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Pavel [41]
3 years ago
11

Match the words in the left-hand column to the appropriate blank in the sentences in the right-hand column. use each word only o

nce. a solar eclipse that occurs when the new moon is too far from earth to completely cover the sun can be either a partial solar eclipse or a(n)
Physics
1 answer:
shepuryov [24]3 years ago
7 0

Answer:

An annular Solar Eclipse

Explanation:

Solar eclipse is an event that occurs naturally on Earth when the moon in its orbit is positioned between the Earth and the Sun.Solar Eclipse can be total ,partial or annular.In the total solar eclipse, the moon completely covers the sun where as in the annular solar eclipse the moon covers the center of the Sun leaving outer edges of the Sun to be visible forming the<em> ring of fire.</em>In partial solar eclipse the Earth moves through the lunar penumbra as the moon moves between Earth and Sun.The moon blocks only some parts of the solar disk.Annular solar eclipse happens during new moon and the moon is at its farthest position from the Earth called Apogee.

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If the voltage is 120 v, and the desired current is 5 a, what resistance must be in the circuit?
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Find the moment of force of 60 Newton about an axis of rotation at distance 20 cm from the force​
777dan777 [17]

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shows a conical pendulum, in which the bob (the small object at the lower end of the cord) moves in a horizontal circle at const
Contact [7]

Answer:

a) T=0.40 N

b) T=1.9 s

Explanation:

Let's find the radius of the circumference first. We know that bob follows a circular path of circumference 0.94 m, it means that the perimeter is 0.94 m.

The perimeter of a circunference is:

P=2\pi r=0.94

r=\frac{0.94}{2\pi}=0.15 m

Now, we need to find the angle of the pendulum from vertical.

tan(\alpha)=\frac{r}{L}=\frac{0.15}{0.90}=0.17

\alpha=9.44 ^{\circ}

Let's apply Newton's second law to find the tension.

\sum F=ma_{c}=m\omega^{2}r

We use centripetal acceleration here, because we have a circular motion.

The vertical equation of motion will be:

Tcos(\alpha)=mg (1)

The horizontal equation of motion will be:

Tsin(\alpha)=m\omega^{2}r (2)

a) We can find T usinf the equation (1):

T=\frac {mg}{cos(\alpha)}=\frac{0.04*9.81}{cos(9.44)}=0.40 N

We can find the angular velocity (ω) from the equation (2):

\omega=\sqrt{\frac{Tsin(\alpha)}{mr}}=3.31 rad/s

b) We know that the period is T=2π/ω, therefore:

T=\frac{2\pi}{\omega}=\frac{2\pi}{3.31}=1.9 s

I hope it helps you!

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