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Blababa [14]
4 years ago
6

How does remove coloration interface "light-shadow"?

Physics
1 answer:
tekilochka [14]4 years ago
5 0
Normal force, weight, Kinetic friction, and air resistance are a few I think of the top of my head.
I hope this helps
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How much of the Moon is always illuminated one time? Explain your answer.
Natalija [7]

Answer:

50% of it .

Explanation:

50% of it is illuminated by the Sun.

6 0
3 years ago
(a) what is the system of interest if the acceleration of the child in the wagon is to be calculated? (select all that apply.)
Leno4ka [110]

since child is moving along with the wagon and we need to find the acceleration of child inside that wagon then in this case the system of interest must be child + wagon

System of interest will be the system that is used to find the force or acceleration using Newton's law

Here we have to assume that system on which if we will calculate the forces then the net value of force on that system will help to calculate the unknown quantities

So here our system will be boy + wagon

6 0
3 years ago
3. What is the power of a derby race car that uses a 50 N force to travel 20 meters in 10 seconds?
makkiz [27]
I think it is B. 100 W
7 0
3 years ago
What is the lens equation?
Strike441 [17]
The lens equation gives d relation between focal length, object distance n image distance.

1/f = 1/v + 1/u

seldon
7 0
3 years ago
Read 2 more answers
A red cross helicopter takes off from headquarters and flies 120 km at 70 degrees south of west. There it drops off some relief
fiasKO [112]

Answer:

130 km at 35.38 degrees north of east

Explanation:

Suppose the HQ is at the origin (x = 0, y = 0)

So the coordinates of the helicopter after the 1st flight is

x_1 = -120cos70^o = -41.04 km

y_1 = -120sin70^o = -112.763 km

After the 2nd flight its coordinate would be:

x_2 = x_1 - 75sin60^o = -41.04 - 64.95 = -106km

y_2 = y_1 + 75cos60^o = -112.763 + 37.5 = -75.263 km

So in order to fly back to its HQ it must fly a distance and direction of

s = \sqrt{y_2^2 + x_2^2} = \sqrt{75.263^2 + 106^2} = \sqrt{5664.519169 + 11236} = \sqrt{16900.519169} = 130 km

tan\theta = \frac{y_2}{x_2} = \frac{75.263}{106} = 0.71

\theta = tan^{-1}0.71 = 0.62 rad \approx 35.38^o north of east

3 0
3 years ago
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