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3 years ago

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:

Did a quiz on this not to long ago

5 0

3 years ago

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)
vf^2=18.0 m/s)^2 + (-19.62 m/s^2)(11 m)
vf^2=(324 m^2/s^2) - (215.82 m^2/s^2)
vf^2=108.18 m^2/s^2
vf=sqrt(108.18 m^2/s^2)
vf=10.4 m/s

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

Read 2 more answers

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