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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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Rearrange the equation for density to solve for mass.
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Density = mass/volume

To rearrange the equation for density to solve for MASS

you will get
mass = density x volume
5 0
3 years ago
A vacuum gage attached to a power plant condenser gives a reading of 27.86 in. of mercury. The surrounding atmospheric pressure
Nadusha1986 [10]

Answer:

absolute pressure =  1.07 lbft/in^2

Explanation:

given data:

vaccum gauge reading h = 27.86 inch = 2.32 ft

we know that

gauge pressure p is given as

p = \rho gh

p = 848 \ lb/ft^3 * 32 \ ft/s^2 *2.32 ft  = 62955.52 \ lbft/s^2 * 1/ft^2

we know that  1\  lb ft\s^2 = \frac{1}{32.174}\  lbft

1 ft = 12 inch

therefore p = 62955.52 * \frac{1}{32.174} * \frac{1}{12^2}\ lbf/in^2

               p = 13.59\  lbft/in^2

P_{atm} = 14.66\  lbf/in^2

so absolute pressure = P_{atm} - p

                                   = 14.66 - 13.59 = 1.07 lbft/in^2

4 0
3 years ago
A car starts moving after waiting for a traffic light to turn green. It is able to travel a distance of 300 meters in 10 seconds
Marizza181 [45]

Explanation: you will nee to divide 300 divided by 10 and you will have your answer but be sure to add m/s at the end of the problem......

5 0
3 years ago
Read 2 more answers
A chain 72 meters long whose mass is 29 kilograms is hanging over the edge of a tall building and does not touch the ground. How
dangina [55]

Answer:

Work done required is 3567.2 J

Explanation:

Given :

Length of chain, l = 72 m

Mass of chain, M = 29 kg

Linear mass density of chain, μ = \frac{Mass\ of\ chain }{Length\ of\ chain} = \frac{29}{72}  = 0.40 kg/m

Let x be the length of the chain which lift to the top of the building.

Work done required to lift the chain is equal to the potential energy of the chain.

W = ∫μg (72 - x ) dx

Here g is acceleration due to gravity.

The limit of integration is from 0 to 14.

W = μg ( 72x - x²/2)

Substitute 0.40 kg/m for μ, 9.8 m/s² for g and 14 m for x in the above equation.

W = 0.40\times9.8\times(72\times14\ - \frac{14^{2} }{2})

W = 3567.2 J

5 0
3 years ago
If you divide a speed in miles per minute by 60, you get the same speed
Mariulka [41]
Yes so x multiply by the hour but u add 59 for each hour to get the exath speed per minute
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