Answer:
Explanation:
Given
Initial reading on scale =40 N
So, we can conclude that weight of the sack is 40 N
After this a 10 N force is applied upward on the sack such that the net force becomes (40-10) N downward (because downward force is more)
This net downward force is the resultant of earth graviational pull and the applied upward force.
So, this downward force acts on the machine which inturn applies an upaward force of same magnitude called Normal reaction.
This situation can be diagramatically represented by figure given below
Answer:

Explanation:
A 6.0-cm-diameter parallel-plate capacitor has a 0.46 mm gap.
What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000V/s?
Let given is,
The diameter of a parallel plate capacitor is 6 cm or 0.06 m
Separation between plates, d = 0.046 mm
The potential difference across the capacitor is increasing at 500,000 V/s
We need to find the displacement current in the capacitor. Capacitance for parallel plate capacitor is given by :
, r is radius
Let I is the displacement current. It is given by :

Here,
is rate of increasing potential difference
So

So, the value of displacement current is
.
The springs stored energy is transferred to the cube as kinetic energy and then by the slop the KE is converted to height energy.
<span>0.5 . k . x^2 = 0.5 . m . v^2 = m . g . ∆h </span>
<span>0.5 . 50 . (0.1^2) = 0.05 . 9.8 . ∆h </span>
<span>∆h = 0.51 m = 51 cm </span>
<span>This is the height gained </span>
<span>Distance along the slope = ∆h / sin 60 = 0.589 = 59 cm </span>
<span>In the second case, the stored spring energy is converted into height energy AND frictional heat energy. </span>
<span>The height energy is m . g . d sin 60 where d is the distance the cube moves along the slope. </span>
<span>The Frictional energy converted is F . d </span>
<span>F ( the frictional force ) = µ . N </span>
<span>N ( the reaction to the component of the gravity force perpendicular to the surface of the slope ) = m . g . cos60 </span>
<span>Total energy converted </span>
<span>0.5 . k . x^2 = (m . g . dsin60) + (µ . m . g . cos60 . d ) </span>
<span>Solve for d </span>
<span>d = 0.528 = 53 cm</span>
The focal point of a concave mirror is halfway along the radius, therefore the radius would be 2•16= 32 cm