You're driving your new sports car at 85 mph over the top of a hill that has a radius of curvature of 525 m.
what fraction of your normal weight is your apparent weight as you crest the hill?
express your answer in percent.

The apparent weight is  71% of the normal weight while going through the crest of hill.

Given data:

The speed of car at top of hill is, v = 85 mph = 37.99 m/s.

The radius of curvature of path is, r = 525 m.

The expression for the apparent weight of car moving in a curved path is,

Here, is the centripetal force and is the normal weight.

Solving as,

Thus, we can conclude that the apparent weight is  71% of the normal weight while going through the crest of hill.

Learn more about the apparent weight here:

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Explanation:

It is given that,

Speed of the sports car, v = 85 mph = 37.99 m/s

The radius of curvature, r = 525 m

Let is the normal weight and is the apparent weight of the person. Its apparent weight is given by :

So,

or

Hence, this is the required solution.

Electric field will be greater when the battery is connected by the factor of 2.

Explanation:

Solution:

I will be doing some algebraic calculations to answer this question:

As we know that,

Q = CV

and

C =

So, when separation = d/2, then,

= by rearranging we get

So,

 = 2C

We further know that, Voltage will remain same if the battery is  connected.

This further implies that,

Q = CV

So,

= V

 = 2CV

= 2Q

and we also know that,

Electric field E =

So, the new E or =

Hence,

= = 2E

= 2E

when battery is disconnected, Q remain the same.

So,

When disconnected

E = E

E =  = Same

Hence, we can see that the magnitude of the electric does not depend upon the distance of separation. Instead it does depend upon the magnitude of charge.

So, when battery is disconnected, Q is same, so the Electric field.

But when it is connected, = 2Q and the = 2E

So,

= = 2

Electric field will be greater when the battery is connected by the factor of 2.