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3 years ago

During which month does the sun rise north of due east in New York State?

Physics

1 answer:

Lubov Fominskaja [6]3 years ago

7 0

Answer:(2) July

Explanation:

We all know the sun rises in the east and sets in the west.

No matter where you are on Earth excluding the North and South Poles, you have a due east and due west point on your horizon. That point marks the intersection of your horizon with the celestial equator, the imaginary great circle above the true equator of the Earth.

This is to say that the sun rises close to due east and sets close to due west, for everyone at the equinox. The equinox sun is on the celestial equator. The celestial equator intersects your horizon at due east and due west irrespective of where you are on earth.

The sun rises due east and sets due west during the spring and fall equinoxes. Other times, the sun rises either north or south due east. There are slight changes in the rising and setting of the sun each day. At summer solstice, the sun rises far to the north east and set to the north west. Everyday, the sun rises a bit in furtherance to the south.

Therefore in Newyork state, during summer in July, the sun rise north of due east.

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A man-made satellite of mass 6105 kg is in orbit around the earth, making one revolution in 430 minutes. What is the magnitude o

blondinia [14]

Answer:

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

Explanation:

At first we assume that Earth is represented by an uniform sphere, such that the man-made satellite rotates in a circular orbit around the planet. Hence, the following condition must be satisfied:

$\left(\frac{4\pi^{2}}{T^{2}} \right)\cdot r = \frac{G\cdot M}{r^{2}}$ (1)

Where:

$T$ - Period of rotation of the satellite, measured in seconds.

$r$ - Distance of the satellite with respect to the center of the planet, measured in meters.

$G$ - Gravitational constant, measured in newton-square meters per square kilogram.

$M$ - Mass of the Earth, measured in kilograms.

Now we clear the distance of the satellite with respect to the center of the planet:

$r^{3} = \frac{G\cdot M\cdot T^{2}}{4\pi^{2}}$

$r = \sqrt[3]{\frac{G\cdot M\cdot T^{2}}{4\pi^{2}} }$ (2)

If we know that $G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}$ , $M = 6.0\times 10^{24}\,kg$ and $T = 25800\,s$ , then the distance of the satellite is:

$r = \sqrt[3]{\frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6.0\times 10^{24}\,kg)\cdot (25800\,s)^{2}}{4\pi^{2}} }$

$r \approx 18.897\times 10^{6}\,m$

The gravitational force exerted on the satellite by the Earth is determined by the Newton's Law of Gravitation:

$F = \frac{G\cdot m\cdot M}{r^{2}}$ (3)

Where:

$m$ - Mass of the satellite, measured in kilograms.

$F$ - Force exerted on the satellite by the Earth, measured in newtons.

If we know that $G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}$ , $M = 6.0\times 10^{24}\,kg$ , $m = 6105\,kg$ and $r \approx 18.897\times 10^{6}\,m$ , then the gravitational force is:

$F = \frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6105\,kg)\cdot (6\times 10^{24}\,kg)}{(18.897\times 10^{6}\,m)^{2}}$

$F = 6841.905\,N$

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

4 0

3 years ago

A 5.67-kg block of wood is attached to a spring with a spring constant of 150 N/m. The block is free to slide on a horizontal fr

nikklg [1K]

Answer:

v=0.94 m/s

Explanation:

Given that

M= 5.67 kg

k= 150 N/m

m=1 kg

μ = 0.45

The maximum acceleration of upper block can be μ g.

a= μ g ( g = 10 m/s²)

The maximum acceleration of system will ω²X.

ω = natural frequency

X=maximum displacement

For top stop slipping

μ g =ω²X

We know for spring mass system natural frequency given as

$\omega=\sqrt{\dfrac{k}{M+m}}$

By putting the values

$\omega=\sqrt{\dfrac{150}{5.67+1}}$

ω = 4.47 rad/s

μ g =ω²X

By putting the values

0.45 x 10 = 4.47² X

X = 0.2 m

From energy conservation

$\dfrac{1}{2}kX^2=\dfrac{1}{2}(m+M)v^2$

$kX^2=(m+M)v^2$

150 x 0.2²=6.67 v²

v=0.94 m/s

This is the maximum speed of system.

7 0

2 years ago

Read 2 more answers

Which part of the wave has the highest frequency?

trapecia [35]

Answer:

The last part on the right side of the diagram

Explanation:

Im on plato and just got it right :)

6 0

2 years ago

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_____ of rough surfaces reduces friction

Radda [10]

Answer:

Hey shaikaadil700 !

Lubricating  of rough surfaces reduces friction.

Explanation:

• Lubricating is the smoothening or polishing of the surfaces

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3 0

2 years ago

What is the difference between a mechanical wave and an electromagnetic wave

ivann1987 [24]

Answer:

1. Electromagnetic waves travel in a vacuum whereas mechanical waves do not.

2. The ripples made in a pool of water after a stone is thrown in the middle are an example of mechanical wave. Examples of electromagnetic waves include light and radio signals.

3. Mechanical waves are caused by wave amplitude and not by frequency. Electromagnetic Waves are produced by vibration of the charged particles.

4. While an electromagnetic wave is called just a disturbance, a mechanical wave is considered a periodic disturbance.

Explanation:

5 0

2 years ago