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ratelena [41]
3 years ago
6

The passing of the Moon directly between Earth and the Sun is a/an

Physics
2 answers:
erastova [34]3 years ago
8 0

it's the "solar eclipse"

vladimir1956 [14]3 years ago
6 0
<span>The correct answer for this question is that the passing of the moon directly between the earth and the sun is a solar eclipse. Solar eclipses cause the earth to temporarily become slightly darker, and can be observed by looking through specialist glasses so as not to damage the eyes.</span>
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The smallest possible part of an element that can still be identified as the element is called
andreyandreev [35.5K]

Answer:

BLM

Explanation:

ACAB

4 0
3 years ago
How much work is needed to push a 117- kg packing crate a distance of 2.75 m up an inclined plane that makes an angle of 28 o wi
Nataly_w [17]

Answer:

W_{tot,min} = 1481.372\,J

Explanation:

The minimum total work is the work needed to counteract the work associated with the weight:

W_{tot, min} = W_{F}

W_{tot,min} = m\cdot g\cdot \sin \theta \cdot \Delta s

W_{tot,min} = (117\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (\sin 28^{\textdegree}) \cdot (2.75\,m)

W_{tot,min} = 1481.372\,J

6 0
3 years ago
Read 2 more answers
What's the diameter of a dish antenna that will receive 10−20W of power from Voyager at this time? Assume that the radio transmi
Murrr4er [49]

Complete Question:

The Voyager 1 spacecraft is now beyond the outer reaches of our solar system, but earthbound scientists still receive data from the spacecraft s 20-W radio transmitter. Voyager is expected to continue transmitting until about 2025, when it will be some 25 billion km from Earth.

What s the diameter of a dish antenna that will receive 10−20W of power from Voyager at this time? Assume that the radio transmitter on Voyager transmits equally in all directions(isotropically).  In fact, the antenna on Voyager focuses the signal in a beam aimed at the earth, so this problem over-estimates the size of the receiving dish needed.

Answer:

d = 2,236 m.

Explanation:

The received power on Earth, can be calculated as the product of the intensity (or power density) times the area that intercepts the power radiated.

As we assume that  the transmitter antenna is ominidirectional, power is spreading out over a sphere with a radius equal to the distance to the source.

So, we can get the power density as follows:

I = P /A = P / 4*π*r², where P = 20 W, and r= 25 billion km = 25*10¹² m.

⇒ I = 20 W / 4*π* (25*10¹²)² m²

The received power, is just the product of this value times the area of the receiver antenna, which we assumed be a circle of diameter d:

Pr = I. Ar =( 20W / 4*π*(25*10¹²)² m²) * π * (d²/4) = 10⁻²⁰ W

Simplifying common terms, we can solve for d:

d= √(16*(25)²*10⁴/20) = 2,236 m.

3 0
3 years ago
A .5kg bird is perched on its nest so that it has 50J of potential energy. how far is it off the of the ground?
pshichka [43]

It is 10.20 m from the ground.

<u>Explanation:</u>

<u>Given:</u>

m = 0.5 kg

PE = 50 J

We know that the Potential energy is calculated by the formula:

P. E = m \times g \times h

where m is the is mass in kg;  g  is acceleration due to gravity which is 9.8 m/s  and  h  is height in meters.

PE is the Potential Energy.

Potential Energy is the amount of energy stored when an object is stationary.

Here, if we substitute the values in the formula, we get

P. E = m \times g \times h

50 = 0.5 × 9.8 × h

50 = 4.9 × h

h = \frac {50} {4.9}

h = 10.20 m

3 0
3 years ago
If the accepted value of a wave is 121 m/s, who has the most accurate method of measuring the speed of a wave?
qaws [65]
The answer would be erin out of all of them thank me later :)
5 0
3 years ago
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