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steposvetlana [31]
3 years ago
10

Which part of the sun-layers will you be able to see Northern lights or auroras?. A. Ozonosphere B. Ionosphere C. Exosphere D. T

roposphere
Physics
2 answers:
mina [271]3 years ago
6 0
The ionosphere is the part of the sun-layers which allows us to see the northern lights or auroras. The ionosphere is the most concentrated with ions and charged electrons and reflects radio waves.
BabaBlast [244]3 years ago
4 0

Answer

The answer is ionosphere

Explanation:

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Divide the change in speed by the time for the change.
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A ball rolls up a ramp with a velocity of 8.0 m/s. How high up the ramp does it travel?
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The greatest height the ball will attain is 3.27 m

<h3>Data obtained from the question</h3>
  • Initial velocity (u) = 8 m/s
  • Final velocity (v) = 0 m/s (at maximum height)
  • Acceleration due to gravity (g) = 9.8 m/s²
  • Maximum height (h) =?

The maximum height to which the ball can attain can be obtained as follow:

v² = u² – 2gh (since the ball is going against gravity)

0² = 8² – (2 × 9.8 × h)

0 = 64 – 19.6h

Collect like terms

0 – 64 = –19.6h

–64 = –19.6h

Divide both side by –19.6

h = –64 / –19.6h

h = 3.27 m

Thus, the greatest height the ball can attain is 3.27 m

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Which of the following statements best describes the second law of
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3 years ago
Calculate the net force on the right charge due to the other two. Enter a positive value if the force is directed to the right a
lbvjy [14]

Answer:

Answer:

A. - 0.017N. It acts to the left.

B. - 0.043N. It acts to the left.

C. 0.060N. It acts to the right.

Explanation:

A. For the +65μC charge, we consider it to be the origin. Hence, the two other charges are on the +x axis.

The net coulombs force on the charge is

F = [KQ(1)Q(2)]/(r^2) + [KQ(1)Q(3)]/(r^2)

Where K = Coloumbs constant =

Q(1) = charge on the leftmost side.

Q(2) = charge in the middle.

Q(3) = charge on the rightmost side.

F = [(8.988 × 10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(-95×10^-6)×(65×10^-6)]/(40^2)

F = 0.01753 - 0.03469

F = -0.017N

It has a negative sign, hence, it acts to the left.

B. For the +48μC charge, we consider it to be the origin. Hence, the leftmost charge is on the - x axis and the rightmost charge is on the +x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) + [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = -0.017 - 0.02562

F = - 0.043N

It has a negative sign, hence, it acts to the left.

C. For the -95μC charge, we consider it to be the origin. Hence, the two other charges are on the - x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) - [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(-95×10^-6)]/(40^2) - [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = +0.03469 + 0.02562

F = +0.060N

It has a positive sign, hence, it acts to the right.

Read more on Brainly.com - brainly.com/question/14592748#readmore

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