The answer would be that they are close to water hope this helps!
Answer:
The magnitude of Electric Field is 
Explanation:
Given:
- Radius of the solid sphere=R
- Total charge of the sphere=Q
Let consider a Gaussian surface at a distance of r such that 0<r>R in the shape of sphere such that the electric Field due to this E and it is radially outwards.
The charge inside this Gaussian surface volume we have , 
Now using Gauss Law we have
Answer:
The angular velocity I would have to rotate it in order to generate an emf of amplitude 1.0 V is 254.65 rad/s
Explanation:
given information:
B = 0.5 mT = 0.0005 T
N = 1000
r = 5 cm = 0.05 m
emf, ε = 1 V
according to Faraday's law
ε = -N dΦ/dt, Φ = B A
= - N d( B A)/dt
= - N d( B A cos ωt)/dt
= - N B A d(cos ωt)/dt
= N B A ω sin ωt
A = πr², so
ε = N B πr² ω sin ωt
where
ε = emf
N = number of coil turn
B = magnetic field
r = radius
ω = angular velocity
Φ = magnetic flux
emf maximum, sin ωt = 1. So,
ε = N B πr² ω
ω = ε/N B πr²
= 1/[(1000) (0.0005) π (0.05)²
= 254.65 rad/s
Answer:
Exposure time limitation, shielding and distance.
Explanation:
- Limitation of exposure time, since the dose received is directly proportional to the exposure time, so that, at a shorter time, lower dose. For this reason, planning is suggested, to reduce time.
-
Use of shields. This allows a reduction in the dose received by the technician when filtered by the barrier (screen). There are two types of shields or screens, the primary barriers (attenuate the radiation of the primary beam) and the secondary barriers (avoid diffuse radiation).
-
Distance to the radioactive source. The dose received is inversely proportional to the square of the distance to the radioactive source. Therefore, if the distance is doubled, the dose received will decrease by a quarter. Reason for this, it is advisable to use devices or remote controls whenever possible.
