An X-Ray machine delivers a radiation dose of 5mRem/hr. at 3ft from the machine. How far will the X-Ray technician have to move
to reduce his exposure to 2mRem/hr.? I1/I2 = (D2)2/(D1)2 -------> I1(D1)2 = I2(D2)2
1 answer:
You might be interested in
Answer:
(a) 37.5 kg
(b) 4
Explanation:
Force, F = 150 N
kinetic friction coefficient = 0.15
(a) acceleration, a = 2.53 m/s^2
According to the newton's second law
Net force = mass x acceleration
F - friction force = m a
150 - 0.15 x m g = m a
150 = m (2.53 + 0.15 x 9.8)
m = 37.5 kg
(b) As the block moves with the constant speed so the applied force becomes the friction force.
Answer:
a) the capacitance is the same for both capacitors.
b) The potential difference between the plates for the capacitor with charge +2q, is double of the one for the capacitor with charge +q.
c) The electric field magnitude between the plates for the capacitor with charge +2q, is double of the one for the capacitor with charge +q
d) The energy stored between the plates for the capacitor with charge +2q, is 4 times the value for the one with charge +q
Explanation:
a) The capacitance of a capacitor, by definition, is as follows:
Appying Gauss' Law to one of plates, it can be showed, that the capacitance (for a parallel plates capacitor) can be expressed as follows:
C = ε*A / d
As it can be seen, it does not depend on the charge. so we conclude that the capacitance must be the same for both capacitors, due to they are identical except for the value of the charge on the plates.
b) By definition, as we said above, the capacitance is equal to the proportion between the charge of one of the plates, and the potential difference between them.
If this proportion must remain the same, and one of the capacitors has the double of the charge than the other, the potential difference must be the double also.
c) Applying Gauss' law, to the surface of one of the plates, and assuming a constant surface charge density σ, it can be showed that the electric field can be calculated as follows:
E*A = Q/ε₀ as σ=Q/A
⇒ E = σ/ε₀
As σ is directly proportional to the charge (being the area A the same), we conclude that the electric field for the capacitor with charge +2q must be the double than the one for the capacitior with charge +q.
c) The electric potential energy, stored between plates of a capacitor, can be written as follows:
If the capacitance remains the same, we can conclude that the electric potential energy for the capacitor with charge +2q, as the charge is raised to the 2nd power, must be 4 times the one for the capacitor with charge +q.