Which of the following is not a an example of dissipated energy?
b. kinetic
When energy is changed from one form to another, ____.
b. all of the energy can be accounted for
Answer:

Upward
Explanation:
We are given that
Mass of scarp paper,
1mg=
Distance,d =8 mm=
Magnitude of electric force =
Where 
Substitute the values


Gravitational force act in downward direction.
The electric force acts in opposite direction and magnitude of electric force is equal to gravitational force.
Hence, the direction of electric force is upward.
Answer:
The sea level will be 5m higher in 1667 y (years)
Explanation:
From the question, the rate at which the ocean's level is currently rising is about 3mm per year.
First, we will convert mm (millimeter) to m (meter)
1 mm = 0.001 m
Then,
3 mm = 3 × 0.001 m
= 0.003m
That is, the rate at which the ocean's level is currently rising is about 0.003m per year.
Now, to determine how long it will take for the ocean's level be 5 m higher than now at the given rate,
If the ocean rises 0.003 m in 1 year, then
the ocean will rise 5 m in x years
x = (5 m × 1 year) / 0.003 m
x = 5 / 0.003
x = 1666.67 years
x ≅ 1667 years
Hence, the sea level will be 5m higher in 1667 y (years)
False, some adults will know more than your friends as they are already grown up and may have experienced the problem already
<span>1.57 seconds.
The rod hanging from the nail constructs a physical pendulum. The period of such a pendulum follows the formula
T = 2*pi*sqrt(L/g)
where
T = time
L = length of pendulum
g = local gravitational acceleration
So the problem becomes one of determining L. It's tempting to consider L to be the distance between the center of mass and the pivot, but that isn't the right value. The correct value is the distance between the pivot and the center of percussion. So let's determine what that is. We can treat the uniform thin rod as an uniform beam and for an uniform beam the distance between the center of mass and the center of percussion is expressed as
b = L^2/(12A)
where
b = distance between center of mass and center of percussion
L = length of beam
A = distance between pivot and center of mass
Since the rod is uniform, the CoM will be midway from either end, or 0.962 m / 2 = 0.481 m from the end. The pivot will therefore be 0.481 m - 0.048 m = 0.433 m from the CoM
Now let's calculate the distance the CoP will be from the CoM:
b = L^2/(12A)
b = (0.962 m)^2/(12 * 0.433 m)
b = (0.925444 m^2)/(5.196 m)
b = 0.178107005 m
With the distance between the CoM and CoP known, we can now calculate the effective length of the pendulum. So:
0.433 m + 0.178107005 m = 0.611107005 m
And finally, with the effective length known, let's calculate the period.
T = 2*pi*sqrt(L/g)
T = 2*pi*sqrt((0.611107005 m)/(9.8 m/s^2))
T = 2*pi*sqrt(0.062357858 s^2)
T = 2*pi*0.249715554 s
T = 1.569009097 s
Rounding to 3 significant figures gives 1.57 seconds.
Let's check if this result is sane. Looking up "Seconds Pendulum", I get a length of 0.994 meters which is longer than the length of 0.611 meters calculated. But upon looking closer at the "Seconds Pendulum", you'll realize that it's period is actually 2 seconds, or 1 second per swing. So the length of the calculated pendulum is sane.</span>