We need to consider for this exercise the concept Drag Force and Torque. The equation of Drag force is
Where,
F_D = Drag Force
= Drag coefficient
A = Area
= Density
V = Velocity
Our values are given by,
(That is proper of a cone-shape)
Part A ) Replacing our values,
Part B ) To find the torque we apply the equation as follow,
Answer:
It will take 10 seconds to travel 200m at a speed of 10m/s
Explanation:
HOPE THAT THIS IS HELPFUL.
HAVE A GREAT DAY.
Answer:
T = 2.83701481512 seconds
Explanation:
Hi!
The formula that you will want to use to solve this question is:
T--> period
L --> length of the pendulum
g --> acceleration due to gravity (9.8m/s^2)
since we know that the mass of the bob at the end of the pendulum does not affect the period of the pendulum, we can go ahead and ignore that bit of information (unless, of course, the weight causes the pendulum to stretch)
so now we can plug in our given info into the formula above and solve!
T = 2*pi * sqrt(2/9.8)
T = 2.83701481512 seconds
*Note*
- I used 3.14 to pi, if you need to use a different value for pi (a longer version, etc) your answer will be slightly different
I hope this helped!
1)
The electrostatic force between two charges is given by
(1)
where
k is the Coulomb's constant
q1, q2 are the two charges
r is the separation between the charges
When the two spheres are brought in contact with each other, the charge equally redistribute among the two spheres, such that each sphere will have a charge of
where Q is the total charge between the two spheres.
So we can actually rewrite the force as
And since we know that
r = 1.41 m (distance between the spheres)
(the sign is positive since the charges repel each other)
We can solve the equation for Q:
So, the final charge on the sphere on the right is
2)
Now we know the total charge initially on the two spheres. Moreover, at the beginning we know that
(we put a negative sign since the force is attractive, which means that the charges have opposite signs)
r = 1.41 m is the separation between the charges
And also,
So we can rewrite eq.(1) as
Solving for q1,
Since , we can substituting all numbers into the equation:
which gives two solutions:
Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is