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Anna11 [10]
3 years ago
13

Most of an atom is empty space.

Physics
1 answer:
yan [13]3 years ago
3 0

Answer:

I found this on the internet: Atoms are not mostly empty space because there is no such thing as purely empty space. Rather, space is filled with a wide variety of particles and fields. ... It's true that a large percentage of the atom's mass is concentrated in its tiny nucleus, but that does not imply that the rest of the atom is empty.

Explanation:

So i am going to go with false.

Hope i helped!

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Swampy areas can be found in many Florida locations. What weather conditions are most frequently found in this environment? *
Black_prince [1.1K]

Answer:

Heavy rain, extreme weather conditions, humid, water spouts, tornadoes, heavy lightening.

Explanation:

Swampy areas are a transition area formed where water and dry lands tend to meet. Swampy areas are actually forested areas with contains moist and humid weather conditions which has necessitated the wet soil around such an environment.

Florida is a very typical example of a weather that supports the creation of swampy areas due to its humid, heavy rains, water spouts,turnadoes etc and other extreme weather conditions.

7 0
3 years ago
Suppose that A’, B’ and C’ are at rest in frame S’, which moves with respect to S at speed v in the +x direction. Let B’ be loca
Inga [223]

Answer:

1) an observer in B 'sees the two simultaneous events

2)observer B sees that the events are not simultaneous

3)  Δt = Δt₀ /√ (1 + v²/c²)

Explanation:

This is an exercise in simultaneity in special relativity. Let us remember that the speed of light is the same in all inertial systems

1) The events are at rest in the reference system S ', so as they advance at the speed of light which is constant, so it takes them the same time to arrive at the observation point B' which is at the point middle of the two events

Consequently an observer in B 'sees the two simultaneous events

2) For an observer B in system S that is fixed on the Earth, see that the event in A and B occur at the same instant, but the event in A must travel a smaller distance and the event in B must travel a greater distance since the system S 'moves with velocity + v. Therefore, since the velocity is constant, the event that travels the shortest distance is seen first.

Consequently observer B sees that the events are not simultaneous

3) let's calculate the times for each event

        Δt = Δt₀ /√ (1 + v²/c²)

where t₀ is the time in the system S' which is at rest for the events

8 0
3 years ago
A commuter airplane starts from an airport and takes theroute. The plane first flies to city A, located 175 km away in a directi
bearhunter [10]

Answer:

245.45km in a direction 21.45° west of north from city A

Explanation:

Let's place the origin of a coordinate system at city A.

The final position of the airplane is given by:

rf = ra + rb + rc    where ra, rb and rc are the vectors of the relative displacements the airplane has made. If we separate this equation into its x and y coordinates:

rfX = raX+ rbX + rcX = 175*cos(30)-150*sin(20)-190 = -89.75km

rfY = raY + rbY + rcT = 175*sin(30)+150*cos(20) = 228.45km

The module of this position is:

rf = \sqrt{rfX^2+rfY^2} = 245.45km

And the angle measure from the y-axis is:

\alpha =atan(rfX/rfY) = 21.45\°

So the answer is 245.45km in a direction 21.45° west of north from city A

6 0
4 years ago
[O.04H]The table below shows the use of some energy production methods over time.
mr Goodwill [35]

I think The coastal areas were highly polluted

8 0
3 years ago
Read 2 more answers
The current theory of the structure of the
irina1246 [14]

Answer:

pt 1: m=1.66698*10^{21} kg

Pt 2: KE=1212.23531 J

Explanation:

Information Given: (p = density)

l = 5200km  d = 35km p = 2700kg/m^{2}

Part 1: Mass

  • Find volume
  1. V=(l)^2(d)
  2. V=(4.2*10^6)^2(35*10^3)
  3. V=61.74*10^{16}
  • Find Mass
  1. m=Vp
  2. m=(61.74*10^{16})(2700)
  3. m=1.66698*10^{21}

Part 2: Kinetic Energy

  1. v=\frac{3.8cm}{yr}*\frac{m}{100cm}*\frac{yr}{365d}*\frac{d}{24hr}*\frac{hr}{3600s}
  2. v=1.20497*10^{-9}

KE=\frac{1}{2}mv^2

KE=\frac{1}{2} (1.66698*10^{21})(1.20497*10^{-9})^2

KE=1212.23531 J

Part 3: Jogger Speed

set up, because I don't have the mass :(

Information given:

KE_{jogger}

  1. KE=\frac{1}{2}mv^2
  2. v_{jogger} =\sqrt{\frac{2KE}{m_{jogger} } }
  • Input the values

Hope it helps :)

6 0
3 years ago
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