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horsena [70]
3 years ago
13

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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Determine the energy in joules of a photon whose frequency is 3.55 x10^17 hz
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By using the Plancks-Einstein equation, we can find the energy;
E = hf
where h is the plancks constant = 6.63 x 10⁻³⁴
f = frequency = 3.55 x 10¹⁷hz
E = (6.63 x 10⁻³⁴) x (3.55 x 10¹⁷)
E = 2.354 x 10⁻¹⁶J
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3 years ago
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The head of a grass string trimmer has 100 g of cord wound in a light, cylindrical spool with inside diameter 3.00 cm and outsid
faust18 [17]

Answer:

a).11.546J

b).2.957kW

Explanation:

Using Inertia and tangential velocity

a).

w=2250*2\pi *\frac{1}{60}\\ w=235.61

I=\frac{1}{2}*m*((\frac{d_{i} }{2})^{2} +(\frac{d_{e} }{2})^{2})\\m=100g *\frac{ikg}{1000g}=0.1kg\\ d_{i}=3cm*\frac{1m}{100cm}=0.03m \\ d_{e}=18cm*\frac{1m}{100cm}=0.18m\\I=\frac{1}{2}*0.1kg*((\frac{0.03m}{2})^{2} +(\frac{0.18m}{2})^{2})\\I=0.41625x10^{-3}kg*m^{2}

Now using Inertia an w

E=\frac{1}{2}*I*(w)^{2} \\ E=\frac{1}{2}*0.416x10^{-3}*(235.61)^{2} \\E=11.54J

average power=\frac{11.4J}{0.230s}=50.2 W

b).

power=t*w

P=11.5465*0.25*235.61

P=2.957 kW

4 0
3 years ago
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A coin 15.0 mm in diameter is placed 15.0 cm from a spherical mirror. The coin's image is 5.0 mm in diameter and is erect. Is th
s344n2d4d5 [400]

Answer:

15 cm

Explanation:

h_{o} = Diameter of the coin = 15 mm

h_{i} = Diameter of the image of coin =  5 mm

d_{o} = distance of the coin from mirror = 15 cm

d_{i} = distance of the image of coin from mirror = ?

Using the equation

\frac{d_{i}}{d_{o}} = \frac{- h_{i}}{h_{o}}

\frac{d_{i}}{15} = \frac{- (5)}{15}

d_{i} = - 5 cm

R = radius of curvature

Using the mirror equation

\frac{1}{d_{o}} + \frac{1}{d_{i}} = \frac{2}{R}

\frac{1}{15} + \frac{1}{- 5} = \frac{2}{R}

R = - 15 cm

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0<br> What is 33 C in absolute temperature? hs
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