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Katyanochek1 [597]
3 years ago
11

Earth is slightly closer to the sun in january than in july. How does the area swept out by earth’s orbit around the sun during

the 31 days of january compare to the area swept out during the 31 days of july?
Physics
1 answer:
mr Goodwill [35]3 years ago
6 0

This is because both areas are the same.

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The sun's _____ is the layer that emits visible light.
galben [10]
The Photosphere is the right answer
7 0
3 years ago
The primary source of evidence proposed by many scientist to support the theory of an ancient earth is ____ dating.
vova2212 [387]
It depends on your definition of “ancient.” Radiometric dating using Carbon-14 can reliably date back to about 50,000 years, uranium-lead or lead-lead dating can date back multiple millions, potassium-argon dating can reach 1.5 billion, and rubidium-strontium can reach 50 billion (nearly 4x the age of the universe). It depends on the context in which this question is being asked.
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3 years ago
A gas expands against a constant external pressure of 2.00 atm until its volume has increased from 6.00 to 10.00 L. During this
mars1129 [50]

Answer:

ΔU = - 310.6 J (negative sign indicates decrease in internal energy)

W = 810.6 J

Explanation:

a.

Using first law of thermodynamics:

Q = ΔU + W

where,

Q = Heat Absorbed = 500 J

ΔU = Change in Internal Energy of Gas = ?

W = Work Done = PΔV =

P = Pressure = 2 atm = 202650 Pa

ΔV = Change in Volume = 10 L - 6 L = 4 L = 0.004 m³

Therefore,

Q = ΔU + PΔV

500 J = ΔU + (202650 Pa)(0.004 m³)

ΔU = 500 J - 810.6 J

<u>ΔU = - 310.6 J (negative sign indicates decrease in internal energy)</u>

<u></u>

b.

The work done can be simply calculated as:

W = PΔV

W = (202650 Pa)(0.004 m³)

<u>W = 810.6 J</u>

7 0
3 years ago
A kayaker needs to paddle north across a 100-m-wide harbor. The tide is going out, creating a tidal current that flows to the ea
Savatey [412]

Answer:

41.81^{\circ}

Explanation:

The tidal current flows to the east at 2.0 m/s and the speed of the kayaker is 3.0 m/s.

Let Vector \overrightarrow{OA} is the tidal current velocity as shown in the diagram.

In order to travel straight across the harbor, the vector addition of both the velocities (i.e the resultant velocity, \vec {R} must be in the north direction.

Let \overrightarrow{AB} is the speed of the kayaker having angle \theta measured north of east as shown in the figure.

For the resultant velocity in the north direction, the tail of the vector \overrightarrow {OA} and head of the vector \overrightarrow{AB} must lie on the north-south line.

Now, for this condition, from the triangle OAB

|\overrightarrow{AB}|\sin \theta=|\overrightarrow{OA}|

\Rightarrow \sin\theta=\frac{|\overrightarrow{OA}|}{|\overrightarrow{AB}|}=\frac 2 3

\Rightarrow \theta=\sin^{-1}\frac23

\Rightarrow \theta=41.81^{\circ}

Hence, the kayaker must paddle in the direction of 41.81^{\circ}  in the north of east direction.

3 0
3 years ago
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8 0
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