Answer:
On the magnitude of the charges, on their separation and on the sign of the charges
Explanation:
The magnitude of the electric force between two charges is given by

where
k is the Coulomb's constant
q1, q2 are the magnitudes of the two charges
r is the separation between the charges
From the formula, we see that the magnitude of the force depends on the following factors:
- magnitude of the two charges
- separation between the charges
Moreover, the direction of the force depends on the sign of the two charges. In fact:
- if the two charges have same sign, the force is repulsive
- if the two charges have opposite signs, the force is attractive
Answer:
(5g/cm³)*(10cm³) = 50g
Explanation:
This is just a conversion formula. Easy to find using dimensional analysis.
(5g/cm³)*(10cm³) = 50g
Answer:
kinetic friction may be greater than 400 N or smaller than 400 N
Explanation:
As we know that maximum value of static friction on the rough surface is known as limiting friction and the formula of this limiting friction is known as

now when object is sliding on the rough surface then the friction force on that surface is known as kinetic friction and the formula of kinetic friction is known as

now we know that

so here value of limiting static friction force is always more than kinetic friction
also we know that
initially when body is at rest then static friction value will lie from 0 N to maximum limiting friction
and hence kinetic friction may be greater than static friction or if the static friction is maximum limiting friction then kinetic friction is smaller than static friction
so kinetic friction may be greater than 400 N or smaller than 400 N
The answer is wheel and axle
Answer:
a

b

Explanation:
From the question we are told that
The diameter of the Ferris wheel is 
The period of the Ferris wheel is 
The mass of the passenger is 
The apparent weight of the passenger at the lowest point is mathematically represented as

Where
is the centripetal force on the passenger, which is mathematically represented as

Where
is the angular velocity which is mathematically represented as

substituting values


and r is the radius which is evaluated as 
substituting values


So


W is the weight which is mathematically represented as


So


The apparent weight of the passenger at the highest point is mathematically represented as

substituting values

