With acceleration

and initial velocity

the velocity at time <em>t</em> (b) is given by




We can get the position at time <em>t</em> (a) by integrating the velocity:

The particle starts at the origin, so  .
.



Get the coordinates at <em>t</em> = 8.00 s by evaluating  at this time:
 at this time:


so the particle is located at (<em>x</em>, <em>y</em>) = (64.0, 64.0).
Get the speed at <em>t</em> = 8.00 s by evaluating  at the same time:
 at the same time:


This is the <em>velocity</em> at <em>t</em> = 8.00 s. Get the <em>speed</em> by computing the magnitude of this vector:

 
        
             
        
        
        
Answer:
288.0 units; that is the electrostatic force of attraction become quadruple of its initial value.
Explanation:
If all other parameters are constant, 
Electrostatic Force of attraction ∝ (1/r²)
F = (k/r²) = 72.0
If r₁ = r/2, what happens to F₁
F₁ = (k/r₁²) = k/(r/2)² = (4k/r²) = 4F = 4 × 72 = 288.0 units
 
        
                    
             
        
        
        
Answer:
  P₂ = 138.88 10³ Pa
Explanation:
This is a problem of fluid mechanics, we must use the continuity and Bernoulli equation
Let's start by looking for the top speed
         Q = A₁ v₁ = A₂ v₂
We will use index 1 for the lower part and index 2 for the upper part, let's look for the speed in the upper part (v2)
          v₂ = A₁ / A₂ v₁
They indicate that A₂ = ½ A₁ and give the speed at the bottom (v₁ = 1.20 m/s)
          v₂ = 2  1.20
          v₂ = 2.40 m / s
Now let's write the Bernoulli equation
         P₁ + ½ ρ v₁² + ρ g y₁ = P2 + ½ ρ v₂² + ρ g y₂
Let's clear the pressure at point 2
        P₂ = P₁ + ½ ρ (v₁² - v₂²) + ρ g (y₁-y₂)
we put our reference system at the lowest point
         y₁ - y₂ = -20 cm
Let's calculate
        P₂ = 143 10³ + ½ 1000 (1.20² - 2.40²) + 1000 9.8 (-0.200)
        P₂ = 143 103 - 2,160 103 - 1,960 103
        P₂ = 138.88 10³ Pa