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3 years ago

7

In a double-slit experiment, light and dark regions are observed on a screen. what causes a dark region to be observed between t

wo brighter regions?

1 answer:

Colt1911 [192]3 years ago

7 0

According to a source, fringes is the answer. These fringes are what causes dark regions in the double -slit experiment conducted that can be observed in the screen.
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Water drips from a shower head (the sprayer at the top of the shower) and falls onto the floor 2.4 m below. The droplets are fal

expeople1 [14]

Answer:

The third drop is 0.26m

Explanation:

The drop 1 impacts at time T is given by:

T=sqrt(2h/g)

T= sqrt[(2×2.4)/9.8]

T= sqrt(4.8/9.8)

T= sqrt(0.4898)

T= 0.70seconds

4th drops starts at dT=0.70/3= 0.23seconds

The interval between the drops is 0.23seconds

Third drop will fall at t= 0.23

h=1/2gt^2

h= 1/2×9.81×(0.23)^2

h= 0.26m

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3 years ago

Lunar missions have revealed that the moon has:

myrzilka [38]

That the moon has soil within its shadowy craters rich and useful material

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3 years ago

Which of the following is a physical property of wood

Black_prince [1.1K]

The density, hard, strong, and rough.

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3 years ago

Read 2 more answers

A small loop of area A = 5.5 mm2 is inside a long solenoid that has n = 919 turns/cm and carries a sinusoidally varying current

Tema [17]

Answer: $410.90\times 10^{-6} V$

Explanation:

Given

$A=5.5 mm^2$

$n=919 turns/cm\approx 85400 turns/m$

$\omega =226 rad/s$

According to the Faraday law of induction, induced emf is given by

$E=-\frac{\mathrm{d} \phi _B}{\mathrm{d} t}$

Magnetic field $\phi _B=B\cdot A$

$B=\mu _0ni=\mu _0ni_osin\left ( \omega t\right )$

$B=4\pi \times 10^{-7}\times 85400\times 3.08\times \sin (226t)$

$B=0.3305\sin (226t)$

$\phi _{B}=0.3305\times \sin (226t)\times 5.5\times 10^{-6}$

$\phi _{B}=1.818\times 10^{-6}\sin (226t)$

$E=-\frac{\mathrm{d} \phi _B}{\mathrm{d} t}$

$E=-1.818\times 10^{-6}\times 226\cos (226t)$

$E=-410.90\times 10^{-6}\cos (226t)$

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8 0

4 years ago

A car is designed to get its energy from a rotating flywheel with a radius of 1.50 m and a mass of 430 kg. Before a trip, the fl

fiasKO [112]

Answer:

a

$KE = 7.17 *10^{7} \ J$

b

$t = 6411.09 \ s$

Explanation:

From the question we are told that

The radius of the flywheel is $r = 1.50 \ m$

The mass of the flywheel is $m = 430 \ kg$

The rotational speed of the flywheel is $w = 5,200 \ rev/min = 5200 * \frac{2 \pi }{60} =544.61 \ rad/sec$

The power supplied by the motor is $P = 15.0 hp = 15 * 746 = 11190 \ W$

Generally the moment of inertia of the flywheel is mathematically represented as

$I = \frac{1}{2} mr^2$

substituting values

$I = \frac{1}{2} ( 430)(1.50)^2$

$I = 483.75 \ kgm^2$

The kinetic energy that is been stored is

$KE = \frac{1}{2} * I * w^2$

substituting values

$KE = \frac{1}{2} * 483.75 * (544.61)^2$

$KE = 7.17 *10^{7} \ J$

Generally power is mathematically represented as

$P = \frac{KE}{t}$

=> $t = \frac{KE}{P}$

substituting the value

$t = \frac{7.17 *10^{7}}{11190}$

$t = 6411.09 \ s$

3 0

3 years ago