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2 years ago

Find the voltage across the 15 Q resistor. [?] V No links please

Physics

1 answer:

Margaret [11]2 years ago

4 0

Answer:

Explanation:

same idea as before Liam, first, find the parallel resistance in 35 || 20

(35*20) / (35+20) = 700 / 55 = 12.727272 ohms

now add the 12.727272 + 15 = 27.727272 ohms total resistance

V = IR

10 = I * 27.727272

10 / 27.727272 = I

0.360655 = I

V = IR (again, but across the 15 ohm resistor)

V = 0.360655 * 15

V = 5.4098

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A toroidal coil of N turns has a central radius b and a square cross section of side a. Find its self-inductance.

Xelga [282]

Answer:

$L = \frac{\mu_0 N^2 (a^2)}{2\pi b}$

Explanation:

As we know that magnetic field due to torroid is given as

$B = \frac{\mu_0 N i}{2\pi b}$

this is approximately constant magnetic field along the axis of the torroid

now the flux linked with one coil of the torroid is given as

$\phi = B.A$

$\phi = \frac{\mu_0 N i}{2\pi b}(a^2)$

now total flux of N number of coils is given as

$\phi_{total} = \frac{\mu_0 N^2 i(a^2)}{2\pi b}$

now we know that self inductance is the property of coil in which flux of the coil will link with the current in the coil

So we know that

$L = \frac{\phi}{i}$

$L = \frac{\mu_0 N^2 (a^2)}{2\pi b}$

3 0

2 years ago

Diesel fuel is used in the engine of trucks that carry dirt fruit and other cargo. Fuel is burned in engines to make the motor m

Marina86 [1]

Answer: Fuel is burned in engines to make the motor move.

( Chemical to Mechanical )

Explanation:

during combustion of the diesel ( when the fuel is burnt in the engine of the vehicle, the diesel ( chemical energy ) is transformed or converted to Mechanical energy. This mechanical energy is what the truck uses in moving. Without the combustion of the fuel the vehicle won’t move and the combustion of diesel is achieved through compression unlike that of fuel.

4 0

2 years ago

What is the voltage in a circuit that has a current of 10.0 amps and a resistance of 28.5 ohms?

marin [14]

Voltage = Current (I) × Resistance (R)
V = 10 × 28.5 = 285v

6 0

3 years ago

I need help ASAP. This is for 15 points

kramer

Answer:

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7 0

2 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

1 year ago