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3 years ago

8

Three resistors are connected into the section of a circuit described by the diagram. At which labeled point or points of the ci

rcuit could a fourth resistor be added in series with R1?

1 answer:

tangare [24]3 years ago

5 0

Answer:

Pont z only

Explanation:

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At height h above the surface of Earth, the gravitational acceleration is What is h? Note: The radius of Earth is 6380 km.

rusak2 [61]

The acceleration of gravity is inversely proportional to
the square of the distance from Earth's center.

The acceleration of gravity is 9.8 m/s² on the Earth's surface ...
6380 km from the center.

If the acceleration of gravity at 'h' is 4.9 m/s² ... 1/2 of what it is
on the surface, then the distance from the center is

(6380 x √2) = 9,023 km (rounded) ,

and 'h' is the distance above the surface

= (9,023 - 6,380) = 2,643 km (rounded) .

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3 years ago

Four distinguishable particles move freely in a room divided into octants (there are no actual partitions). Let the basic states

mafiozo [28]

Answer:

Explanation:

Since the door that leads to the room is opened, this gives room for particles to move into the next identical room and divided into octants. Now the amount of space that can be occupied becomes double, the number of basic states has increased by 404916

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3 years ago

What happens when a person’s immune system is very weak?

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3 years ago

Suppose light from a 632.8 nm helium-neon laser shines through a diffraction grating ruled at 520 lines/mm. How many bright line

Leya [2.2K]

Answer:

1 bright fringe every 33 cm.

Explanation:

The formula to calculate the position of the m-th order brigh line (constructive interference) produced by diffraction of light through a diffraction grating is:

$y=\frac{m\lambda D}{d}$

where

m is the order of the maximum

$\lambda$ is the wavelength of the light

D is the distance of the screen

d is the separation between two adjacent slit

Here we have:

$\lambda=632.8 nm = 632.8\cdot 10^{-9} m$ is the wavelength of the light

D = 1 m is the distance of the screen (not given in the problem, so we assume it to be 1 meter)

$n=520 lines/mm$ is the number of lines per mm, so the spacing between two lines is

$d=\frac{1}{n}=\frac{1}{520}=1.92\cdot 10^{-3} mm = 1.92\cdot 10^{-6} m$

Therefore, substituting m = 1, we find:

$y=\frac{(632.8\cdot 10^{-9})(1)}{1.92\cdot 10^{-6}}=0.330 m$

So, on the distant screen, there is 1 bright fringe every 33 cm.

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3 years ago

I need HELP Please !!

aliya0001 [1]

You have to figure it out

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3 years ago