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swat32
3 years ago
9

A total solar eclipse is a rare event. Although they occur somewhere on earth every 18 months on average, it is estimated that t

hey recur at any given place only once every 370 years. Why are reoccurring solar eclipses so rare?
Physics
2 answers:
WITCHER [35]3 years ago
6 0

Answer:

D

Explanation:

The Moon passes slightly above or below the line between the Sun and the Earth because of its tilted orbit. I also had it for my USA Test Prep

Setler79 [48]3 years ago
3 0
Because the tip of the moon's shadow ... the area of "totality" ... is never more than a couple hundred miles across, It never covers a single place for more than 7 minutes, and can never stay on the Earth's surface for more than a few hours altogether during one eclipse.

If you're not inside that small area, you don't see a total eclipse.
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What did j.J. Thomson discover about the composition of atoms?
nordsb [41]

Answer:

J.J. Thomson discovered <u><em>the electron</em></u> by experimenting with a Crookes, or cathode ray, tube. He demonstrated that cathode rays were negatively charged. In addition, he also studied positively charged particles in neon gas.

Explanation:

the answer is the underlined part of the answer section. I hoped this helped you a lot!! Study hard for whatever you are doing!!

5 0
3 years ago
Why does Farm Bureau and other advocacy organization oppose any mandated labeling of biotech crops?
Step2247 [10]

Answer:

I’m. Nog sure

Explanation:

3 0
3 years ago
The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles
miss Akunina [59]

Formula of the gravitational force between two particles:

F=G\frac{m_1 m_2}{r^2}

where

G=6.67 \cdot 10^{-11} Nm^2 kg^{-2} is the gravitational constant

m1 and m2 are the masses of the two particles

r is their distance


(a) particle A

The gravitational force exerted by particle B on particle A is

F_B=G\frac{m_A m_B}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(567 kg)}{(0.500 m)^2}=5.14 \cdot 10^{-5} N to the right

The gravitational force exerted by particle C on particle A is

F_C=G\frac{m_A m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(139 kg)}{(0.500 m+0.250m)^2}=5.6 \cdot 10^{-6} N to the right

So the net gravitational force on particle A is

F_A = F_B + F_C =5.14 \cdot 10^{-5} N+5.6 \cdot 10^{-6} N=5.7 \cdot 10^{-5} N to the right


(b) Particle B

The gravitational force exerted by particle A on particle B is

F_A=G\frac{m_A m_B}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(567 kg)}{(0.500 m)^2}=-5.14 \cdot 10^{-5} N to the left

The gravitational force exerted by particle C on particle B is

F_C=G\frac{m_B m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(567 kg)(139 kg)}{(0.250 m)^2}=8.41 \cdot 10^{-5} N to the right

So the net gravitational force on particle B is

F_B = F_A + F_C =-5.14 \cdot 10^{-5} N+8.41 \cdot 10^{-5} N=3.27 \cdot 10^{-5} N to the right


(c) Particle C

The gravitational force exerted by particle A on particle C is

F_A=G\frac{m_A m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(139 kg)}{(0.500 m+0.250m)^2}=-5.6 \cdot 10^{-6} N to the left

The gravitational force exerted by particle B on particle C is

F_B=G\frac{m_B m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(567 kg)(139 kg)}{(0.250 m)^2}=-8.41 \cdot 10^{-5} N to the left

So the net gravitational force on particle C is

F_C = F_B + F_A =-8.41 \cdot 10^{-5} N-5.6 \cdot 10^{-6} N=-8.97 \cdot 10^{-5} N to the left



3 0
3 years ago
The illustration shows the basic unit of life it is
vlabodo [156]
The answer is A, Cell. This is the basic structure of a cell.
6 0
3 years ago
Read 2 more answers
2) The position of a particle is given r(t) = A(cos wt i + sin wt f), where w is a constant. (a) Show that the particle moves in
olya-2409 [2.1K]

Answer:

Explanation:

r(t) = A(cos wt i + sin wt f)

= A cos wt i + A sin wt j

x = A cos wt

y = A sin wt

radius r

r² = x² + y² ( This is equation of a circle with radius r  )

=  A² cos² wt +A² sin² wt

= A²

r = A

radius r = A

b )

speed = dr/dt

v = - Aw sinwt i + Aw coswt j

magnitude of velocity

I v I= Aw √(sin²wt + cos²wt)

= Aw ( constant )

acceleration

= dv / dt = - Aw² cos wt - Aw² sinwt

magnitude of acceleration

I a I = Aw²

= r w²

d ) centripetal force = m acceleration

m w² A  

=

7 0
3 years ago
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