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8090 [49]
3 years ago
9

A beam of microwaves with λ = 0.9 mm is incident upon a 9 cm slit. At a distance of 1.5 m from the slit, what is the approximate

width of the slit's image (cm)?
Physics
1 answer:
liq [111]3 years ago
5 0

Answer:

3 cm

Explanation:

According to the question,

D=1.5 m.

d=9 cm.

\lambda =0.9 mm.

Now the approximate slit's image width is equal to width of central maxima.

And width of central maxima is twice the width from center to first maxima

So,

y=2\frac{\lambda (D)}{d}.

Substitute all the variable in above equation.

y=\frac{(2)0.9\times 10^{-2} m(1.5 m) }{0.09 m}.

y=3 cm.

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Trucks 1 and 2 are traveling at the same constant velocity but truck 1's energy due to motion is two times less than that of tru
brilliants [131]
Kinetic energy = 0.5 * m * v²

m mass
v velocity

If the velocity stays the same and the kinetic energy goes down by a factor of 2, the mass must go down by a factor of 2 also.
7 0
3 years ago
A 5 cm spring is suspended with a mass of 3.8589 g attached to it which extends the spring by 1.5747 cm. The same spring is plac
lana66690 [7]

Answer:

charges of the beads is 1.173 ×10^{-15} C

Explanation:

given data

mass = 3.8589 g = 0.003859 kg

spring length = 5 cm = 0.05 m

extend spring x = 1.5747 cm = 0.15747 m

spring's extension = 0.0116 m

to find out

charges of the beads

solution

we know that force is

force = mass × g

force = 0.003859 × 9.8

force = 0.03782 N

so we know  force for mass

force  = -kx

so k = force / x

put here force and x value

k = -0.03782 / 0.1575

k = -0.24 N/m

and

force for spring's extension

force = -kx

force = -0.24 ( 0.0116) = 0.002784 N

so here

total length L = 0.05 + 0.0116 = 0.0616

so charges of the beads = force × L² / ke

charges of the beads = 0.002784 × (0.0616)² / (9 ×10^{9} )

so charges of the beads = 1.173 ×10^{-15} C

3 0
3 years ago
A Force Play is when the defense...
ivolga24 [154]
Hi i play softball so the answer is the letter:

A or D but i think D
6 0
2 years ago
Optical astronomers need a clear, dark sky to collect good data. Explain why radio astronomers have no problem observing in the
gregori [183]

Answer:

In the clarification portion elsewhere here, the definition of the concern is mentioned.

Explanation:

So like optical telescopes capture light waves, introduce it to concentrate, enhance it, as well as make it usable through different instruments via study, so radio telescopes accumulate weak signal light waves, introduce that one to focus, enhance it, as well as make this information available during research. To research naturally produced radio illumination from stars, galaxies, dark matter, as well as other natural phenomena, we utilize telescopes.

Optical telescopes detect space-borne visible light. There are some drawbacks of optical telescopes mostly on the surface:

  • Mostly at night would they have been seen.
  • Unless the weather gets cloudy, bad, or gloomy, they shouldn't be seen.

Although radio telescopes monitor space-coming radio waves. Those other telescopes, when they are already typically very massive as well as costly, have such an improvement surrounded by optical telescopes. They should be included in poor weather and, when they travel through the surrounding air, the radio waves aren't obscured by clouds. Throughout the afternoon and also some at night, radio telescopes are sometimes used.

3 0
2 years ago
Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150
Phoenix [80]

Answer:

The temperature of the core raises by 2.8^{o}C every second.

Explanation:

Since the average specific heat of the reactor core is 0.3349 kJ/kgC

It means that we require 0.3349 kJ of heat to raise the temperature of 1 kg of core material by 1 degree Celsius

Thus reactor core whose mass is 1.60\times 10^{5}kg will require

0.3349\times 1.60\times 10^{5}kJ\\\\=0.53584\times 10^{5}kJ

energy to raise it's temperature by 1 degree Celsius in 1 second

Hence by the concept of proportionately we can infer 150 MW of power will increase the temperature by

\frac{150\times 10^{6}}{0.53584\times 10^{8}}=2.8^{o}C/s

5 0
3 years ago
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