The initial force between the two charges is given by:
where k is the Coulomb's constant, q1 and q2 the two charges, d their separation. Let's analyze now the other situations:
1. F
In this case, q1 is halved, q2 is doubled, but the distance between the charges remains d.
So, we have:
So, the new force is:
So the force has not changed.
2. F/4
In this case, q1 and q2 are unchanged. The distance between the charges is doubled to 2d.
So, we have:
So, the new force is:
So the force has decreased by a factor 4.
3. 6F
In this case, q1 is doubled and q2 is tripled. The distance between the charges remains d.
So, we have:
So, the new force is:
So the force has increased by a factor 6.
The plant that is closest to the sun is murcury. Then it is venus, then earth, and then mars. Then it is jupiter, then saturn, then uranus, then neptune.
<span>A major characteristic of both volcanoes and earthquakes is that they are located in the same geographic area. Most earthquakes are along the edges of tectonic plates. This is where most volcanoes are too. Most earthquakes directly beneath a volcano are caused by the movement of magma.</span>
The chaotic nature of the Solar System excluding Pluto was established by the numerical computation of the maximum Lyapunov exponent of its secular system over 200 myr.
<h3>What is chaotic motion of the solar system ?</h3>
There has been an increase in awareness of chaotic dynamics in the solar system over the past 20 years. The orbits of tiny objects in the solar system, such as asteroids, comets, and interplanetary dust, are now known to be chaotic and to experience significant variations across geological time periods.
- a completely unpredictable orbit, or one where significant changes in the orbit can result from even small changes in the position and/or velocity of the orbiting entity.
Learn more about Chaotic motion here:
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Given:
Area of pool = 3m×4m
Diameter of orifice = 0.076m
Outlet Velocity = 6.3m/s
Accumulation velocity = 1.5cm/min
Required:
Inlet flowrate
Solution:
The problem can be solved by this general formula.
Accumulation = Inlet flowrate - Outlet flowrate
Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice
First, we need to convert the units of the accumulation velocity into m/s to be consistent.
Accumulation velocity = 1.5cm/min × (1min/60s)×(1m/100cm)
Accumulation velocity = 0.00025 m/s
We then calculate the area of the pool and the area of the orifice by:
Area of pool = 3 × 4 m²
Area of pool = 12m²
Area of orifice = πd²/4 = π(0.076m)²/4
Area of orifice = 0.00454m²
Since we have all we need, we plug in the values to the general equation earlier
Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice
0.00025 m/s × 12m² = Inlet flowrate - 6.3m/s × 0.00454m²
Transposing terms,
Inlet flowrate = 0.316 m³/s