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ludmilkaskok [199]
3 years ago
15

An ideal air-filled parallel-plate capacitor has round plates and carries a fixed amount of equal but opposite charge on its pla

tes. All the geometric parameters of the capacitor (plate diameter and plate separation) are now DOUBLED. If the original energy density between the plates was u0 , what is the new energy density?
Physics
1 answer:
Vikentia [17]3 years ago
8 0

Answer:

The new energy density is same as initial energy density u0 as it is independent of plate dimensions

Explanation:

As we know that energy density is total energy per unit volume

so we know that isolated plates of capacitor is placed here

C = \frac{\epsilon_0 A}{d}

here we have energy density given as

u = \frac{Q^2}{2(\epsilon_0 A/d)} Ad

so we have

u = \frac{Q^2d^2}{2\epsilon_0}

so it is independent of the dimensions of the plate while total charge on the plates is constant always

so energy density will not change and remains the same

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3 years ago
A low orbit satellite at 100 km altitude had a camera that resolved a car location to within 0.3 meter. Find the minimum diamete
NeX [460]

Answer:

Minimum diameter of the camera lens is 22.4 cm

The focal length of the camera's lens is 300cm

Explanation:

y = Resolve distance = 0.3 m

h = Height of satellite = 100 km

λ = Wavelength = 550 nm

Angular resolution

tan\theta\approx \theta =\frac{y}{h}\\\Rightarrow \theta=\frac{0.3}{100\times 10^3}=3\times 10^{-6}

From Rayleigh criteria

sin\theta=1.22\frac{\lambda}{D}\\\Rightarrow D=1.22\frac{\lambda}{sin\theta}\\\Rightarrow D=1.22\frac{550\times 10^{-9}}{sin3\times 10^{-6}}=0.2236\ m=22.4\ cm

Minimum diameter of the camera lens is 22.4 cm

Relation between resolvable feature, focal length and angular resolution

d=f\Delta \theta\\\Rightarrow f=\frac{d}{\Delta \theta}\\\Rightarrow f=\frac{9\times 10^{-6}}{3\times 10^{-6}}=3\ m=300\ cm

The focal length of the camera's lens is 300cm

3 0
3 years ago
A boat of mass 225 kg drifts along a river at a speed of 21 m/s to the west. what impulse is required to decrease the speed of t
zzz [600]
The impulse required to decrease the speed of the boat is equal to the variation of momentum of the boat:
J=\Delta p=m \Delta v
where
m=225 kg is the mass of the boat
\Delta v=v_f-v_i=15 m/s-21 m/s=-6 m/s is the variation of velocity of the boat
By substituting the numbers into the first equation, we find the impulse:
J=m\Delta v=(225 kg)(-6 m/s)=-1350 N s
and the negative sign means the direction of the impulse is against the direction of motion of the boat.
8 0
3 years ago
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Answer:

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When object goes under acceleration
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when object goes under acceleration

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★ Acceleration: Rate of increase in velocity.

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