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4 years ago

How can the distance between the first bright band and the central band be increased in a double-slit experiment?

Physics

1 answer:

Strike441 [17]4 years ago

8 0

Answer:

Decrease the slit separation, increase the distance of the screen from the slits, and increase the wavelength.

Explanation:

The distance $y$ from the central band to the first bright band is given by

$y = \dfrac{ \lambda D}{d}$

where $\lambda$ is the wavelength of light (or any particle), $D$ is the distance to the screen, and $d$ is the slit separation.

From this equation we see that, by increasing the wavelength $\lambda$ , increasing the distance from the screen $D$ , and decreasing the slit separation $d$ , we increase the distance between the first bright band and the central band.

Therefore, the 2nd choice "<em>Decrease the slit separation, increase the distance of the screen from the slits, and increase the wavelength.</em>" is correct.

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Which group of stars is represented by the line image

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Answer:

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A satellite, orbiting the earth at the equator at an altitude of 400 km, has an antenna that can be modeled as a 1.76-m-long rod

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Answer:

The inducerd emf is 1.08 V

Solution:

As per the question:

Altitude of the satellite, H = 400 km

Length of the antenna, l = 1.76 m

Magnetic field, B = $8.0\times 10^{- 5}\ T$

Now,

When a conducting rod moves in a uniform magnetic field linearly with velocity, v, then the potential difference due to its motion is given by:

$e = - l(vec{v}\times \vec{B})$

Here, velocity v is perpendicular to the rod

Thus

e = lvB (1)

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$v = \sqrt{\frac{Gm_{E}}{R_{E}} + H}$

where

G = Gravitational constant

$m_{e} = 5.972\times 10^{24}\ kg$ = mass of earth

$R_{E} = 6371\ km$ = radius of earth

$v = \sqrt{\frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}}{6371\times 1000 + 400\times 1000} = 7670.018\ m/s$

Using this value value in eqn (1):

$e = 1.76\times 7670.018\times 8.0\times 10^{- 5} = 1.08\ V$

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3 years ago

If a player through a basketball to the target with an initial velocity of 17 m/s making an angle of 30 degrees with the horizon

Svetllana [295]

Answer:

The final position made with the vertical is 2.77 m.

Explanation:

Given;

initial velocity of the ball, V = 17 m/s

angle of projection, θ = 30⁰

time of motion, t = 1.3 s

The vertical component of the velocity is calculated as;

$V_y = Vsin \theta\\\\V_y = 17 \times sin(30)\\\\V_y = 8.5 \ m/s$

The final position made with the vertical (Yf) after 1.3 seconds is calculated as;

$Y_f = V_yt - \frac{1}{2}g t^2\\\\Y_f = (8.5 \times 1.3 ) - (\frac{1}{2} \times 9.8 \times 1.3^2)\\\\Y_f = 11.05 \ - \ 8.281\\\\Y_f = 2.77 \ m$

Therefore, the final position made with the vertical is 2.77 m.

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3 years ago

How and why planets orbit the sun and how and why a planets speed changes orbit

anygoal [31]

the gravitonal pull

Explanation:

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3 years ago

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The average speed between earth and the sun is 1.50 x10^8 km. Calculate the average speed of the Earth in its orbit in kilometer

cluponka [151]

Answer:

The average speed of the earth in its orbit is $29.86km/s$

Explanation:

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The average speed of the earth in its orbit can be found by the next equation :

$v = \frac{2 \pi r}{T}$ (1)

Where r is the radius and T is the period.

In this case, the orbit of the Earth can be considered as a circle

( $r = 1.50x10^{8}km$ ) instead of an ellipse.

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Then, equation 1 can be used:

$v = \frac{2 \pi (1.50x10^{8}km)}{31557600s}$

$v = 29.86km/s$

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3 years ago

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