The normal force acting on the object is 500 N in the upward direction
<u>Explanation:</u>
As George is applying a downward force, the normal force will be in the upward direction. The normal force will be exerted due to the acceleration due to gravity exerted on the object.
So, as per Newton's second law, the normal force acting on the object can be measured by the product of mass of the object and the acceleration due to gravity acting on the object.
But as the acceleration due to gravity is a downward acting acceleration and the normal force is a upward acting force, so the acceleration will be having a negative sign in the formula.

Here, acceleration due to gravity g = -10 m/s² and mass is given as 50 kg, then
Normal force = 50 × (-10) = -500 N
So, the normal force acting on the object is 500 N in the upward direction.
If the boat's speed is s, and the river's speed is r, and the boat is traveling east (0 degrees),
(0,r) + (s cos297,s sin297) = (6,0)
now just solve for r and s.
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Answer:
The strength of the magnetic field that the line produces is
.
Explanation:
From Biot-Savart law, the equation to determine the strength of the magnetic field for any straight wire can be deduced:
(1)
Where
is the permiability constant, I is the current and r is the distance from the wire.
Notice that it is necessary to express the current, I, from kiloampere to ampere.
⇒ 
Finally, equation 1 can be used:
Hence, the strength of the magnetic field that the line produces is
.
Answer:

Explanation:
Given:
- Three identical charges q.
- Two charges on x - axis separated by distance a about origin
- One on y-axis
- All three charges are vertices
Find:
- Find an expression for the electric field at points on the y-axis above the uppermost charge.
- Show that the working reduces to point charge when y >> a.
Solution
- Take a variable distance y above the top most charge.
- Then compute the distance from charges on the axis to the variable distance y:

- Then compute the angle that Force makes with the y axis:
cos(Q) = sqrt(3)*a / 2*r
- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:
F_1,2 = 2*F_x*cos(Q)
- The total net force would be:
F_net = F_1,2 + kq / y^2
- Hence,

- Now for the limit y >>a:

- Insert limit i.e a/y = 0

Hence the Electric Field is off a point charge of magnitude 3q.