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valentinak56 [21]
3 years ago
10

2

Physics
1 answer:
Yuliya22 [10]3 years ago
8 0
As wavelength increase, frequency decrease
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What is NOT a major factor affecting climate?
professor190 [17]

B Longitude

Explanation:

The line of longitude or longitude does not affects climates.

Climate is the weather condition over a place for a long period of time. It usually takes several years to delineate the climate of an area.

  •  Longitude is the distance on the earth surface that runs form east to west.
  • All lines of longitude are all great circles.
  • Latitude are lines from north to south on the earth surface. Only the equator is a great circle.
  • Latitude affects climate. In fact based on the division of lines of latitude on can predict climatic patterns on earth.
  • Proximity to water bodies also affects climatic condition because of the effect of land and sea breeze and other climatic factors.
  • Elevation affects climate a whole lot. Physiography of a place determines the weather to a very large extent.

learn more:

Climate brainly.com/question/10856870

#learnwithBrainly

8 0
3 years ago
Your teacher's SUV can go from rest to 25 m/s (roughly 55mph) in 10 seconds, The car's velocity changes at a uniform rate.
maw [93]
Heidhebd bdjdishrhr bdjjdhdhrhr heheheh hdudjdhhf hdjdhdhf
4 0
3 years ago
During a demonstration of the gravitational force on falling objects to her class, Sarah drops an 11 lb. bowling ball from the t
leonid [27]

The instant it was dropped, the ball had zero speed.

After falling for 1 second, its speed was 9.8 m/s straight down (gravity).

Its AVERAGE speed for that 1 second was (1/2) (0 + 9.8) = 4.9 m/s.

Falling for 1 second at an average speed of 4.9 m/s, is covered <em>4.9 meters</em>.

ANYTHING you drop does that, if air resistance doesn't hold it back.

7 0
3 years ago
Read 2 more answers
Consider an insulating sphere of radius 6 cm surrounded by a conducting sphere of inner radius 18 cm and outer radius 26 cm. Fur
dimulka [17.4K]

Answer:

-1.7908787542\times 10^{-9}\ C

3.4260289211\times 10^{-9}\ C

1.7908787542\times 10^{-9}\ C

1.6351501669\times 10^{-9}\ C

Explanation:

r = Radius

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

Electric field is given by

E=-\dfrac{kq}{r^2}\\\Rightarrow q=-\dfrac{Er^2}{k}\\\Rightarrow q=-\dfrac{1610\times 0.1^2}{8.99\times 10^9}\\\Rightarrow q=-1.7908787542\times 10^{-9}\ C

The charge is -1.7908787542\times 10^{-9}\ C

Q+q=\dfrac{Er^2}{k}\\\Rightarrow Q=\dfrac{Er^2}{k}-q\\\Rightarrow Q=\dfrac{120\times 0.35^2}{8.99\times 10^9}-(-1.7908787542\times 10^{-9})\\\Rightarrow Q=3.4260289211\times 10^{-9}\ C

The charge is 3.4260289211\times 10^{-9}\ C

The charge inside will have the polarity changed

q=+1.7908787542\times 10^{-9}\ C

Outside the charge will be

3.4260289211\times 10^{-9}-1.7908787542\times 10^{-9}\\ =1.6351501669\times 10^{-9}\ C

3 0
3 years ago
16. The energy absorbed in 10minutes by an electrical heater is 1.5 MJ. The supply voltage is 240 V. calculate: a) The current d
dsp73

Answer:

If by 1.5 MJ you mean 1.5E6 Joules then

W = P t    = power X time

W / t = P   power

P = 1.5E6 J / 600 sec = 2500 J / s

P = I V

a) I = 2500 J/s / (240 J/c) = 10.4 C / sec  = 10.4 amps

b) Q = I t = 10.4 C / sec * 300 sec = 3120 Coulombs

c)  E = P * t = 2500 J / sec * 100 hr * 3600 sec / hr = 9.0E8 Joules

5 0
2 years ago
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