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sergejj [24]
3 years ago
14

Which of the following describes a lunar eclipse?

Physics
2 answers:
Vesnalui [34]3 years ago
5 0

Answer:

From our perspective on Earth, two types of eclipses occur: lunar, the blocking of the Moon by Earth's shadow, and solar, the obstruction of the Sun by the Moon. When the Moon passes between Sun and Earth, the lunar shadow is seen as a solar eclipse on Earth.

Hopes this helps

melomori [17]3 years ago
4 0

Answer:

c

Explanation:

       when earth casts it shadow on moon then moon is not visible from earth.

and lunar eclipse occurs when earth casts its shadow on moon

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Answer:

A

Explanation:

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2 years ago
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The 45-g arrow is launched so that it hits and embeds in a 1.40 kg block. The block hangs from strings. After the arrow joins th
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Question: How fast was the arrow moving before it joined the block?

Answer:

The arrow was moving at 15.9 m/s.

Explanation:

The law of conservation of energy says that the kinetic energy of the arrow must be converted into the potential energy of the block and arrow after it they join:

\dfrac{1}{2}m_av^2 = (m_b+m_a)\Delta Hg

where m_a is the mass of the arrow, m_b is the mass of the block, \Delta H of the change in height of the block after the collision, and v is the velocity of the arrow before it hit the block.

Solving for the velocity v, we get:

$v = \sqrt{\frac{2(m_b+m_a)\Delta Hg}{m_a} } $

and we put in the numerical values

m_a = 0.045kg,

m_b = 1.40kg,

\Delta H = 0.4m,

g= 9.8m/s^2

and simplify to get:

\boxed{ v= 15.9m/s}

The arrow was moving at 15.9 m/s

6 0
3 years ago
Consider a steel guitar string of initial length L=1.00 meter and cross-sectional area A=0.500 square millimeters. The Young's m
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Answer:

\Delta l=0.015m

Explanation:

We have given initial length of the steel guitar l = 1 m

Cross sectional area A=0.5mm^2=0.5\times 10^{-6}m^2

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So stress =\frac{force}{area}=\frac{1500}{0.5\times 10^{-6}}=3000\times 10^{-6}=3\times 10^{9}Pa

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So 2\times 10^{11}=\frac{3\times 10^{9}}{strain}

strain=1.5\times 10^{-2}=0.015m

Now strain =\frac{\Delta l}{l}

0.015=\frac{\Delta l}{1}

\Delta l=0.015m

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