The magnitude of the force<span> a 1.5 x 10-3 C charge exerts on a 3.2 x 10-4 C charge located 1.5 m away is 1920 Newtons. The formula used to solve this problem is:
F = kq1q2/r^2
where:
F = Electric force, Newtons
k = Coulomb's constant, 9x10^9 Nm^2/C^2
q1 = point charge 1, C
q2 = point charge 2, C
r = distance between charges, meters
Using direct substitution, the force F is determined to be 1920 Newtons.</span>
Answer:
I believe the correct answer would be A :)
Explanation:
Answer:
a) p = m1 v1 + m2 v2
, b) dp / dt = m1 a1 + m2 a2
, c) It is equivalent to force
dp / dt = 0
Explanation:
In this problem we have two blocks and the system is formed by the two bodies.
Part A. Initially they ask us to find the moment of the whole system
p = m1 v1 + m2 v2
Part B.
Find the derivative
dp / dt = m1 dv1dt + m2 dv2 / dt
dp / dt = m1 a1 + m2 a2
Part C.
Let's analyze the dimensions
m a = [kg] [m / s2] = [N]
It is equivalent to force
Part d
Acceleration is due to a net force applied
Part e
The acceleration of block 1 is due to the force exerted by block 2 during the moment change
Part f
Force of block 1 on block 2
True f12 = m1a1 f21 = m2a2
Part g
By the law of action and reaction are equal magnitude F12 = f21
Part H
dp / dt = 0
Isolated system F12 = F21 and the masses are constant. The total moment is only redistributed
To solve this problem it is necessary to apply the concepts related to Newton's second law and the kinematic equations of movement description.
Newton's second law is defined as
Where,
m = mass
a = acceleration
From this equation we can figure the acceleration out, then
From the cinematic equations of motion we know that
Where,
Final velocity
Initial velocity
a = acceleration
x = displacement
There is not Final velocity and the acceleration is equal to the gravity, then
From the equation of motion where acceleration is equal to the velocity in function of time we have
Therefore the time required is 0.0705s
When the system is experiencing a uniformly accelerated motion, there are a set of equations to work from. In this case, work is energy which consist solely of kinetic energy. That is, 1/2*m*v2. First, let's find the final velocity.
a = (vf - v0)/t
2.6 = (vf - 0)/4
vf = 10.4 m/s
Then W = 1/2*(2100 kg)*(10.4 m/s)2
W = 113568 J = 113.57 kJ
