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Zielflug [23.3K]
3 years ago
6

What season is it when the earth axis is tilted so that it is facing toward the sun?

Physics
2 answers:
iVinArrow [24]3 years ago
5 0
Summer (but only in the northern hemisphere)
Tpy6a [65]3 years ago
3 0
C summer because of the tilt but it could be spring due to the rotation.
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Which of the following planets were the ancient Greeks able to see and make careful observations of?
zepelin [54]

Answer:

Neptune and Uranus

Explanation:

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A stone is thrown vertically upward with a speed of 18.0 m/s. How fast is it moving when it reaches a height of 11.0m? How long
steposvetlana [31]
Given: v0= 18.0 m/s, y0=0m, yf=11m, g=-9.81 m/s^2 

v0= initial velocity, vf= final velocity, y0= initial height, yf= final height, g= gravity, sqrt()= square root, ^2=squared 

vf^2=v0^2 + (2)(g)(yf-y0) 
vf^2=(18.0 m/s)^2+(2)(-9.81 m/s^2)(11 m-0m) 
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vf^2=(324 m^2/s^2) - (215.82 m^2/s^2) 
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8 0
3 years ago
N object has a charge of -2.0 μc. how many electrons must be removed so that the charge becomes +2.2 μc?
balandron [24]
Electrical Charge is a property of matter that causes it to experience a force when place in an electromagnetic field. Charge is measured in coulombs.
The change in the charge is 2.2- (2.0)= 4.2 μ c.
1 electron has a charge of 1.6 ×10∧-19 coulombs
Therefore, the number of electrons will be;
        = (4.2 ×10∧-6)/ (1.6 ×10∧-19)
        = 2.625 × 10∧13 electrons

4 0
3 years ago
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Which is the best description of a constructive inference of light?
maxonik [38]
Two waves have the same direction of displacement.
3 0
3 years ago
A point charge q1 = 4.10 nC is placed at the origin, and a second point charge q2 = -2.90 nC is placed on the x-axis at x = + 20
kaheart [24]

Answer:

1)  U = -4.436 10-7 J ,  2)   r13 = 0.6937 m

Explanation:

The electric potential energy is given by the equation

      U = k Σ q_{i} q_{j}  / r_{ij}

1. Let's apply this equation to our case

      U = k (q1q3 / r13 + q1 q2 / r12 + q2 q3 / r23)

Let's reduce all magnitudes to the SI system

    q1 = 4.10 nC = 4.10 10-9 C

    q2 = -2.90 nC = -2.90 10-09 C

    q3 = 2.00 nC = 2.00 10-9 C

Let's look for the distances

    r13 = 11-0 = 11 cm = 11 10-2 m

    r12 = 20-0 = 20 cm = 20 10-2 m

    r23 = 20 -11 = 9 cm = 9.0 10-2 m

Let's calculate the electric potential

     U = 8.99 109 [(4.10 2.00) 10-18 / 11 10-2 + ​​(4.10 (-2.90)) 10-18 / 20 10-2 + ​​(-2.90 2.00) 10-18 / 9.0 10-2]

     U = 8.99 [0.7455 - 0.5945 - 0.6444) 10-7

     U = -4.436 10-7 J

2.  ask to find the position of q3 for the energy to be zero

    U = 0 = k (q1q3 / r13 + q1 q2 / r12 + q2 q3 / r23)

In this case the distance between 1 and 2 is fixed, since the load that is placed is 3

     q1 q2 / r12 = - q1 q3 / r13 - q2 q3 / r23

     -11.89 10-18 / 20 10-2 = -8.20 10-18 / r13 + 5.810-18 / r23

     -0.5945 102 = -8.20 / r13 + 5.810 / r23

Let's relate the distances, the maximum separation esr12 which is 20 10-2 m

     r12 = r13 + r23

     r13 = r12 -r23

    r13 = 2010-2 - r23

we replace

    -0.5945 102 = -8.20 / (20 10-2 - r23) + 5.810 / r23

Let's solve this equation

    -0.5945 102 (2010-2 - r23) = -8.20 r23 + 5.810 (2010-2 - r23)

    - 11.89 + 0.5945 r23 = -8.20 r23 + 1,162 - 5,810 r23

     -11.89 - 1,162 = - 8.20 r23 -5,810 r23 - 0.5945 r23

     -13.052 = r23 (-14.6045)

     r23 = 13.052 / 14.6045

    r23 = 0.8937 m

The distance from the origin is

    r13 = -r12 + r23

    r13 = -20 10-2 + ​​0.8937

    r13 = 0.6937 m

3 0
3 years ago
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