An air-filled capacitor stores a potential energy of 6.00 mJ due to its charge. It is accidentally filled with water in such a way as not to discharge its plates. How much energy does it continue to store after it is filled

Answer:

There are many kinds of waves all around us. There are waves in the ocean and in lakes. Did you also know that there are also waves in the air? Sound travels through the air in waves and light is made up of waves of electromagnetic energy.

The wavelength of a wave describes how long the wave is. The distance from the "crest" (top) of one wave to the crest of the next wave is the wavelength. Alternately, we can measure from the "trough" (bottom) of one wave to the trough of the next wave and get the same value for the wavelength.

The frequency of a wave is inversely proportional to its wavelength. That means that waves with a high frequency have a short wavelength, while waves with a low frequency have a longer wavelength.

Light waves have very, very short wavelengths. Red light waves have wavelengths around 700 nanometers (nm), while blue and purple light have even shorter waves with wavelengths around 400 or 500 nm. Some radio waves, another type of electromagnetic radiation, have much longer waves than light, with wavelengths ranging from millimeters to kilometers.

Sound waves traveling through air have wavelengths from millimeters to meters. Low-pitch bass notes that humans can barely hear have huge wavelengths around 17 meters and frequencies around 20 hertz (Hz). Extremely high-pitched sounds that are on the other edge of the range that humans can hear have smaller wavelengths around 17 mm and frequencies around 20 kHz (kilohertz, or thousands of Hertz).

Explanation:

Explanation:

Hope This Helped U

Brqinliest

The force acts on this bullet : 4.8456 N

Given

m=0.09 kg

Δt=0.025 s

vo=0(from rest)

vt=1.346 m/s

Required

Force

Solution

Impulse is a change in momentum

I=ΔP

F.Δt=m(vt-vo)

Input the value

F x 0.025 = 0.09(1.346-0)

F=4.8456 N

Answer:

0.4

Explanation:

Given that a particular inductor is connected to a circuit where it experiences a change in current of 0.8 amps every 0.10 sec. If the inductor has a self-inductance of 2.0 V, what is the inductance

Using the power formula

P = IV

Substitute all the parameters

P = 0.8 × 2

P = 1.6 W

But P = I^2 R

Substitute power and current

1.6 = 0.8^2 R

R = 1.6 / 0.64

R = 2.5 ohms

Inductance = reciprocal of resistance

Inductance = 1 / 2.5

Inductance = 0.4

Answer:

Friction is the force that opposes movement between moving objects.

The angle at which one object starts to slip on the other is directly related to the coefficient. When the two objects are horizontal there is no frictional force. So, the coefficient of static friction is equal to the tangent of the angle at which the objects slide. A similar method can be used to measure μk.

Explanation:

Answer:

brainliest plsssssss

Explanation:

The resistance that one surface or object encounters when moving over another

Answer: tennis ball

Explanation:

For every complete revolution the boy makes around the center of the carousel, he travels a distance of 2π (5.0 m) = 10π m, which gives a linear speed of

v = (10π m) / (8.0 s) ≈ 3.927 m/s

Then his centripetal acceleration would be

a = v ² / (5.0 m) ≈ 3.084 m/s²

so that the centripetal force exerted on him has magnitude

F = (45 kg) a ≈ 138.791 N ≈ 140 N

(rounded to 2 significant digits)

Answer:

Puck 1 = 12 m/s

Explanation:

Conservation of energy

KE_1a + KE_2a = KE_1b + KE_2b

(1/2)(0.1kg)(15m/s)^2 + (1/2)(0.1kg)(12m/s)^2 = (1/2)(0.1kg)(V)^2 + (1/2)(0.1kg)(15m/s)^2

V = 12m/s

Answer:

12 m/s west

Explanation:

just did it

Explanation:

Such spheres are intimately connected. Many animals (biosphere), for example, migrate through to the sky, while groundwater (hydrosphere) also flows through the ground (lithosphere). The domains are actually so closely related that a shift in one globe always results in a shift in one or both of some other spheres.

Answer:

The average power delivered by the elevator motor during this period is 6.686 kW.

Explanation:

Given;

mass of the elevator, m = 636 kg

initial speed of the elevator, u = 0

time of motion, t = 4.5 s

final speed of the elevator, v = 2.05 m/s

The upward force of the elevator is calculated as;

F = m(a + g)

where;

m is mass of the elevator

a is the constant acceleration of the elevator

g is acceleration due to gravity = 9.8 m/s²

[tex]a = \frac{v-u}{t} \\\\a = \frac{2.05 -0}{4.5} \\\\a = 0.456 \ m/s^2[/tex]

F = (636)(0.456 + 9.8)

F = (636)(10.256)

F = 6522.816 N

The average power delivered by the elevator is calculated as;

[tex]P_{avg} = \frac{1}{2} (Fv)\\\\P_{avg} = \frac{1}{2} (6522.816 \ \times \ 2.05)\\\\P_{avg} = 6685.89 \ W\\\\P_{avg} = 6.68589 \ kW\\\\P_{avg} = 6.686 \ k W[/tex]

Therefore, the average power delivered by the elevator motor during this period is 6.686 kW.

Answer:

 Q = 4 Q₀

Explanation:

This is an exercise on capacitors, where the capacitance is

         C =  [tex]\epsilon_{o} \ \frac{A}{d}[/tex]

if we apply the given conditions

         C = \epsilon_{o} \    \frac{2A}{0.5d}

         C = 4 \epsilon_{o} \    \frac{A}{d}

let's call the capacitance Co with the initial values

         C₀ = \epsilon_{o} \    \frac{A}{d}

         C = 4 C₀

The charge on each plate of a capacitor is

           Q = C ΔV

If the potential difference is maintained, the new charge is

       Q = 4 C₀ ΔV

let's call

        Q₀ = C₀ ΔV

we substitute

        Q = 4 Q₀

The satellite is moved to a new circular orbit that is 2R from the center of the moon, then the gravitational field strength of the moon at this new distance would be one-fourth of the initial gravitational field.

It can be defined as the force by which a body attracts another body toward its center as the result of the gravitational pull of one body and another.

As given in the problem A satellite of mass m orbits a moon of mass M in a uniform circular motion with a constant tangential speed of v. The gravitational field strength at a distance R from the center of the moon is gR. The satellite is moved to a new circular orbit that is 2R from the center of the moon.

The gravitational field strength is inversely proportional to the square of the distance from the center of the planet.

Thus, the gravitational field strength of the moon at this new distance would be one-fourth of the initial gravitational field.

To learn more about gravity here, refer to the link given below ;

brainly.com/question/4014727

#SPJ1

Answer:

Explanation:

B

Answer:

Laws

Explanation:

Laws are a primary way discourage drinking and driving

A primary way to discourage drinking and driving is Law.

The person who takes in alcohol and then drives on the road. This is strictly prohibited.

Laws against the  'drinking and driving' will make people get scared of getting charged or sentenced to jail for some years. Lot of accidents have caused when there were no laws against the action.

Thus, Laws are a primary way to discourage drinking and driving.

Learn more about drinking and driving.

https://brainly.com/question/11317786

#SPJ2

question: is this actually hypothetical?

Explanation:

also just leave the car there go get some McDonald's or sum and come back and if they're still breathing then go ahead and move the car .

Answer:

the same thing the last guy said

Answer:

0.8%

Explanation:

We are given;

Number of oscillations; n = 20

Time taken; t = 25 s

Formula for period of oscillation;

T = t/n = 25/20 = 1.25 s

We are told that the least count is 0.2 s. Thus, error is; ΔT = 0.2 s

percentage error in the measurement of time is given by;

(0.2/(20 × 1.25)) × 100% = 0.8%

Answer:

4:1

Explanation:

Given that the initial speed of the first car, u = 2v while the initial speed of the second car, u = v. To find the distance travelled, we are going to apply one of the equations of motion. The equation chosen is

v² = u² - 2as, where

s = the distance needed

a = acceleration due to gravity

u = initial velocity which is v & 2v

v = final velocity which is 0

For the first car with initial velocity, 2v, on substituting into the equation, we have

v² = u² - 2as(1)

0 = 4v - 2as(1)

4v = 2as(1)

2v = as(1), making s(1) subject of formula we have

s(1) = 2v/a

Taking the second car, we have u = v

v² = u² - 2as(2)

0 = v - 2as(2)

v = 2as(2), making s(2) subject of formula, we have

s(2) = v/2a

Not, ratio of s1 : s2 =

2v/a : v/2a

s1/s2 = 2v/a ÷ v/2a

s1/s2 = 2v/a * 2a/v

s1/s2 = 4av/av

s1/s2 = 4/1

Therefore, the ratio of the first car to the second car is 4:1

Please leave a like if it helped you

Answer:

3N

Explanation:

The frictional force always acts directly opposite to the force of motion, that is it opposes motion. According to the Newton second law of motion:

Sum of horizontal forces = 0

Hence:

5N - 4(0.5) - F = 0

5 - 2 = F

F = 3N

Answer:

The time it will take the mega-ray blaster to hit the ground is 2.57 s.

Explanation:

Given;

initial velocity of Optimus Prime, u = 24 m/s

height of fall of the mega-ray blaster, h = 94 m

The time of fall of the mega-ray blaster is calculated using the following kinematic equation;

[tex]h = ut + \frac{1}{2}gt^2\\\\94 = 24t + \frac{1}{2}(9.8)t^2\\\\94 = 24t + 4.9t^2\\\\4.9t^2 +24t -94 = 0\\\\Use \ formula \ method \ to \ solve \ for \ "t"\\\\a = 4.9 , b = 24, c = -94\\\\t = \frac{-b \ +/- \ \sqrt{b^2 -4ac} }{2a} \\\\t = \frac{-24 \ +/- \ \sqrt{(24)^2 -4(-94 \times4.9)} }{2(4.9)} \\\\t = \frac{-24 \ +/- \ \sqrt{2418.4} }{9.8}\\\\t = \frac{-24 \ +/- \ 49.177 }{9.8}\\\\t = \frac{-24 \ +\ 49.177 }{9.8} \ \ or \ \ t = \frac{-24 \ -\ 49.177 }{9.8} \\\\[/tex]

[tex]t = 2.57 \ s \ \ or \ \ t = -7.47 \ s[/tex]

t = 2.57 s

Therefore, the time it will take the mega-ray blaster to hit the ground is 2.57 s.

Answer:

h = 9.83 cm

Explanation:

Let's analyze this interesting exercise a bit, let's start by comparing the density of the ball with that of water

let's reduce the magnitudes to the SI system

         r = 10 cm = 0.10 m

         m = 10 g = 0.010 kg

         A = 100 cm² = 0.01 m²

the definition of density is

          ρ = m / V

the volume of a sphere

         V = [tex]\frac{4}{3} \ \pi r^{3}[/tex]

          V = [tex]\frac{4}{3}[/tex] π 0.1³

          V = 4.189 10⁻³ m³

let's calculate the density of the ball

           ρ = [tex]\frac{0.010}{4.189 \ 10^{-3} }[/tex]

           ρ = 2.387 kg / m³

the tabulated density of water is

         ρ_water = 997 kg / m³

we can see that the density of the body is less than the density of water. Consequently the body floats in the water, therefore the water level that rises corresponds to the submerged part of the body. Let's write the equilibrium equation

            B - W = 0

            B = W

where B is the thrust that is given by Archimedes' principle

           ρ_liquid  g V_submerged = m g

           V_submerged = m / ρ_liquid

we calculate

            V _submerged = 0.10 9.8 / 997

             V_submerged = 9.83 10⁻⁴ m³

The volume increassed of the water container

           V = A h

            h = V / A

let's calculate

            h = 9.83 10⁻⁴ / 0.01

            h = 0.0983  m

this is equal to h = 9.83 cm

Answer:

B

Explanation:

But molecules at a higher temperature have more energy.

Answer: B. Raising the temperature of the air

Answer:0.4 m

Explanation:

Given

Maximum displacement A=0.49

The sum of kinetic and elastic potential energy is [tex]\frac{1}{2}kA^2[/tex]

where k=spring constant

U+K.E.=[tex]\frac{1}{2}kA^2[/tex]

when K.E.=U/2

K.E.=kinetic energy

U=Elastic potential Energy

[tex]\rightarrow \ U+\frac{U}{2}=\frac{1}{2}KA^2\\\rightarrow \ \frac{3U}{2}=\frac{1}{2}KA^2\\\rightarrow \ U=\frac{1}{3}KA^2\\\rightarrow \ \frac{Kx^2}{2}=\frac{1}{3}KA^2\\\\x=\sqrt{\frac{2}{3}}A\\x=0.4\ m[/tex]

Answer:

the molecules will begin to move slowly and will turn to ice

Explanation:

hope this was good or not not sure if am right but yeah

Answer:

The answer is temperature lol

Explanation:

:)

Answer:

=101343.62N/m^2

Explanation:

absolute pressure on the bottom of a swimming pool= atmospheric pressure +( 2 ×ρ ×g)

( 2 ×ρ ×g)= guage pressure

atmospheric pressure= 101325pa

h= height= 1.9 mm = 1.9×10^-3m

ρ = density of water

= 1000kg/m^3

g= acceleration due to gravity= 9.8m/s^2

Then substitute, we have

absolute pressure on the bottom of a swimming pool= 101325+ [0.0019 ×1000 × 9.8)]

=101343.62N/m^2

Hence, the absolute pressure on the bottom of a swimming pool is =101343.62N/m^2