Answer:
1000 N
Explanation:
The magnitude of the electrostatic force between two charged object is given by
where
k is the Coulomb constant
q1, q2 is the magnitude of the two charges
r is the distance between the two objects
Moreover, the force is:
- Attractive if the two forces have opposite sign
- Repulsive if the two forces have same sign
In this problem:
are the two charges
r = 3000 m is their separation
Therefore, the electric force between the charges is:
Answer:
t = 2.77 s
Explanation:
The ball in its movement describes a curved line called a semiparabola, therefore two coordinates are required to fix the position at each instant of time, since the movement is performed in the X-Y plane:
Equation of movement of the ball in the X axis
X = v₀x*t Equation (1)
Equation of movement of the ball in the Y axis
Y = y₀+ v₀y*t -½ g*t² Equation (2)
Where
X : horizontal position in meters (m)
Y : vertical position in meters (m)
y₀ : initial vertical position in meters (m)
v₀x : X-initial speed in m/s
v₀y : Y-initial speed in m/s
g: acceleration due to gravity in m/s²
t : time to position (X,Y)
Data
y₀ = 37.5 m
v₀y = 0
g = 9.8 m/s²
Problem development
The time the ball remains in the air is the same as the ball takes to touch the floor, that is, Y = 0
We apply the Equation (2):
Y = y₀+ (v₀y)*t - (½) g*t²
0 = 37.5 +(0)*t- (1/2)*(g)*t²
0 = 37.5 - (1/2)*(9.8)*t²
(1/2)*(9.8)*t² = 37.5
t² = (2)(37.5)/(9.8)
t = 2.77 s