Answer:
The electric field inside the wire will remain the same or constant, while the drift velocity will by a factor of four.
Explanation:
Electron mobility, μ = 
where
= Drift velocity
E = Electric field
Given that the electric field strength = 1.48 V/m,
Therefore since the electric potential depends on the length of the wire and the attached potential difference, then when the electron mobility is increased 4 times the Electric field E will be the same but the drift velocity will increase four times. That is
4·μ = 
Answer:
D The temperature, pressure and volume will all increase
Answer:
Option C
Explanation:
Given:
- Depth of tissue d = 12 cm
- frequency of ultrasound f = 1 MHz
- Input Intensity I_i = 1000 W/m^2
- attenuation coefficient soft tissue a = 0.54
Find:
- Out-put intensity at the required depth
Solution
- The amount of attenuation in (dB) with the progression of depth is given by:
Attenuation = a*f*d
Attenuation = 0.54*12*1
Attenuation = 6.48 dB
- The relation with attenuation and ratio of input and output intensity is given by:
Attenuation = 10*log_10 (I_i / I_o)
6.48 dB = 10*log_10 (I_i / I_o)
I_i / I_o = 10^(0.648)
I_o = 1000 / 10^(0.648)
I_o = 225 W/m^2
- Hence the answer is option C: I_o = 250 W/m^2
