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wlad13 [49]
3 years ago
13

In the diagram, the Sun, Earth, and Moon are in perfect alignment. Which two conclusions can be drawn based on the diagram?

Physics
2 answers:
mihalych1998 [28]3 years ago
8 0

Answer:An observer on the Moon sees a solar eclipse. and An observer on Earth sees a lunar eclipse.

Explanation:from the moon the sun is completly if not just mostly invisible(solar eclipse) and from earth the moon would be in lunar eclipse from the earth blocking the sun.

AysviL [449]3 years ago
4 0

Answer:

I know one answer is an observer on earth sees a lunar eclipse

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Does anyone know the answer
scoundrel [369]
No put false I am right
7 0
3 years ago
Q6) A speed skater moving to the left across frictionless ice at 8.0 m/s hits a 5.0-m-wide patch of rough ice. She slows steadil
77julia77 [94]

Answer:

-2.8 m/s²

Explanation:

Acceleration: This can be defined as the rate of change of velocity The S. I unit of acceleration is m/s²

Using the equation of motion,

v² = u² + 2as................... Equation 1

Where v = Final velocity, u = initial velocity, a = acceleration, s = distance,

Given: v = 6.0 m/s, u = 8.0 m/s, s = 5.0 m.

Substituting into equation 1

6² = 8²+2(a)5

36 = 64 + 10a

10a = 36-64

10a = -28

10a/10 = -28/10

a = -2.8 m/s²

Note: a is negative because because the skater decelerate on the rough ice

Hence the magnitude of her acceleration is  = -2.8 m/s²

6 0
3 years ago
6. Given that white light contains all colors of the spectrum, what growth result would you expect under white light?
lesya692 [45]

You could probably expect normal plant growth, as a white light is similar to the sun in the respect that it contains all colors of the spectrum.

4 0
3 years ago
At which location would you expect the LOWEST TEMPERATURE?
MissTica

Answer:

c

Explanation:

3 0
2 years ago
Read 2 more answers
Ml(d^2θ/dt^2) =-mgθ
Nata [24]

The equation of motion of a pendulum is:

\dfrac{\textrm{d}^2\theta}{\textrm{d}t^2} = -\dfrac{g}{\ell}\sin\theta,

where \ell it its length and g is the gravitational acceleration. Notice that the mass is absent from the equation! This is quite hard to solve, but for <em>small</em> angles (\theta \ll 1), we can use:

\sin\theta \simeq \theta.

Additionally, let us define:

\omega^2\equiv\dfrac{g}{\ell}.

We can now write:

\dfrac{\textrm{d}^2\theta}{\textrm{d}t^2} = -\omega^2\theta.

The solution to this differential equation is:

\theta(t) = A\sin(\omega t + \phi),

where A and \phi are constants to be determined using the initial conditions. Notice that they will not have any influence on the period, since it is given simply by:

T = \dfrac{2\pi}{\omega} = 2\pi\sqrt{\dfrac{g}{\ell}}.

This justifies that the period depends only on the pendulum's length.

4 0
3 years ago
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