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3 years ago

Please help in finding the Total Resistence, Total Current, Req Current Center of Parallel and V40 olms

Physics

1 answer:

Blizzard [7]3 years ago

3 0

Total Current = 2 Amps
Req total = 60 ohms
Current on EF mesh = 1/3 Amp
Current on 24 ohms resistor = 1/6 Amp

Voltage on the 40ohms resistor is lefting.

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What is the acceleration of a racing car if it's speed is increased uniformly from 44 m /s to 66 m / s over an 11 second period

jolli1 [7]

Answer:

the acceleration of the race car is 2 m/s²

Explanation:

Given;

initial velocity of the race car, u = 44 m/s

final velocity of the race car, v = 66 m/s

time of motion of the race car, t = 11 s

The acceleration of the race car is calculated as;

$a = \frac{v-u}{t} \\\\a = \frac{66-44}{11} \\\\a = 2 \ m/s^2$

Therefore, the acceleration of the race car is 2 m/s²

7 0

2 years ago

Coherent light with wavelength 598 nm passes through two very narrow slits, and the interference pattern is observed on a screen

Stella [2.4K]

Answer:

1.196 μm

Explanation:

D = Screen distance = 3 m

$\lambda$ = Wavelength = 598 m

y = Distance of first-order bright fringe from the center of the central bright fringe = 4.84 mm

d = Slit distance

$tan\theta=\frac{y}{D}\\\Rightarrow \theta=tan^{-1}{\frac{y}{D}}\\\Rightarrow \theta=tan^{-1}{\frac{4.84\times 10^{-3}}{3}}\\\Rightarrow \theta=0.09243\ ^{\circ}$

$sin\theta=\frac{\lambda}{d}\\\Rightarrow d=\frac{\lambda}{sin\theta}\\\Rightarrow d=\frac{598\times 10^{-9}}{sin0.09243}\\\Rightarrow d=0.00037066\ m$

For first dark fringe

$dsin\theta=\frac{\lambda'}{2}\\\Rightarrow \lambda'=2dsin\theta\\\Rightarrow \lambda'=2\times 0.00037066\times sin0.09243\\\Rightarrow \lambda'=1.196\times 10^{-6}\\\Rightarrow \lambda'=1.196\ \mu m$

Wavelength of first-order dark fringe observed at this same point on the screen is 1.196 μm

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3 years ago

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uysha [10]

Answer:

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