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amm1812
3 years ago
9

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?
a. Both areas are the same.
b. The area swept out in January is larger.
c. The area swept out in July is larger.
Physics
1 answer:
suter [353]3 years ago
3 0

Answer: Option <em>a.</em>

Explanation:

Kepler's 2nd law of planetary motion states:

<em>A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.</em>

It tells us that it doesn't matter how far Earth is from the Sun, at equal times, the area swept out by Earth's orbit it's always the same independently from the position in the orbit.

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Which of the following is an example of a conductor?
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(8c7p26) During spring semester at MIT, residents of the parallel buildings of the East Campus dorms battle one another with lar
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Answer: 1175 J

Explanation:

Hooke's Law states that "the strain in a solid is proportional to the applied stress within the elastic limit of that solid."

Given

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from the question, x(f) = 0 and x(i) = maximum elongation = 4.8 m

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W = 1/2 * 102 * 4.8²

W = 1/2 * 102 * 23.04

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7 0
3 years ago
An astronaut finds herself in a predicament in which she has become untethered from her shuttle. She figures that she could get
Blizzard [7]

In order to solve the problem, it is necessary to apply the concepts related to the conservation of momentum, especially when there is an impact or the throwing of an object.

The equation that defines the linear moment is given by

mV_i = (m-m_O)V_f - m_OV_O

where,

m=Total mass

m_O = Mass of Object

V_i = Velocity before throwing

V_f = Final Velocity

V_O = Velocity of Object

Our values are:

m_1=5.3kgm_2=7.9kg\\m_3=10.5kg\\m_A=75kg\\m_{Total}=m=98.7Kg

Solving to find the final speed, after throwing the object we have

V_f=\frac{mV_0+m_TV_O}{m-m_O}

We have three objects. For each object a launch is made so the final mass (denominator) will begin to be subtracted successively. In addition, during each new launch the initial speed will be given for each object thrown again.

That way during each section the equations should be modified depending on the previous one, let's start:

A) 5.3Kg\rightarrow 15m/s

V_{f1}=\frac{mV_0+m_TV_O}{m-m_O}

V_{f1}=\frac{(98.7)*0+5.3*15}{98.7-5.3}

V_{f1}=0.8511m/s

B) 7.9Kg\rightarrow 11.2m/s

V_{f2}=\frac{mV_{f1}+m_TV_O}{m-m_O}

V_{f2}=\frac{(98.7)(0.8511)+(7.9)(11.2)}{98.7-5.3-7.9}

V_{f2} = 2.0173m/s

C) 10.5Kg\rightarrow 7m/s

V_{f3}=\frac{mV_{f2}+m_TV_O}{m-m_O}

V_{f3}=\frac{(98.7)(2.0173)+(10.5)(7)}{98.7-5.3-7.9-10.5}

V_{f3} = 3.63478m/s

Therefore the final velocity of astronaut is 3.63m/s

7 0
3 years ago
It takes you 8.3 min to walk with an average velocity of 1.6 m/s to the north from the bus stop to the museum entrance. How far
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solution:

1.6 m/s = 96 m/min (in other words, 1.6 m/s x 60 s/min)  

96 m/min x 8.3 min = 796.8 m

s=ut +\frac{1}{2}at^2\\there is no accleration mentioned so,\\s= uv\\8.3\times60=498(s)\\510\times1.6=816(m)


3 0
3 years ago
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