Hi there!
Recall Newton's Second Law:

∑F = net force (N)
m = mass (kg)
a = acceleration (m/s²)
We must begin by solving for the acceleration using the following:
a = Δv/t
In this instance:
Δv = 3 m/s
t = 2.5 sec
a = 3/2.5 = 1.2 m/s²
Now, plug this value along with the mass into the equation for net force:

Answer:
See Explanation Below
Explanation:
This question is incomplete.
I'll answer this question on general terms. You'll get your result if you apply the steps I'll highlight below.
To start with; what's Ampere law;
It states that for any closed loop path, the sum of the length elements times the magnetic field in the direction of the length element is equal to the permeability times the electric current enclosed in the loop.
The simple translation of this law is that, the sum of current in the close loop gives us the desired result.
Rephrasing your question;
Three currents, (I1 = +3A, I2 = +4A and I3 = -5A) are passing through a surface bounded by a closed path. The currents have different values and directions. According to Ampere’s law, what is the value of I on the right side of this equation?
First, we take note of the signs in front of the given currents.
The negative sign in front of I3 means that; it is moving in opposite direction of I1 and I2.
To calculate the value of I.
The value of I is the sum of the three currents:
i.e. 3A + 4A - 5A
I = 2 A
Given that
Work = 600,000 J ,
distance(S) = 500 m ,
mass (m) = 250 Kg ,
Determine the velocity of car (v) = ?
We know that,
Work = Force × distance
=> Force = Work ÷ distance
= 600,000 ÷ 500
= 500 N .
Also Force F = m.a ; from Newtons II law
500 = 250 × a
a = 2 m/s.
<em>Final Velocity from the given formula </em>
V² = u² + 2.a.s
= 0 + 2 × 2 × 500
= \sqrt{2000}
<em> v = 44.7 m/s</em>
In order to develop this problem it is necessary to use the concepts related to the conservation of both potential cinematic as gravitational energy,


Where,
M = Mass of Earth
m = Mass of Object
v = Velocity
r = Radius
G = Gravitational universal constant
Our values are given as,



Replacing we have,




Therefore the speed of the object when striking the surface of earth is 4456 m/s