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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 small ball of mass 2.00 kilograms is moving at a velocity 1.50 meters/second. It hits a larger, stationary ball of mass 5.00 k

rewona [7]

The kinetic energy of the small ball before the collision is

   KE = (1/2) (mass) (speed)²

   = (1/2) (2 kg) (1.5 m/s)

   =    (1 kg) (2.25 m²/s²)

   =    2.25 joules.

Now is a good time to review the Law of Conservation of Energy:

   Energy is never created or destroyed.
   If it seems that some energy disappeared,
   it actually had to go somewhere.
   And if it seems like some energy magically appeared,
   it actually had to come from somewhere.

The small ball has 2.25 joules of kinetic energy before the collision.
If the small ball doesn't have a jet engine on it or a hamster inside,
and does not stop briefly to eat spinach, then there won't be any
more kinetic energy than that after the collision. The large ball
and the small ball will just have to share the same 2.25 joules.

3 0

3 years ago

A block lies on a frictionless floor. a force of 5 n pulls toward the east while a force of 4 n pulls toward the north. what is

sveticcg [70]

Given below the arrangement of loading on the larger boat by two tug boats.

F₁ = 5 N

F₂ = 4 N

Angle between them θ = 90⁰

Resultant between two vectors, $F=\sqrt{F_1^2+F_2^2+2F_1F_2cos\theta }$

Substituting

$F = \sqrt{5^2+4^2+2*5*4*cos 90} \\ \\ = 6.403 N$

So magnitude of the net force on the block = 6.403 N

3 0

3 years ago

Electric resistance ____________ with a(n) ____________in the length of a wire and as a result current flow ___________.

maria [59]

The answer is D. Electric resistance increases with an increase in the length of a wire and as a result current flow decreases. There is a direct relationship between the length of the wire and the resistance. The longer the wire, the more resistance there will be. Additionally, from Ohm's Law, current is inversely proportional to resistance. This means as the current increases, resistance decreases or vice versa.

3 0

3 years ago

Two asteroids identical to those above collide at right angles and stick together; i.e, their initial velocities were perpendicu

11111nata11111 [884]

Answer:

velocity = 62.89 m/s in 58 degree measured from the x-axis

Explanation:

Relevant information:

Before the collision, asteroid A of mass 1,000 kg moved at 100 m/s, and asteroid B of mass 2,000 kg moved at 80 m/s.

Two asteroids moving with velocities collide at right angles and stick together. Asteroid A initially moving to right direction and asteroid B initially move in the upward direction.

Before collision Momentum of A = 1000 x 100 = $10^5$ kg - m/s in the right direction.

Before collision Momentum of B = 2000 x 80 = 1.6 x $10^5$ kg - m/s in upward direction.

Mass of System of after collision = 1000 + 2000 = 3000 kg

Now applying the Momentum Conservation, we get

Initial momentum in right direction = final momentum in right direction = $10^5$

And, Initial momentum in upward direction = Final momentum in upward direction = 1.6 x $10^5$

So, $V_x = \frac{10^5}{3000}$ = $\frac{100}{3}$ m/s

and $V_y=\frac{160}{3}$ m/s

Therefore, velocity is = $\sqrt{V_x^2 + V_y^2}$

= $\sqrt{(\frac{100}{3})^2 + (\frac{160}{3})^2}$

= 62.89 m/s

And direction is

tan θ = $\frac{V_y}{V_x}$ = 1.6

therefore, $\theta = \tan^{-1}1.6$

= $58 ^{\circ}$ from x-axis

4 0

3 years ago

To take off from the ground, an airplane must reach a sufficiently high speed. The velocity required for the takeoff, the takeof

il63 [147K]

<h2> The potential and kinetic energy of airplane are affected by these factors </h2>

Explanation:

When airplane rises up , it requires potential energy . This potential energy can be taken from the kinetic energy of airplane .

Thus if the speed of wind is larger , it can either oppose the motion of velocity or can favour the velocity of airplane . By which its kinetic energy is effected .

If the weight of airplane is changed , it will effect the potential energy required . Thus heavier plane requires higher potential energy for attaining the same height .

Thus these two factor has important role in the flight of airplane .

6 0

3 years ago