The correct answer to the question is: A) miles/hour and B) metre/ second.
EXPLANATION:
Before answering this question, first we have to understand speed.
The speed of a body is defined as the rate of distance travelled or the distance travelled by a body per unit time.
Hence, it is a derived quantity which is obtained from distance and time.
The unit of distance can be metre, miles, and the unit of time can be second, minutes or hour.
As speed is the distance covered per unit time, the perfect units will be miles/hour and metre/second.
Hence, the correct options are first and second.
The half life is 30 minutes.
30 mins- 1/2 left
60 mins- 1/4 left
90 mins- 1/8 left
120 mins- 1/16 left
120 mins= 2 hours
Answer:
a) The current density ,J = 2.05×10^-5
b) The drift velocity Vd= 1.51×10^-15
Explanation:
The equation for the current density and drift velocity is given by:
J = i/A = (ne)×Vd
Where i= current
A = Are
Vd = drift velocity
e = charge ,q= 1.602 ×10^-19C
n = volume
Given: i = 5.8×10^-10A
Raduis,r = 3mm= 3.0×10^-3m
n = 8.49×10^28m^3
a) Current density, J =( 5.8×10^-10)/[3.142(3.0×10^-3)^2]
J = (5.8×10^-10) /(2.83×10^-5)
J = 2.05 ×10^-5
b) Drift velocity, Vd = J/ (ne)
Vd = (2.05×10^-5)/ (8.49×10^28)(1.602×10^-19)
Vd = (2.05×10^-5)/(1.36 ×10^10)
Vd = 1.51× 10^-5
Answer:
b. Constant magnitude, but varying direction, perpendicular to the equipotential.
Explanation:
As we know that the relation between electric field and electric potential is given as
here if we say that potential is constant because electric field sensor is moving along equi-potential line.
Then we will say
V = constant
so we have
so electric field will remain constant always in magnitude and always remains perpendicular to the surface
so we have
b. Constant magnitude, but varying direction, perpendicular to the equipotential.
