The electric field is always perpendicular to the surface outside of a conductor. TRUE
<span> If an electron were placed on an electric field line, it would move in a direction perpendicular to the field. FALSE, it would move in an anti-parallel direction because its charge is negative </span>
<span>Electric field lines originate on positive charge and terminate on negative charge. TRUE ; but they can also go to infinity </span>
It is possible for two electric field lines to cross each other.
<span> Usually FALSE; though technically possible at special points where field is zero. </span>
If an electron and a positron were in the presence of a very strong electric field, they would move away from each other.
<span> TRUE; one is positive, and one is negative. If the field is strong enough, the action of the field will overcome the mutual attraction between them </span>
It is not possible for the electric field to ever be zero. FALSE: it IS possible, inside a conductor for instance
If a proton were placed on an electric field line, it would move in a direction anti-parallel to the field.
<span> FALSE: being positive, it would move in the SAME direction as the field</span>ic
Answer:
Explanation:
Heat capacity A = 3 x heat capacity of B
initial temperature of A = 2 x initial temperature of B
TA = 2 TB
Let T be the final temperature of the system
Heat lost by A is equal to the heat gained by B
mass of A x specific heat of A x (TA - T) = mass of B x specific heat of B x ( T - TB)
heat capacity of A x ( TA - T) = heat capacity of B x ( T - TB)
3 x heat capacity of B x ( TA - T) = heat capacity of B x ( T - TB)
3 TA - 3 T = T - TB
6 TB + TB = 4 T
T = 1.75 TB
Answer:
Given that
D= 4 mm
K = 160 W/m-K
h=h = 220 W/m²-K
ηf = 0.65
We know that
For circular fin
m = 37.08
By solving above equation we get
L= 36.18 mm
The effectiveness for circular fin given as
ε = 23.52
Answer:
F₁ = 4,120.2 N
F₂ = 3,924N
Explanation:
1) Balance of angular momentum around the end where F₁ is applied.
F₂ × 0.5m - F₁ × 0 = mass × g × 1m
⇒ F2 × 0.5 m= 20 kg × 9.81 m/s² × 1 m = 1,962 N×m
F₂ = 196.2 Nm / 0.5m = 3,924 N
2) Balance of forces
F₁ - F₂ = mg
F₁ = F₂ + mg = 3,924N + 20kg (9.81 m/s²) = 4,120.2 N
