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Juli2301 [7.4K]
4 years ago
12

A little girl slid down a playground slide, decreasing her potential energy by 1000 J while increasing her kinetic energy by onl

y 900 J. Why did she lose 100 J of mexhanical energy
Physics
2 answers:
Kaylis [27]4 years ago
8 0

Answer:

Due to friction, 100 joules were lost in the form of heat energy

Explanation:

The law of conservation of energy states that

<em>"Energy can neither be created or nor destroyed but rather it changes from one form to another"</em>

As the little girl was sliding down the playground slide, her potential energy was decreasing and at the same time her kinetic energy was accumulated. As we know there is friction in the slide and because of this not all of the of potential energy is converted into kinetic energy. Some of the energy will be lost due to friction taking the form of heat. Therefore, the 100 joules of energy was lost in the form of heat due to friction of the slide.

koban [17]4 years ago
5 0

Answer:

Explanation:

This is because The 100 J of potential energy that doesn't go into increasing her kinetic energy goes into thermal energy—heating her bottom and the slide.

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The inner cylinder of a long, cylindrical capacitor has radius r and linear charge density +λ. It is surrounded by a coaxial cyl
Ulleksa [173]

Hi there!

a)

We can begin by using the equation for energy density.

U = \frac{1}{2}\epsilon_0 E^2

U = Energy (J)

ε₀ = permittivity of free space

E = electric field (V/m)

First, derive the equation for the electric field using Gauss's Law:
\Phi _E = \oint E \cdot dA = \frac{Q_{encl}}{\epsilon_0}

Creating a Gaussian surface being the lateral surface area of a cylinder:
A = 2\pi rL\\\\E \cdot 2\pi rL = \frac{Q_{encl}}{\epsilon_0}\\\\Q = \lambda L\\\\E \cdot 2\pi rL = \frac{\lambda L}{\epsilon_0}\\\\E = \frac{\lambda }{2\pi r \epsilon_0}

Now, we can calculate the energy density using the equation:
U = \frac{1}{2} \epsilon_0 E^2

Plug in the expression for the electric field and solve.

U = \frac{1}{2}\epsilon_0 (\frac{\lambda}{2\pi r \epsilon_0})^2\\\\U = \frac{\lambda^2}{8\pi^2r^2\epsilon_0}

b)

Now, we can integrate over the volume with respect to the radius.

Recall:
V = \pi r^2L \\\\dV = 2\pi rLdr

Now, we can take the integral of the above expression. Let:
r_i = inner cylinder radius

r_o = outer cylindrical shell inner radius

Total energy-field energy:

U = \int\limits^{r_o}_{r_i} {U_D} \, dV =   \int\limits^{r_o}_{r_i} {2\pi rL *U_D} \, dr

Plug in the equation for the electric field energy density and solve.

U =   \int\limits^{r_o}_{r_i} {2\pi rL *\frac{\lambda^2}{8\pi^2r^2\epsilon_0}} \, dr\\\\U = \int\limits^{r_o}_{r_i} { L *\frac{\lambda^2}{4\pi r\epsilon_0}} \, dr\\

Bring constants in front and integrate. Recall the following integration rule:
\int {\frac{1}{x}} \, dx  = ln(x) + C

Now, we can solve!

U = \frac{\lambda^2 L}{4\pi \epsilon_0}\int\limits^{r_o}_{r_i} { \frac{1}{r}} \, dr\\\\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(r)\left \| {{r_o} \atop {r_i}} \right. \\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} (ln(r_o) - ln(r_i))\\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(\frac{r_o}{r_i})

To find the total electric field energy per unit length, we can simply divide by the length, 'L'.

\frac{U}{L} = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(\frac{r_o}{r_i})\frac{1}{L} \\\\\frac{U}{L} = \boxed{\frac{\lambda^2 }{4\pi \epsilon_0} ln(\frac{r_o}{r_i})}

And here's our equation!

3 0
2 years ago
What type of rock is limestone? Describe how a limestone rock is likely to change over a long period of time.
tatuchka [14]

Answer:

Limestone is a sedimentary rock

Explanation:

1. Limestone is a carbonate sedimentary rock made up of calcite(CaCO₃).This rock type forms calcareous shells and tests of organisms that was deposited in a sedimentary basin.

Limestone is used in building and construction works. It also finds application in chemical industries.

2. Over a long period of time, we would take a look at the rock "limestone" through the rock cycle.

Limestone being a sedimentary rock would be converted to marble, a metamorphic rock if subjected to metamorphic conditions over an extensive period of time. With series of metamorphic transformation, marble can grade to higher metamorphic facies of rocks as it combines with other minerals in the crust. The minerals would eventually change and as the changed rock approaches its melting temperature, melt would result.

From the other spectrum, limestone can be weathered if subjected extensively to denudation forces such as wind, water and glaciation. Water is more potent for the chemical weathering of limestone. Limestone would easily and readily dissolve in it over a long period of time.

7 0
3 years ago
Whats the measure of how much energy a sound wave carries, the loudness of a sound?
musickatia [10]
Decibels I believe? I’m not 100% sure
6 0
3 years ago
Which term describes how fast the atoms or molecules inside an object are moving?
IrinaVladis [17]
Im pretty sure it’s A. temperature
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4 years ago
Now find the electromotive force E2(t) induced across the entirety of solenoid 2 by the change in current in solenoid 1. Remembe
lions [1.4K]

Answer:

E_{2} (t) = -\pi\mu_o} \rho^{2} n_{1}n_{2}L\frac{d }{dt}I_{1}(t)

Explanation:

Consider two solenoids that are wound on a common cylinder as shown in fig. 1. Let the cylinder have radius 'ρ' and length 'L' .

No. of turns of solenoid 1 = n₁

No. of turns of solenoid 1 = n₂

Assume that length of  solenoid is much longer than its radius, so that its field can be determined from Ampère's law throughout its entire length:

\oint \overrightarrow {B}\overrightarrow {(r)}.\overrightarrow {dl}= \mu_{o}I

We will consider the field that arises from solenoid 1, having n₁  turns per unit length. The magnetic field due to solenoid 1 passes through solenoid 2, which has n₂ turns per unit length.

Any change in magnetic flux from the field generated by solenoid 1 induces an EMF in solenoid 2 through Faraday's law of induction:

\oint \overrightarrow {B}\overrightarrow {(r)}.\overrightarrow {dl}= -\frac{d}{dt} \phi _{M}(t)

Consider B₁(t) magnetic feild generated in solenoid 1 due to current I₁(t)

Using:

                                  B_{1}(t) =\mu _{o} nI(t)\\ --- (2)

                           

Flux generated due to magnetic field B₁

                      \phi _{1}(t) = \oint \overrightarrow {B_{1}}.dA\\ ---(3)

area of solenoid = A = \pi \rho^{2}

substituting (2) in (3)

                       \phi _{1}(t) = \mu_{o} \pi \rho^{2} n_{1}I_{1}(t) ----(4)

We have to find electromotive force E₂(t) induced across the entirety of solenoid 2 by the change in current in solenoid 1, i.e.

                       E_{2} (t) = -n_{2}L\frac{d \phi_{1}}{dt} ---- (5)

substituting (4) in (5)

E_{2} (t) = -n_{2}L\frac{d }{dt}(\mu_o} \pi \rho^{2} n_{1}I_{1}(t))\\E_{2} (t) = -\pi\mu_o} \rho^{2} n_{1}n_{2}L\frac{d }{dt}I_{1}(t)

5 0
3 years ago
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