Question: A. The state highway patrol radar guns use a frequency of 9.15 GHz. If you're approaching a speed trap driving 30.1 m/s, what frequency shift will your FuzzFoiler 2000 radar detector see?
B. The radar gun measures the frequency of the radar pulse echoing off your car. By what percentage is the measured frequency different from the original frequency? (Enter a positive number for a frequency increase, negative for a decrease. Just enter a number, without a percent sign.)?
Answer:
The frequency change percentage is 9.94%
Explanation:
The frequency shift can be calculated as follows.
= 
=9.95 GHz
So the frequency change seen by the detector is 9.95 - 9.05
% difference
= 9.94%
Answer:
Explanation:
A point charge of +2q centered in a conductive spherical shell of inner diameter a and outer diameter b will induce - 2q charge on the inner surface and +2q charge on the outer surface of the shell. Since 8q charge has been added to the shell , this charge will reside on the outer surface of the shell. so total charge on the outer surface will be 10q. At a point less than a , the electric field will be due to +2q charge situated at the centre . The electric field will be as follows
E = k .2q / r² for r < a
= 8kq/ a²
electric field at a point r = a>b
total charge lying inside is +2q - 2q = 0 . So in the thickness of the shell , electric field will be zero as total charge inside is nil.
For a point at r > b total charge inside is 2q-2q+10q = 10q , so electric field at r which is lying outside the shell .
E = k 10 q / r² for r > b
Answer:
No.
Explanation:
Given that Kevin decides to soup up his car by replacing the car's wheels with ones that have 1.4 times the diameter of the original wheels. Note that the speedometer in a car is calibrated based on the tire's diameter and on the distance the tire covers in each revolution. (a) Will the reading of the speedometer change ?
Considering the formula
V = wr
Where
V = linear speed
W = angular speed
r = radius of the wheel.
But W = 2πrf
Where the the 2 and pi are constant. The radius of the first wheel will be small but counter balance with the larger frequency.
While the radius of the second wheel may be large but it will be of a small frequency.
We can therefore conclude that the reading on the speedometer will not change. Because speedometer will read the linear speed V.
Answer: magnitude of applied force is FA = mg + F
Where F is the resultant force downward that the rope moves with
Explanation:
Force downwards F is,
F = FA - T
T is the upwards tension force on the rope
FA is the actual applied force in pulling the rope down.
Therefore, T = FA - F .....equ. (1)
For the box to move up with force ma ( it's mass times its acceleration upwards) upwards tension on the roap must exceed its own weight mg ( it's mass times acceleration due to gravity 9.8m/s^2)
Therefore, ma = T - mg
T = ma + mg ..... equ. (2)
Equating equ. 1 and 2
T = FA - F = ma + mg
Therefore FA = ma + mg + F
But at constant velocity a = 0
Magnitude of applied force becomes
FA = mg + F
See image below
