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3 years ago

5

A 2.0 ???????? capacitor and a4.0 ???????? capacitor are connected in parallel across a 300 V potential difference. Calculate th

e total energy stored in the capacitor?

1 answer:

kolezko [41]3 years ago

7 0

Answer:

0.27J

Explanation:

$C_eq= C_1 + C_2\\= 2+4\\= 6UF\\U = (1/2) CV^2\\= (1/2)(6 - 6)(300 * 300)\\= 0.27J$

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Conduction is the transfer of

mina [271]

<em>Here is your Answer:-</em>

<em>Conduction is how heat transfers through direct contact with objects that are touching. Any time that two objects or substances touch, the hotter object passes heat to the cooler object. (That hot sand passed the heat energy right into my poor feet)!</em>

8 0

2 years ago

A physics lab group rolls a ball off a horizontal table with a speed of 1.6 m/s. the table is 1.2 m high. how far in the horizon

Brilliant_brown [7]

Answer:

0.784 m

Explanation:

The motion of the ball is a projectile motion, which consists of two independent motions:

- a horizontal uniform motion, with constant speed of $v_0 = 1.6 m/s$

- a vertical accelerated motion, with constant acceleration $g=-9.8 m/s^2$ towards the ground and initial position $y_0 = 1.2 m$

By looking at the vertical motion, we can find the time t at which the ball hits the ground, which is the time t at which the vertical position y becomes zero:

$y(t)=0\\y(t)=y_0 +\frac{1}{2}gt^2\\0=y_0 +\frac{1}{2}gt^2\\t=\sqrt{-\frac{2 y_0}{g}}=\sqrt{-\frac{2\cdot 1.2 m}{-9.8 m/s^2}}=0.49 s$

And now we can use the horizontal motion to find the total horizontal distance traveled during this time interval:

$S_x = v t=(1.6 m/s)(0.49 s)=0.784 m$

4 0

3 years ago

Two positive point charges that are equal in magnitude are fixed in place, one at x = 0.00 m and the other at x = 1.00 m, on the

olga55 [171]

Answer:

0.5 m

Explanation:

Two charges each of magnitude q

Let the third charge is Q is placed at a distance x from the origin so that the charge is in equilibrium.

The force on Q due to q at origin is balanced by the charge on Q due to the charge q placed at x = 1 m.

So,

$\frac{KQq}{x^{2}}=\frac{KQq}{\left ( 1-x \right )^{2}}$

1 - x = x

1 = 2x

x = 0.5 m

Thus, the third charge is placed at x = 0.5 m .

6 0

4 years ago

Two point charges are brought closer together, increasing the force between them by a factor of 20. By what factor was their sep

olga55 [171]

The separation between them is $\frac{r}{\sqrt{20} }$

Concept :

If the force increases, distance between charges must decrease. Force is indirectly proportional to the distance squared.

Given,

Two point charges are brought closer together, increasing the force between them by a factor of 20.

Original force is

F = $\frac{kq_{1} q_{2} }{r^{2} }$ -------- ( 1 )

Here, $q_{1} , q_{2}$ are charges and r is the distance between them

New force F' = $\frac{kq_{1q_{2} } }{r'^{2} }$ ----------- (2 )

Divide ( 1 ) and ( 2 )

$\frac{F'}{F}$ = $\frac{\frac{kq_{1}q_{2} }{r'^{2} } }{\frac{kq_{1}_{2} }{r^{2} } }$

20 = $\frac{r^{2} }{r'^{2} }$

r' = $\frac{r}{\sqrt{20} }$

Given that force between them are increasing and therefore distance between them decrease by $\frac{r}{\sqrt{20} }$

Learn more about two point charges here : brainly.com/question/24206363

#SPJ4

8 0

2 years ago

Its science<br> Observations and Inferences <br> Multiple choice

alexandr1967 [171]

Answer:

Like me brainslest medasnad

Explanation:

3 0

3 years ago

Read 2 more answers