When person is observing destructive interference at 0.20 m distance from the equidistant position then we can say that path difference must be equal to half of the wavelength
now we will have
now we know that
y = 0.20 m
d = 2.4 m
L = 10 m
now here we have
now frequency of wave is given as
Answer:
Distance will be 49.34 m
Explanation:
We have given wavelength
Diameter of the antenna d = 2.7 m
Range L = 7.8 km = 7800 m
We have to find the smallest distance hat two speedboats can be from each other and still be resolved as two separate objects D
We know that distance is given by
So distance D will be 49.34 m
Answer:
t = 23.9nS
Explanation:
given :
Area A= 10 cm by 2 cm => 2 x 10^-2m x 10 x 10^-2m
distance d= 1mm=> 0.001
resistor R= 975 ohm
Capacitance can be calculated through the following formula,
C = (ε0 x A )/d
C = (8.85 x 10^-12 x (2 x 10^-2 x 10 x 10^-2))/0.001
C = 17.7 x 10^-12 (pico 'p' = 10^-12)
C = 17.7pF
the voltage between two plates is related to time, There we use the following formula of the final voltage
Vc = Vx (1-e^-(t/CR))
75 = 100 x (1-e^-(t/CR))
75/100 = (1-e^-(t/CR))
.75 = (1-e^-(t/CR))
.75 -1 = -e^-(t/CR)
-0.25 = -e^-(t/CR) --->(cancelling out the negative sign)
e^-(t/CR) = 0.25
in order to remove the exponent, take logs on both sides
-t/CR = ln (0.25)
t/CR = -ln(0.25)
t = -CR x ln (0.25)
t = -(17.7 x 10^-12 x 975) x (-1.38629)
t = 23.9 x
t = 23.9ns
Thus, it took 23.9ns for the potential difference between the deflection plates to reach 75 volts