Answer: High pass filters, and in particular LC high pass filters are used in many RF applications where they block the lower frequencies and allow through higher frequency signals. Typically LC filters are used for the higher radio frequencies where active filters are not so manageable, and inductors more appropriate.
The electric field between plates is 4000V/m.
An electric field (sometimes E-field) is the physical field that surrounds electrically charged particles and exerts force on all other charged particles in the field, either attracting or repelling them. It also refers to the physical field for a system of charged particles.
The value of the electric field has dimensions of force per unit charge. In the metre-kilogram-second and SI systems, the appropriate units are newtons per coulomb, equivalent to volts per metre.
The voltage between points A and B is
V=E.d
E =V/d (uniform E- field only)
where d is the distance from A to B, or the distance between the plates.
Given:
distance d = 3 cm
voltage V = 120 V
Electric field E = V/d
E = 120 V / 3cm
E = 40 V / 1 cm [ 1 cm = 1/100 m ]
E = 4000 V/m.
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<span>First law of thermodynamics. This conservation law states that energy cannot be created or destroyed but can be changed from one form to another. In essence, energy is always conserved but can be converted from one form into another. Like when an engine burns fuel, it converts the energy stored in the fuel's chemical bonds into useful mechanical energy and then into heat, or more specifically, the melting ice cubes. Yeast breaks down maltose into glucose to produce alcohol and Co2 in the fermentation process. This is a prime example of the 1st law of thermodynamics. No form of usable energy is really lost; it only changes from one form to another</span>
<span>The lighter object is in orbit around the heaviest, and the Sun is the heaviest object in the Solar System</span>
Answer:
a) V_f = 25.514 m/s
b) Q =53.46 degrees CCW from + x-axis
Explanation:
Given:
- Initial speed V_i = 20.5 j m/s
- Acceleration a = 0.31 i m/s^2
- Time duration for acceleration t = 49.0 s
Find:
(a) What is the magnitude of the satellite's velocity when the thruster turns off?
(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis.
Solution:
- We can apply the kinematic equation of motion for our problem assuming a constant acceleration as given:
V_f = V_i + a*t
V_f = 20.5 j + 0.31 i *49
V_f = 20.5 j + 15.19 i
- The magnitude of the velocity vector is given by:
V_f = sqrt ( 20.5^2 + 15.19^2)
V_f = sqrt(650.9861)
V_f = 25.514 m/s
- The direction of the velocity vector can be computed by using x and y components of velocity found above:
tan(Q) = (V_y / V_x)
Q = arctan (20.5 / 15.19)
Q =53.46 degrees
- The velocity vector is at angle @ 53.46 degrees CCW from the positive x-axis.
