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galben [10]
3 years ago
11

Does the North Pole receive more daylight hours during Winter or Summer?

Physics
2 answers:
Mandarinka [93]3 years ago
5 0

Answer:

Summer

Explanation:

The North Pole stays in full sunlight all day long throughout the entire summer (unless there are clouds), and this is the reason that the Arctic is called the land of the "Midnight Sun"

MrMuchimi3 years ago
3 0

Answer:

Summer begins in the Northern Hemisphere on June 20 or 21 when the North Pole is tilted a full 23.5° toward the sun. On this day, the Northern Hemisphere has the most hours of daylight, while the Southern Hemisphere has the least hours of daylight.

Explanation:

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Dos cargas Q1=2pc y Q2=4pc estan separadas por una distancia de 6cm ¿con que fuerza se atraen?
noname [10]

Here we can use coulomb's law to find the force between two charges

As per coulombs law

]tex]F = \frac{kq_1q_2}{r^2}[/tex]

here we have

k = 9 * 10^9

q_1 = 2pC

q_2 = 4pC

r = 6cm = 0.06 m

now by using the above equation we have

F = \frac{9*10^9 * 2*10^{-12} * 4*10^{-12}}{0.06^2}

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so here the force between two charges is of above magnitude and this will be repulsive force between them as both charges are of same sign.

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3 years ago
The overall length of a piccolo is 32.0 cm. The resonating air column vibrates as in a pipe that is open at both ends. (a) Find
lana66690 [7]

Answer:

lowest frequency = 535.93 Hz

distance  between adjacent anti nodes is 4.25 cm

Explanation:

given data

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to find out

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solution

we consider here speed of sound through air at room temperature 20 degree is  approximately  v = 343 m/s

so

lowest frequency will be = \frac{v}{2L}   ..............1

put here value in equation 1

lowest frequency will be = \frac{343}{2(0.32)}

lowest frequency = 535.93 Hz

and

we have given highest frequency f = 4000Hz

so

wavelength =  \frac{v}{f}   ..............2

put here value

wavelength =  \frac{343}{4000}  

wavelength = 0.08575 m

so distance =  \frac{wavelength}{2}   ..............3

distance =  \frac{0.08575}{2}  

distance = 0.0425 m

so distance  between adjacent anti nodes is 4.25 cm

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