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Helga [31]
3 years ago
6

Same diagram... At which location would it be possible for a LUNAR ECLIPSE to happen? *

Physics
2 answers:
STALIN [3.7K]3 years ago
5 0
The answer would be a

Because a solar eclipse is where the moon covers the sun from the earth so it hat to be in front of the sunlight
Radda [10]3 years ago
5 0
A but for Lunar eclipse the shadows have to line up
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5. A current of 3.00 A flows through a resistor when it is connected
viktelen [127]
I think it is D
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7 0
2 years ago
An object of mass m travels along the parabola yequalsx squared with a constant speed of 5 ​units/sec. What is the force on the
Dmitriy789 [7]

Explanation:

The object is moving along the parabola y = x² and is at the point (√2, 2).  Because the object is changing directions, it has a centripetal acceleration towards the center of the circle of curvature.

First, we need to find the radius of curvature.  This is given by the equation:

R = [1 + (y')²]^(³/₂) / |y"|

y' = 2x and y" = 2:

R = [1 + (2x)²]^(³/₂) / |2|

R = (1 + 4x²)^(³/₂) / 2

At x = √2:

R = (1 + 4(√2)²)^(³/₂) / 2

R = (9)^(³/₂) / 2

R = 27 / 2

R = 13.5

So the centripetal force is:

F = m v² / r

F = m (5)² / 13.5

F = 1.85 m

7 0
3 years ago
Margaret, a researcher, desires to conduct a field experiment to determine the effects of a shopping mall's ambience on consumer
vichka [17]

Answer:

Extraneous

Explanation:

Extraneous variables are any variables that you are not intentionally studying in your experiment or test

6 0
3 years ago
Pls help my science teacher is a horrible teacher...
Anettt [7]

First of all, looks like your teacher is indeed pretty horrible. Secondly, the constraints to consider would be proper weight distribution, methods to minimize excessive motion of the building structure, and quantities such as volume and density, which would help in determining the optimal structure. Keeping the frequency of oscillation for a building low in case of an earthquake or natural disaster would also be a priority.

4 0
3 years ago
A heat engine accepts 200,000 Btu of heat from a source at 1500 R and rejects 100,000 Btu of heat to a sink at 600 R. Calculate
diamong [38]

To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.

By definition we know that the change in entropy is given by

\Delta S = \frac{Q}{T}

Where,

Q = Heat transfer

T = Temperature

On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is

W = Q_{source}-Q_{sink}

According to the data given we have to,

Q_{source} = 200000Btu

T_{source} = 1500R

Q_{sink} = 100000Btu

T_{sink} = 600R

PART A) The total change in entropy, would be given by the changes that exist in the source and sink, that is

\Delta S_{sink} = \frac{Q_{sink}}{T_{sink}}

\Delta S_{sink} = \frac{100000}{600}

\Delta S_{sink} = 166.67Btu/R

On the other hand,

\Delta S_{source} = \frac{Q_{source}}{T_{source}}

\Delta S_{source} = \frac{-200000}{1500}

\Delta S_{source} = -133.33Btu/R

The total change of entropy would be,

S = \Delta S_{source}+\Delta S_{sink}

S = -133.33+166.67

S = 33.34Btu/R

Since S\neq   0 the heat engine is not reversible.

PART B)

Work done by heat engine is given by

W=Q_{source}-Q_{sink}

W = 200000-100000

W = 100000 Btu

Therefore the work in the system is 100000Btu

4 0
3 years ago
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