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faust18 [17]
3 years ago
8

When the Earth is closest to the Sun, it is at:

Physics
1 answer:
ahrayia [7]3 years ago
5 0

a). Perihelion . . . the point in Earth's orbit that's closest to the Sun.
                            We pass it every year early in January. 


b). Aphelion . . . the point in Earth's orbit that's farthest from the Sun.
                          We pass it every year early in July. 

c). Proxihelion . . . a made-up, meaningless word

d). Equinox . . . the points on the map of the stars where the Sun
                         appears to be on March 21 and September 21.

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A skateboarder shoots off a ramp with a velocity of 7.3 m/s, directed at an angle of 60° above the horizontal. The end of the ra
erica [24]

Answer:

y maximum   3.54 m, value X  2.35 m

Explanation:

We have a projectile launch problem, let's calculate the maximum height of the projectile, where the vertical speed must be zero

                     

        Vyf² = Vyo² - 2g (Y-Yo)

Where Yo is the initial height of the ramp 1.5 m

        0 = Vyo² -2g (Y-Yo)

        Y-Yo = Voy² / 2g

       Y = Yo + Voy² / 2g

Let's calculate the velocity components using trigonometry

       Voy = vo without T

       Vox = Vo cost

       Voy = 7.3 sin 60

       Vox = 7.3 cos 60

       Voy = 6.32 m / s

       Vox = 3.65 m / s

Let's calculate the maximum height

         Y = 1.5 +6.32²/2 9.8

          Y = 3.54 m

This is the maximum height from the ground

b) They ask us for the position of this point horizontally, we can calculate it looking for the time it took for the skateboarder to reach the highest point

                             

          Vfy = Voy - gt

          0 = Voy - gt

           t = Voy / g

           t = 6.32 / 9.8

           t = 0.645 s

     

Since there is no acceleration on the x-axis, we have a uniform movement, we can calculate the distance for this time

          X = Vox t

          X = 3.65 0.645

          X= 2.35 m

8 0
3 years ago
If electromagnetic radiation acted like particles in the double-slit experiment, what would be observed? (1 point)
ludmilkaskok [199]

If electromagnetic radiation acted like particles in the double-slit experiment, we would observe one bright band would appear in the center of the screen.

<h3>Bahavior of particles in double-slit experiment</h3>

In a double-slit experiment, single particles, such as photons, pass one at a time through a screen containing two slits.

The photons behave like wave and the constructive interfernce of the waves of these photons will generate a high amplitude wave seen as a bright band in the center of the screen.

Thus, if electromagnetic radiation acted like particles in the double-slit experiment, we would observe one bright band would appear in the center of the screen.

Learn more about double slit experiment here: brainly.com/question/4449144

6 0
2 years ago
An electron enters a region of uniform perpendicular en and bn fields. it is observed that thevelocity nv of the electron is una
konstantin123 [22]
<span>v is perpendicular to both E and B and has a magnitude E/B</span>
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3 years ago
Which possesses the most gravitational potential energy?
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Answer:

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3 years ago
Read 2 more answers
A speedboat is approaching a dock at 25 m/s (56 mph). When the dock is 150 m away, the driver begins to slow down. a) What accel
trasher [3.6K]

Answer:

a) -2.038 m/s²

b) 40.33 mph

c) 312.5 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-25^2}{2\times 150}\\\Rightarrow a=-2.083\ m/s^2

Acceleration of the boat is -2.083 m/s² if the boat will stop at 150 m.

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -1\times 150+25^2}\\\Rightarrow v=18.03\ m/s

Speed of the boat by when it will hit the dock is 18.03 m/s

Converting to mph

1\ mile=1609.34\ m

1\ h=3600\ seconds

18.03\times \frac{3600}{1609.34}=40.33\ mph

Speed of the boat by when it will hit the dock is 40.33 mph

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-25^2}{2\times -1}\\\Rightarrow s=312.5\ m

The distance at which the boat will have to start decelerating is 312.5 m

5 0
3 years ago
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