All categories

3 years ago

A capacitor has a capacitance of 0. 40 µF at a voltage of 9. 0 V. What is the charge on each plate of the capacitor? µC.

1 answer:

4 0

The capacitor is a device that can store electrical energy. It is a two-conductor configuration. The charge on each plate of the capacitor will be 3.6 µC.

<h3>What is a capacitor?</h3>

A capacitor is a device that can store electrical energy. It is a two-conductor configuration separated by an insulating medium that carries charges of equal size and opposite sign.

An electric insulator or vacuum, such as glass, paper, air, or a semi-conductor termed a dielectric, can be used as the non-conductive zone.

The given data in the problem is;

C is the capicitence of capicitor= 0. 40 µF

V is the voltage = 9. 0 V

Q is a charge on each plate of the capacitor=?µC.

The formula for the capacitor is given as;

$\rm Q=CV \\\\ \rm Q=0. 40 \times 9. 0 \\\\ \rm Q=3.6 \ \mu C.$

Hence the charge on each plate of the capacitor will be 3.6 µC.

To learn more about the capacitor refer to the link;

brainly.com/question/14048432

You might be interested in

A 225-kg object and a 525-kg object are separated by 3.80 m. (a) Find the magnitude of the net gravitational force exerted by th

pav-90 [236]

Answer: $F_{net}=3.383\times 10^{-7}\ N$

Explanation:

Given

Mass of first object $m_1=225\ kg$

Mass of second object $m_2=525\ kg$

Distance between them $d=3.8\ m$

$m_3=61\ kg$ object is placed between them

So force exerted by $m_1$ on $m_3$

$F_{13}=\frac{Gm_1m_3}{1.9^2}$

$F_{13}=\frac{6.674\times 10^{-11}(225\times 61)}{1.9^2}$

$F_{13}=2.5374141274×10^{−7}\ N$

Force exerted by $m_2\ on\ m_3$

$F_{23}=\frac{Gm_2m_3}{1.9^2}$

$F_{23}=\frac{6.674\times 10^{-11}(525\times 61)}{1.9^2}$

$F_{23}=5.920632964\times 10^{-7}\ N$

So net force on $m_3$ is

$F_{net}=F_{23}-F_{13}$

$F_{net}=5.920632964\times 10^{-7}-2.5374141274\times 10^{-7}$

$F_{net}=3.383\times 10^{-7}\ N$

i.e. net force is towards $m_2$

(b)For net force to be zero on $m_3$ , suppose

So force exerted by $m_1$ and $m_2$ must be equal

$F_{13}=F_{23}$

$\Rightarrow \frac{Gm_1m_3}{x^2}=\frac{Gm_2m_3}{(3.8-x)^2}$

$\Rightarrow \frac{m_1}{x^2}=\frac{m_2}{(3.8-x)^2}$

$\Rightarrow (\frac{3.8-x}{x})^2=\frac{m_2}{m_1}$

$\Rightarrow \frac{3.8-x}{x}=\sqrt{\frac{525}{225}}$

$\Rightarrow 3.8-x=1.52752x$

$\Rightarrow 3.8=2.52x$

$\Rightarrow x=1.507\ m$

4 0

3 years ago

The difference in the composition of the terrestrial planets and the gas giants can be explained by the different condensation

miv72 [106K]

The inner planets are usually rocky because the gravitational pull is stronger closer to the star or in this case the sun. The dust and rocky particles that are left over after a super nova or in a nebula will tend to orbit closer to a proto-star when a solar system is in its early days. In our solar system these planets are Mercury, Venus, Earth and Mars. Gases are less dense and will be less affected by the pull of gravity because rocky particles have more mass. The outer planets are gas giants formed from clouds of gas that would be further out in the spinning disk around a proto-star.

6 0

4 years ago

Amadeo the Archer is practicing his archery skills. He shoots both arrows into a field. His first arrow (arrow A) is embedded ha

attashe74 [19]

Answer:

The arrow with greater impart is Arrow B

The both arrows will feel the same impulse

Explanation:

1. Arrow B since $KE= W = F * d$ it used more force to stop itself in a shorter distance.

2. They should feel the same impulse since the both had the same momentum $m * v= F * t$

4 0

3 years ago

jet is flying at 500 mph east relative to the ground. A Cessna is flying at 150 mph 60° north of west relative to the ground. Wh

Greeley [361]

Answer:

C. 590 mph

$\vert v_{cj}\vert=589.49\ mph$

Explanation:

Given:

velocity of jet, $v_j=500\ mph$
direction of velocity of jet, east relative to the ground
velocity of Cessna, $v_c=150\ mph$
direction of velocity of Cessna, 60° north of west

Taking the x-axis alignment towards east and hence we have the velocity vector of the jet as reference.

Refer the attached schematic.

So,

$\vec v_j=500\ \hat i\ mph$

$\vec v_c=150\times (\cos120\ \hat i+\sin120\ \hat j)$

$\vec v_c=-75\ \hat i+75\sqrt{3}\ \hat j\ mph$

Now the vector of relative velocity of Cessna with respect to jet:

$\vec v_{cj}=\vec v_j-\vec v_c$

$\vec v_{cj}=500\ \hat i-(-75\ \hat i+75\sqrt{3}\ \hat j )$

$\vec v_{cj}=575\ \hat i-75\sqrt{3}\ \hat j\ mph$

Now the magnitude of this velocity:

$\vert v_{cj}\vert=\sqrt{(575)^2+(75\sqrt{3} )^2}$

$\vert v_{cj}\vert=589.49\ mph$ is the relative velocity of Cessna with respect to the jet.

8 0

3 years ago

A scientist kept 1,000 grams of a radioactive material in a container. After 66 days, he observed that 125 grams of the radioact

bixtya [17]

The sample appears to have gone through 3 half-lives
1st half life: 1000 to 500 g
2nd half life: 500 to 250 g
3rd half life: 250 to 125 g
The duration of a half-life, therefore, can be inferred to be 66 ÷ (3) = 22 days.

After a 4th half life, there will be 125÷2= 62.5 g.
At this point, an additional 22 days will have passed, for a total of 88 days.
Answer is C.

8 0

4 years ago