If y gets smaller as x gets bigger and y have an INVERSE relationship?
Answer:
Answer is explained in the explanation section below.
Explanation:
Solution:
We know from the Coulomb's Law that, Coulomb's force is directly proportional to the product of two charges q1 and q2 and inversely proportional to the square of the radius between them.
So,
F = 
Now, we are asked to get the greatest force. So, in order to do that, product of the charges must be greatest because the force and product of charges are directly proportional.
Let's suppose, q1 = q
So,
if q1 = q
then
q2 = Q-q
Product of Charges = q1 x q2
Now, it is:
Product of Charges = q x (Q-q)
So,
Product of Charges = qQ - 
And the expression qQ -
is clearly a quadratic expression. And clearly its roots are 0 and Q.
So, the highest value of the quadratic equation will be surely at mid-point between the two roots 0 and Q.
So, the midpoint is:
q =
q = Q/2 and it is the highest value of each charge in order to get the greatest force.
It would have to be the right type of metal like copper to run through it
Answer:
4
Explanation:
as toddler will go away then come closer
Answer:
Airplane speed relative to the ground is 260 km/h and θ = 22.6º direction from north to east
Explanation:
This is a problem of vector composition, a very practical method is to decompose the vectors with respect to an xy reference system, perform the sum of each component and then with the Pythagorean theorem and trigonometry find the result.
Let's take the north direction with the Y axis and the east direction as the X axis
Vy = 240 km / h airplane
Vx = 100 Km / h wind
a) See the annex
Analytical calculation of the magnitude of the speed and direction of the aircraft
V² = Vx² + Vy²
V = √ (240² + 100²)
V = 260 km/h
Airplane speed relative to the ground is 260 km/h
Tan θ = Vy / Vx
tan θ = 100/240
θ = 22.6º
Direction from north to eastb
b) What direction should the pilot have so that the resulting northbound
Vo = 240 km/h airplane
Vox = Vo cos θ
Voy = Vo sin θ
Vx = 100 km / h wind
To travel north the speeds the x axis (East) must add zero
Vx -Vox = 0
Vx = Vox = Vo cos θ
100 = 240 cos θ
θ = cos⁻¹ (100/240)
θ = 65.7º
North to West Direction
The speed in that case would be
V² = Vx² + Vy²
To go north we must find Vy
Vy² = V² - Vx²
Vy = √( 240² - 100²)
Vy = 218.2 km / h