Answer:
Explanation:
i )
When it is disconnected with the battery , the charge stored in it becomes fixed . When the plate distance becomes half , its capacitance becomes twice from C to 2C . Let charge stored in it at the time of disconnection from battery be Q . Let plate separation reduces from d to d / 2
So charged stored in it will remain unchanged .
ii )
Potential difference = charge / capacitance
in the first case potential difference = Q / C
in the second case potential difference = Q / 2C
So potential difference becomes half .
iii ) electric field = potential diff / plate separation
in the first case electric field = Q / (d x C )
in the second case electric field = 2 Q / (d x 2C)
= Q / (d x C )
So electric field remains unchanged .
iv)
energy stored in first case = Q² / 2C
In the second case energy stored = Q² / 2x2C
so energy stored becomes half .
Answer: dakdadakdadakdadakda
Explanation:(sings) blah blah blah middle fingers in the air l-l-l-loser
Answer:
Mass released = 8.6 g
Given data:
Initial number of moles nitrogen= 0.950 mol
Initial volume = 25.5 L
Final mass of nitrogen released = ?
Final volume = 17.3 L
Formula:
V₁/n₁ = V₂/n₂
25.5 L / 0.950 mol = 17.3 L/n₂
n₂ = 17.3 L× 0.950 mol/25.5 L
n₂ = 16.435 L.mol /25.5 L
n₂ = 0.644 mol
Initial mass of nitrogen:
Mass = number of moles × molar mass
Mass = 0.950 mol × 28 g/mol
Mass = 26.6 g
Final mass of nitrogen:
Mass = number of moles × molar mass
Mass = 0.644 mol × 28 g/mol
Mass = 18.0 g
Mass released = initial mass - final mass
Mass released = 26.6 g - 18.0 g
Mass released = 8.6 g
Read more on Brainly.com - brainly.com/question/15623698#readmore
Answer:
Resulting Change in the Magnetic Flux 
Explanation:
Thanks!
Answer:
a)- 1.799 rad/sec²
b)- 17.6 x 10ˉ³Nm
Explanation:
ω₀ = 720 rev/min x (1 min/60 sec) x (2π rad / 1 rev) = 24π rad/s
a) Assuming a constant angular acceleration, the formula will be
α = (ωf -ω₀) / t
As final state of the grindstone is at rest, so ωf =0
⇒ α = (0-24π) / 41.9 = - 1.799 rad/sec²
b)Moment of inertia I for a disk about its central axis
I = ½mr²
where m=2kg and radius 'r'= 0.099m
I = ½(2)(0.099²)
I = 9.8 x 10ˉ³ kgm²
Next is to determine the frictional torque exerted on the grindstone, that caused it to stop, applying the rotational equivalent of the Newton's 2nd law:
τ = I α =>(9.8 x 10ˉ³)(- 1.799)
τ = - 17.6 x 10ˉ³Nm
(The negative sign indicates that the frictional torque opposes to the rotation of the grindstone).