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Trava [24]
3 years ago
12

A 6.00-μf parallel-plate capacitor has charges of 40.0 μc on its plates. how much potential energy is stored in this capacitor?

Physics
1 answer:
dedylja [7]3 years ago
7 0

Thew energy stored in a capacitor of capacitance C and voltage between the plates V is

E=\frac{1}{2} CV^2=\frac{1}{2C} Q^2.

Substituting numerical value

E=\frac{1}{2*6*10^{-6}} (40*10^{-6})^2\\ E=133.33\; \mu J

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Is any of this right?
Amiraneli [1.4K]
You will need to multiply them.
I have physics class and for acceleration we use 9.8m/s^2 but you used 9.5m/s^2.
If that’s what ur teacher wants u to use that’s ok. But yeah you multiply them.
F=ma
You multiply m and a
5 0
3 years ago
What does the slope on a velocity-time graph measure?
Lostsunrise [7]

Answer:The slope in the velocity-time graph represents the acceleration. The slope is defined as the ratio of change in y-axis to change in the x-axis. The slope is represented by the letter m and following is the general formula used for determining the slope:

Explanation:

6 0
2 years ago
Read 2 more answers
. A spring has a length of 0.200 m when a 0.300-kg mass hangs from it, and a length of 0.750 m when a 1.95-kg mass hangs from it
kap26 [50]

Answer:

29.4 N/m

0.1  

Explanation:

a) From the restoring Force we know that :  

F_r = —k*x  

the gravitational force :  

F_g=mg  

Where:

F_r is the restoring force .

F_g is the gravitational force

g is the acceleration of gravity

k is the constant force  

xi , x2 are the displacement made by the two masses.

Givens:

<em>m1 = 1.29 kg</em>

<em>m2 = 0.3 kg  </em>

<em>x1   = -0.75 m  </em>

<em>x2 = -0.2 m </em>

<em>g   = 9.8 m/s^2  </em>

Plugging known information to get :

F_r =F_g

-k*x1 + k*x2=m1*g-m2*g

k=29.4 N/m

b) To get the unloaded length 1:  

l=x1-(F_1/k)

Givens:

m1 = 1.95kg , x1 = —0.75m  

Plugging known infromation to get :

l= x1 — (F_1/k)  

= 0.1  

 

3 0
3 years ago
What is conduction and induction?
Romashka [77]
Conduction involves physical contact to charge, well induction does not.
Learn more at: <span>www.physicsclassroom.com/class/estatics/Lesson-2/Charging-by-Conduction</span>
8 0
3 years ago
Read 2 more answers
A 2.60 kg lion runs at a speed of 5.00 m/s until he sees his prey. The lion then speeds up 8.00 m/s to catch it. How much work d
ELEN [110]

Answer: 50.7 J

Explanation:

Given

mass of lion is  m=2.6\ kg

The initial speed of the lion is v_1=5\ m/s

increased speed of lion is v_2=8\ m/s

Initially, its kinetic energy is K_1=\frac{1}{2}mv_1^2

Final kinetic energy k_2=\frac{1}{2}mv_2^2

work did by lion after speed up is k_2-k_1

\Rightarrow W=\dfrac{1}{2}\times 2.6[8^2-5^2]\\\\\Rightarrow W=1.3\times [39]=50.7\ J

5 0
2 years ago
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