore D=5000\times (t_{1}+120).......(ii)$

Solving equation 'i' and 'ii' we get

$D=5000\times (\frac{D}{9000}+120)\\\\\therefore \frac{D}{5000}=\frac{D}{9000}+120\\\\\frac{D}{5000}-\frac{D}{9000}=120\\\\\therefore D=\frac{120}{\frac{1}{5000}-\frac{1}{9000}}=1350000meters=1350kilomerers$

8 0

3 years ago

what is the energy (in j) of a photon required to excite an electron from n = 2 to n = 8 in a he⁺ ion? submit an answer to three

grin007 [14]

Answer:

Approximately $5.11 \times 10^{-19}\; {\rm J}$ .

Explanation:

Since the result needs to be accurate to three significant figures, keep at least four significant figures in the calculations.

Look up the Rydberg constant for hydrogen: $R_{\text{H}} \approx 1.0968\times 10^{7}\; {\rm m^{-1}$ .

Look up the speed of light in vacuum: $c \approx 2.9979 \times 10^{8}\; {\rm m \cdot s^{-1}}$ .

Look up Planck's constant: $h \approx 6.6261 \times 10^{-34}\; {\rm J \cdot s}$ .

Apply the Rydberg formula to find the wavelength $\lambda$ (in vacuum) of the photon in question:

$\begin{aligned}\frac{1}{\lambda} &= R_{\text{H}} \, \left(\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}\right)\end{aligned}$ .

The frequency of that photon would be:

$\begin{aligned}f &= \frac{c}{\lambda}\end{aligned}$ .

Combine this expression with the Rydberg formula to find the frequency of this photon:

$\begin{aligned}f &= \frac{c}{\lambda} \\ &= c\, \left(\frac{1}{\lambda}\right) \\ &= c\, \left(R_{\text{H}}\, \left(\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}\right)\right) \\ &\approx (2.9979 \times 10^{8}\; {\rm m \cdot s^{-1}}) \\ &\quad \times (1.0968 \times 10^{7}\; {\rm m^{-1}}) \times \left(\frac{1}{2^{2}} - \frac{1}{8^{2}}\right)\\ &\approx 7.7065 \times 10^{14}\; {\rm s^{-1}} \end{aligned}$ .

Apply the Einstein-Planck equation to find the energy of this photon:

$\begin{aligned}E &= h\, f \\ &\approx (6.6261 \times 10^{-34}\; {\rm J \cdot s}) \times (7.7065 \times 10^{14}\; {\rm s^{-1}) \\ &\approx 5.11 \times 10^{-19}\; {\rm J}\end{aligned}$ .

(Rounded to three significant figures.)

6 0

2 years ago

A star has a declination of approximately -90°. in what direction is the Star located from the celestial equator?

nata0808 [166]

The start is located on theWest

5 0

3 years ago