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2 years ago

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A 20-μF capacitor has a charge of 60 μC. How much energy is stored in this capacitor? A) 100 μJ B) 80 μJ C) 90 μJ D) 70 μJ E) 11

0 μJ

1 answer:

IgorLugansk [536]2 years ago

4 0

Answer:

C...90

Explanation:

Energy stored = 1/2 of(Charge × Voltage)

....then substitute Voltage with (Charge ÷ Capacitance)

Then obtain the formula for energy stored as....

E = 1/2 of (Charge)^2 ÷ Capacitance

= 0.5 x (60^2÷20)

;Energy stored = 90

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two students are on a balcony 19.6 m above the street. one student throws a ball vertically downward at 14.7 m:ds. at the same i

NARA [144]

A. The difference in the two ball's time in the air is 3 seconds

B. The velocity of each ball as it strikes the ground is 24.5 m/s

C. The balls 0.500 s after they are thrown are 14.7 m apart

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

Initial Height = H = 19.6 m

Initial Velocity = u = 14.7 m/s

<u>Unknown:</u>

A. Δt = ?

B. v = ?

C. Δh = ?

<u>Solution:</u>

<h2>Question A:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$0 = 19.6 - 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 - 14.7t - 4.9t^2$

$4.9t^2 + 14.7t - 19.6 = 0$

$t^2 + 3t - 4 = 0$

$(t + 4)(t - 1) = 0$

$(t - 1) = 0$

$\boxed {t = 1 ~ second}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$0 = 19.6 + 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 + 14.7t - 4.9t^2$

$4.9t^2 - 14.7t - 19.6 = 0$

$t^2 - 3t - 4 = 0$

$(t - 4)(t + 1) = 0$

$(t - 4) = 0$

$\boxed {t = 4 ~ seconds}$

The difference in the two ball's time in the air is:

$\Delta t = 4 ~ seconds - 1 ~ second$

$\large {\boxed {\Delta t = 3 ~ seconds} }$

<h2>Question B:</h2><h3>First Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (-14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

<h3>Second Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

The velocity of each ball as it strikes the ground is 24.5 m/s

<h2>Question C:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$h = 19.6 - 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 11.025 ~ m}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$h = 19.6 + 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 25.725 ~ m}$

The difference in the two ball's height after 0.500 s is:

$\Delta h = 25.725 ~ m - 11.025 ~ m$

$\large {\boxed {\Delta h = 14.7 ~ m} }$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437
Kinetic Energy : brainly.com/question/692781
Acceleration : brainly.com/question/2283922
The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

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2 years ago

Materials have unique properties because each one is made up of different kinds of which particle?

inysia [295]

D. Atoms.

Explanation:

All the matter is made of elementary particles called "atoms".

Further, an atom is made of electrons, protons and neutrons. The electrons & protons are again made of the fundamental sub-particles, electrons (leptons) and the protons(quarks).

The classification of particles is shown in the figure attached

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2 years ago

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A 1150 kg car is on a 8.70° hill. using x-y axis tilted down the plane, what is the x-component of the weight?

Fed [463]

I assume the x-y axis are tilted such that the x-axis is parallel to the surface of the hill while the y-axis is perpendicular to it.

In this case, the x-component of the weight is given by:
$W_x =mg \sin \theta$
where
m is the mass of the car
g is the acceleration of gravity
$\theta$ is the angle of the hill

Substituting numbers into the formula, we find
$W_x=(1150 kg)(9.81 m/s^2)(\sin 8.70^{\circ})=1706 N$

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2 years ago

A 6.0-μF capacitor charged to 50 V and a 4.0-μF capacitor charged to 34 V are connected to each other, with the two positive pla

ch4aika [34]

Answer:

5702.88 J or 5.7mJ

Explanation:

Given that :

C 1 = 6.0-μF

C 2 = 4.0-μF

V 1 = 50V

V 2 = 34V

Note that : Q = CV

Q 1 = C1 * V1

Q 1 = 50×6 = 300μC

Q 2 = 34×4 = 136μC

Parallel connection = C 1 + C 2

= 6+4 = 10μC

V = Qt/C

Where Qt = Q1+Q2

V = Q1+Q2/C

V = 300+136/10

V = 437/10

V = 43.6volts

Uc1 = 1/2×C1V^2

= 1/2 × 6μF × 43.6^2

= 1/2 × 6μF × 1900.96

= 3μF × 1900.96volts

= 5702.88J

= 5702.88J/1000

= 5.7mJ

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2 years ago

9ma electric current is flowing through a conducting wire , then the number of electron passing through it in 3 min is

Leviafan [203]

Answer:

1.0125 x 10^19

Explanation:

current flowing through conductive wire= 9mA = 9 x 10^ -3 A

charge passing per 3 min

Q = It

= 9 x 10^ -3 x (3 x 60)

= 1.620 C

no of electrons in charge

Q = ne

1.620 = n x 1.6 x 10 ^ -19

n. = 1.0125 x 10 ^19

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2 years ago