8/4 = y/y-x
8y - 8x = 4y
y = 2x
y = 2 x 4
y = 8
Hope this helps
The charge of the object must be 
Answer: Option C
<u>Explanation:</u>
Suppose an electric charge can be represented by the symbol Q. This electric charge generates an electric field; Because Q is the source of the electric field, we call this as source charge. The electric field strength of the source charge can be measured with any other charge anywhere in the area. The test charges used to test the field strength.
Its quantity indicated by the symbol q. In the electric field, q exerts an electric, either attractive or repulsive force. As usual, this force is indicated by the symbol F. The electric field’s magnitude is simply defined as the force per charge (q) on Q.

Here, given E = 4500 N/C and F = 0.05 N.
We need to find charge of the object (q)
By substituting the given values, we get

Answer:
a) 
b) 
c) 
Explanation:
From the exercise we know the initial velocity of the projectile and its initial height

To find what time does it take to reach maximum height we need to find how high will it go
b) We can calculate its initial height using the following formula
Knowing that its velocity is zero at its maximum height



So, the projectile goes 1024 ft high
a) From the equation of height we calculate how long does it take to reach maximum point



Solving the quadratic equation



So, the projectile reach maximum point at t=2s
c) We can calculate the final velocity by using the following formula:


Since the projectile is going down the velocity at the instant it reaches the ground is:

As we know that spring force is given as

here we know that
F = 4 N
x = 2 cm = 0.02 m
now from the above equation we will have


so the elastic constant of the spring will be 200 N/m