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3 years ago

I NEED THIS A SOON AS POSSIBLE

2 answers:

6 0

Work is done when spring is extended or compressed. Elastic potential energy is stored in the spring. Provided inelastic deformation has not happened, the work done is equal to the elastic potential energy stored.

ludmilkaskok [199]3 years ago

5 0

Answer: c) 1.0 N/m constant spring stretched 0.3 m.

Explanation:

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4 years ago

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PLEASE HELP ME ALREADY!!! MAX POINTS RANDOM ANSWERS WILL BE REPORTED!!!

horsena [70]

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A crate filled with delicious junk food rests on a horizontal floor. Only gravity and the support force of the floor act on it,

stealth61 [152]

The words "... as shown ..." tell us that there's a picture that goes along
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3 years ago

The capacitor is now disconnected from the battery, and the dielectric plate is slowly removed the rest of the way out of the ca

Xelga [282]

The question is incomplete. The complete question is :

A dielectric-filled parallel-plate capacitor has plate area A = 10.0 cm2 , plate separation d = 10.0 mm and dielectric constant k = 3.00. The capacitor is connected to a battery that creates a constant voltage V = 15.0 V . Throughout the problem, use ϵ0 = 8.85×10−12 C2/N⋅m2 .

Find the energy U1 of the dielectric-filled capacitor. I got U1=2.99*10^-10 J which I know is correct. Now I need these:

1. The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy U2 of the capacitor at the moment when the capacitor is half-filled with the dielectric.

2. The capacitor is now disconnected from the battery, and the dielectric plate is slowly removed the rest of the way out of the capacitor. Find the new energy of the capacitor, U3.

Solution :

Given :

$A = 10 \ cm^2$

$=0.0010 \ m^2$

d = 10 mm

= 0.010 m

Then, Capacitance,

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

$C=\frac{8.85 \times 10^{12} \times 3 \times 0.0010}{0.010}$

$C=2.655 \times 10^{12} \ F$

$U_1 = \frac{1}{2}CV^2$

$U_1 = \frac{1}{2} \times 2.655 \times 10^{-12} \times (15V)^2$

$U_1=2.987 \times 10^{-10}\ J$

Now,

$C_k=\frac{1}{2} \frac{k \epsilon_0}{d} \times \frac{A}{2}$

And

$C_{air}=\frac{1}{2} \frac{\epsilon_0}{d} \times \frac{A}{2}$

In parallel combination,

$C_{eq}= C_k + C_{air}$

$C_{eq} = \frac{1}{2} \frac{\epsilon_0 A}{d}(1+k)$

$C_{eq} = \frac{1}{2} \times \frac{8.85 \times 10^{-12} \times 0.0010}{0.01} \times (1+3)$

$C_{eq} = 1.77 \times 10^{-12}\ F$

Then energy,

$U_2 =\frac{1}{2} C_{eq} V^2$

$U_2=\frac{1}{2} \times 1.77 \times 10^{-12} \times (15V)^2$

$U_2=1.99 \times 10^{-10} \ J$

b). Now the charge on the $\text{capacitor}$ is :

$Q=C_{eq} V$

$Q = 1.77 \times 10^{-12} \times 15 V$

$Q = 26.55 \times 10^{-12} \ C$

Now when the capacitor gets disconnected from battery and the $\text{dielectric}$ is slowly $\text{removed the rest}$ of the way out of the $\text{capacitor}$ is :

$C_3=\frac{A \epsilon_0}{d}$

$C_3 = \frac{0.0010 \times 8.85 \times 10^{-12}}{0.01}$

$C_3=0.885 \times 10^{-12} \ F$

$C_3 = 0.885 \times 10^{-12} \ F$

Without the dielectric,

$U_3=\frac{1}{2} \frac{Q^2}{C}$

$U_3=\frac{1}{2} \times \frac{(25.55 \times 10^{-12})^2}{0.885 \times 10^{-12}}$

$U_3=3.98 \times 10^{-10} \ J$

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3 years ago

An example of constant velocity

pashok25 [27]

Some examples of constant velocity (or at least almost- constant velocity) motion include (among many others): • A car traveling at constant speed without changing direction. A hockey puck sliding across ice. A space probe that is drifting through interstellar space.

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3 years ago