To determine the Force it is necessary that Samantha starts from the consideration of the dynamic balance between the forces acting on the car. For this, the centripetal force must be equal to the friction force (otherwise the car would lose contact with the ground), the expression of these two forces would be


Here,
= Coefficient of kinematic friction
m = mass
g = Acceleration due to gravity
v = Velocity
r = Radius
From this relationship then she should assume the following
1) The car travels at a constant speed
2) There is indeed that frictional force between the car and the road
3) There is no loss of mass during displacement
4) The centripetal force must be equal to the frictional force so that the car does not lose contact or slip
5) The radius must be constant
Answer : The correct option is, (d) 4 times
Solution :
According to the Coulomb's law, the electrostatic force of attraction or repulsion between two charges is directly proportional to the product of the charges and is inversely proportional to the square of the distance between the the charges.
Formula used :

where,
F = electrostatic force of attraction or repulsion
= Coulomb's constant
and
are the charges
r = distance between two charges
First we have to calculate the force exerted between S and q when the distance between the charge is 1 unit and let us assumed that the charge be 'q'
..........(1)
Now we have to calculate the force exerted between S and p when the distance between the charge is 2 unit at the same charge.
...........(2)
Equation equation 1 and 2, we get


Therefore, the force exerted between S and q is 4 times the force exerted between S and p.
Answer:
the speed of the cruiser relative to the pursuit ship is 0.3846c
Explanation:
the solution is in the attached Word file
The spring has been extended for 3.5 m
<u>Explanation:</u>
We have the formula,
PE =1/2 K X²
Rewrite the equation as
PE=1/2 K d²
multiply both the sides by 2/K to simplify the equation
2/k . PE= 1/2 K d² . 2/K
√d²=√2PE/K
Cancelling the root value and now we have,
d=√2PE/k
d=√2×98 J / 16N/m
d=√12.25
d=3.5 m
The spring has been extended for 3.5 m
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Answer:
200 million years
Explanation:
The equation that describes the decay of a radioactive isotope is

where
is the amount of radioactive isotope left at time t
is the initial amount of isotope
is the half-life of the sample
In this problem, the ratio between unstable isotope and daughter isotope is 1:15; this means that

Because the "total proportion" of original sample was 1+15=16.
Also we know that the half-life is

So we can re-arrange the equation to find t, the age of the rock:

So, 200 million years.