Answer:
Explanation:
Area of electrodes, A = 2 cm x 2 cm = 4 cm²
Separation between electrodes, d = 1 mm
Voltage, V = 9 V
(a)
Let C is the capacitance between the electrodes
C = 3.54 x 10^-12 F
Let q be the charge on each of the electrode
q = C x V
q = 3.54 x 10^-12 x 9 = 3.2 x 10^-11 C
(b)
As, the battery is disconnected the charge on the electrodes remains same.
(c)
As the battery is connected the voltage is same.
capacitance is change.
As the distance is doubled, the capacitance becomes half and the charge is also halved. q' = q/2 = 1.6 x 10^-11 C
Answer:
<em>380 kHz</em>
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Explanation:
The speed of sound is taken as 1500 m/s
The length of the fetus is 1.6 cm long
The condition is that the wavelength used must be at most 1/4 of the size of the object that is to be imaged.
For this 1.6 cm baby, the wavelength must not exceed
λ = 
of 1.6 cm = x 1.6 cm = 0.4 cm =
0.4 cm = 0.004 m this is the wavelength of the required ultrasonic sound.
we know that
v = λf
where v is the speed of a wave
λ is the wavelength of the wave
f is the frequency of the wave
f = v/λ
substituting values, we have
f = 1500/0.004 = 375000 Hz
==> 375000/1000 = 375 kHz ≅ <em>380 kHz</em>
Answer:
Check the explanation
Explanation:
a) in solving the first question, we will be choosing a spherical ball of radius 1 m, Therefore the width of the object is the diameter = 2m.
b)Given that and according to the question, the radius of the electron's orbit = 1x105 x radius of the object = 1x105 x1 = 1x105 m
Answer:
The relative velocity of an object A with respect to another object B.
Explanation:
The relative velocity of an object A with respect to another object B is the velocity that object A would appear to have to an observer situated on object B moving along with it.
