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Lynna [10]
2 years ago
6

Why isn't the earth the same distance from sun all year long?

Physics
2 answers:
miskamm [114]2 years ago
7 0

Answer:

Earth's orbit is elliptical

Explanation:

The earth's orbit is not a straight circle and the sun is not in the very center of it. The orbit is more of a wonky oval with the sun closer to one side of it than the other causing the earth's distance from the sun to vary throughout the year.

attashe74 [19]2 years ago
5 0

Because the planets (and everything else in the solar system) don't travel in circular orbits with the sun at the center of the circle.  

They travel in elliptical orbits, with the sun at one focus of the ellipse.

Different points on the ellipse have different distances from the focus where the sun is.

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A 530-g squirrel with a surface area of 935 cm2 falls from a 4.4-m tree to the ground. Estimate its terminal velocity. (Use the
Agata [3.3K]

Answer with Explanation:

We are given that

Mass of squirrel,m=530 g=\frac{530}{1000}=0.530 kg

1kg=1000 g

Area=A=935 cm^2=935\times 10^{-4} m^2

1 cm^2=10^{-4} m^2

Height,h=4.4 m

C=1

\rho=1.21 kg/m^3

Width of rectangular prism,b=11.6 cm=\frac{11.6}{100}=0.116 m

1 m=100 cm

Length,l=23.2 cm=0.232 m

Area=l\times b=0.116\times 0.232=0.0269 m^2

Terminal velocity,v_t=\sqrt{\frac{2mg}{\rho CA}}

Where g=9.8 m/s^2

Using the formula

v_t=\sqrt{\frac{2\times 0.530\times 9.8}{1.21\times 1\times 0.0269}}

v_t=17.86 m/s

The velocity of person,v=\sqrt{2gh}

Using the formula

v=\sqrt{2\times 9.8\times 4.4}

v=9.29 m/s

4 0
2 years ago
Wave 1 carries more energy than Wave 2. Wave 3 carries more energy than Wave 2. Keeping speed constant, the greatest amount of e
inn [45]

Answer:

<em>The greatest amount of energy is carried by a wave with the  largest frequency.</em>

Explanation:

<em>velocity= frequency X wavelength </em>

<em>If the three waves moves with the same t the speed of light, then those with the largest frequency have more energy. </em>

<em>That is to say,  a wave with a higher frequency and a  lower  wavelength will have the highest energy.</em>

7 0
2 years ago
The diagram below shows eight different positions of the moon around Earth.
KATRIN_1 [288]

Answer:

6 and 8

Explanation:

The different positions of the moon, as seen from Earth, reflect how much light is being reflected off the moon from the sun.

Position 1 indicates the 3rd Quarter.

Position 2 indicates the Waning Gibbous.

Position 3 indicates the Full Moon.

Position 4 indicates the Waxing Gibbous.

Position 5 indicates the 1st Quarter.

Position 6 indicates the Waxing Crescent.

Position 7 indicates the New Moon.

Position 8 indicates the Waning Crescent.

You would read the diagram counterclockwise, with positions 7 to 3 as the moon increases light and positions 3 to 7 as the moon decreases light.

Therefore, we see that our 2 choices where we have a crescent is positions 6 and 8.

6 0
3 years ago
Suppose an extraction with methylene chloride (d = 1.33 g/mL), also called dichloromethane, is performed with the desired compou
bekas [8.4K]

Answer:

Methylene chloride.

Explanation:

Two layers in the separation funnel are the aqueous phase and the organic phase. For solvents lighter than water (i.e., density < 1), the organic phase will rest on top in the separatory funnel, whereas solvents denser than water (density > 1) will sink to the bottom.

Since, methylene chloride is denser than Brine therefore, methylene chloreide sinks at the bottom while methylene chloride stays on top.

3 0
2 years ago
2000, the Millennium Bridge, a new footbridge over the River Thames in London, England, was opened to the public. However, after
sweet [91]

Answer:

n = 1810

A = 25 mm

Explanation:

Given:

Lateral force amplitude, F = 25 N

Frequency, f = 1 Hz

mass of the bridge, m = 2000 kg/m

Span, L = 144 m

Amplitude of the oscillation, A = 75 mm = 0.075 m

time, t = 6T

now,

Amplitude as a function of time is given as:

A(t)=A_oe^{\frac{-bt}{2m}}

or amplitude for unforce oscillation

\frac{A_o}{e}=A_oe^{\frac{-b(6T)}{2m}}

or

\frac{6bt}{2m}=1

or

b=\frac{m}{3T}

Now, provided in the question Amplitude of the driven oscillation

A=\frac{F_{max}}{\sqrt{(k-m\omega_d^2)+(b\omega_d^2)}}

the value of the maximum amplitude is obtained (k=m\omega_d^2)

thus,

A=\frac{F_{max}}{(b\omega_d}

Now, for n people on the bridge

Fmax = nF

thus,

max amplitude

0.075=\frac{nF}{((\frac{m}{3T})2\pi}

or

n = 1810

hence, there were 1810 people on the bridge

b)A=\frac{F_{max}}{(b\omega_d}

since the effect of damping in the millenium bridge is 3 times

thus,

b=3b

therefore,

A=\frac{F_{max}}{(3b\omega_d}

or

A=\frac{1}{(3}A_o

or

A=\frac{1}{(3}0.075

or

A = 0.025 m = 25 mm

6 0
2 years ago
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