Explanation:
when an object speeds up,it has positive acceleration.
hope it helps....
Using Placks’s constant and frequency
Answer:
+ 0.07 C
Explanation:
From the question given above, the following data were obtained:
Potential difference (V) = 12 V
Energy (E) = 0.418 J
Charge (Q) =?
The energy (E) , potential difference (V) and charge (Q) are related by the following equation:
E = ½QV
With the above formula, we can obtain the charge as follow:
Potential difference (V) = 12 V
Energy (E) = 0.418 J
Charge (Q) =?
E = ½QV
0.418 = ½ × Q × 12
0.418 = Q × 6
Divide both side by 6
Q = 0.418 / 6
Q = + 0.07 C
To solve this exercise it is necessary to apply the concepts related to Work and Kinetic Energy. Work from the rotational movement is described as

In the case of rotational kinetic energy we know that

PART A)
is given in revolutions and needs to be in radians therefore


Replacing in the work equation we have to



PART B) From the torque and moment of inertia it is possible to calculate the angular acceleration and the final speed, with which the kinetic energy can be determined.

Rearrange for the angular acceleration,



From the kinematic equations of angular motion we have,




In this way the rotational kinetic energy would be given by


