Answer:
The drill's angular displacement during that time interval is 24.17 rad.
Explanation:
Given;
initial angular velocity of the electric drill, = 5.21 rad/s
angular acceleration of the electric drill, α = 0.311 rad/s²
time of motion of the electric drill, t = 4.13 s
The angular displacement of the electric drill at the given time interval is calculated as;
Therefore, the drill's angular displacement during that time interval is 24.17 rad.
Answer:
Potential energy can transfer into other forms of energy like kinetic energy. Kinetic energy is energy an object has because of its motion. ... If released, as the ball moves faster and faster toward the ground, the force of gravity will transfer the potential energy to kinetic energy.
On a microscopic scale, conduction occurs as rapidly moving or vibrating atoms and molecules interact with neighboring particles, transferring some of their kinetic energy. Heat is transferred by conduction when adjacent atoms vibrate against one another, or as electrons move from one atom to another.
Answer:
the average induced emf in the second winding is
Explanation:
The magnetic field inside the first solenoid is given by,
Where is the permeability of the free space
N is the number of turns of solenoid per unit length
I is the current in the solenoid
A is the cross-sectional area of the wire
Replacing we have,
Thus average emf induced in the second windigs is,
Therefore the average induced emf in the second winding is
Explanation:
The charge (positive or negative) is always on the surface in the conductors. So, if the plates have substantial thickness, the charge is distributed throughout the surface. We know from the question that the bottom plate is charged positively, so an equal amount of negative charge must be distributed on the lower surface of the upper plate.
The direction of the E-field is always directed from positive to the negative charge. Since the plates have flat surfaces, the electric field lines are straight, except the edges. In most college level questions this effect is negligible, that is why in most questions the plates are considered very large.
Answer:
Explanation:
Capacitor of 0.75μF, charged to 70V and connect in series with 55Ω and 140 Ω to discharge.
Energy dissipates in 55Ω resistor is given by V²/R
Since the 55ohms and 140ohms l discharge the capacitor fully, the voltage will be zero volts and this voltage will be shared by the resistor in ratio.
So for 55ohms, using voltage divider rule
V=R1/(R1+R2) ×Vt
V=55/(55+140) ×70
V=19.74Volts is across the 55ohms resistor.
Then, energy loss will be
E=V²/R
E=19.74²/55
E=7.09J
7.09J of heat is dissipated by the 55ohms resistor