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3 years ago

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Physics

1 answer:

igomit [66]3 years ago

6 0

Reactions between atoms occur whenan atom seeks to fill its outer shell of electrons

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In a Young's double-slit experiment, a set of parallel slits with a separation of 0.142 mm is illuminated by light having a wave

Alexandra [31]

Answer:

$1.152\ \mu m$

$1.44\ \mu m$

Explanation:

d = Gap between slits = 0.142 mm

$\lambda$ = Wavelength of light = 576 nm

L = Distance between light and screen = 3.5 m

m = Order = 2

Difference in path length is given by

$\delta=dsin\theta=m\lambda\\\Rightarrow \delta=2\times 576\times 10^{-9}\\\Rightarrow \delta=0.000001152\ m\\\Rightarrow \delta=1.152\times 10^{-6}\ m=1.152\ \mu m$

The difference in path lengths is $1.152\ \mu m$

For dark fringe the difference in path length is given by

$\delta=(m+\dfrac{1}{2})\lambda\\\Rightarrow \delta=(2+\dfrac{1}{2})\times 576\times 10^{-9}\\\Rightarrow \delta=0.00000144\ m=1.44\ \mu m$

The difference in path length is $1.44\ \mu m$

5 0

3 years ago

A football player kicks a field goal from a distance of 45 m from the goalpost. The football is launched at a 35° angle above th

daser333 [38]

Answer: try searching It up on go0gle

Explanation:

7 0

2 years ago

Solved, can someone check it over!

kakasveta [241]

Answer:

its good no need to change anything :))

4 0

2 years ago

what was the acceleration of the cart with Low fan speed cm/s squared? what was the acceleration of the cart with medium fan spe

ArbitrLikvidat [17]

Explanation:

The attached figure shows data for the cart speed, distance and time.

For low fan speed,

Distance, d = 500 cm

Time, t = 7.4 s

Average velocity,

$v=\dfrac{d}{t}\\\\v=\dfrac{500}{7.4}\\\\v=67.56\ cm/s$

Acceleration,

$a=\dfrac{v}{t}\\\\a=\dfrac{67.56}{7.4}\\\\a=9.12\ cm/s^2$

For medium fan speed,

Distance, d = 500 cm

Time, t = 6.4 s

Average velocity,

$v=\dfrac{d}{t}\\\\v=\dfrac{500}{6.4}\\\\v=78.12\ cm/s$

Acceleration,

$a=\dfrac{v}{t}\\\\a=\dfrac{78.12}{6.4}\\\\a=12.2\ cm/s^2$

For high fan speed,

Distance, d = 500 cm

Time, t = 5.6 s

Average velocity,

$v=\dfrac{d}{t}\\\\v=\dfrac{500}{5.6}\\\\v=89.28\ cm/s$

Acceleration,

$a=\dfrac{v}{t}\\\\a=\dfrac{89.28}{5.6}\\\\a=15.94\ cm/s^2$

Hence, this is the required solution.

8 0

3 years ago

Read 2 more answers

A 6 cm long spring extends to 9 cm when a 1 kg load is suspended from it. What would be its length if a 2 kg load were suspended

Ganezh [65]

The answer is D that is the right question

8 0

3 years ago