Describe how day length is affected by the tilt of the earths axis

Which of the following objects has more kinetic energy than it does potential energy? Select the correct answer below:______1.a

Lilit [14]

Answer:

The pendulum of the clock.

Explanation:

Hi there!

The kinetic energy is the energy associated with the velocity of the object. The potential energy is the energy associated with the position of the object. In the objects listed in the question, only one object is moving: the pendulum of the clock (assuming that the clock is functioning). If the clock functions, the pendulum is moving when it is at the lowest point of its arc of motion and with maximum velocity. All potential energy that the pendulum stored when it reached the highest height, is transformed into kinetic energy at the lowest point. Thus, at that point, the object has more kinetic energy than potential energy.

6 0

2 years ago

Molly is looking for a sustainable fuel source that doesn't require technical equipment to harness. She is concerned that trees

notka56 [123]

Answer:

She can consider using agricultural waste or dried dung.

Explanation:

No doubt, biomass has become a crucial source of energy to the society, with almost 90% of the households in rural area now relying on biomass for energy. Biomass has become a great option for household heating and cooking. It is locally available and abundant. It is a clean type of fuel, unlike fossil fuels and it somehow helps in cleaning our environment as it traps carbondioxide. Some common types of biomass include dried dung, agricultural waste, or even charcoal.

8 0

3 years ago

Read 2 more answers

A baseball is thrown through the air. It's initial velocity, described as a vector, is → v ( t = 0 ) = 17.1 ˆ i + 14.7 ˆ j m / s

ludmilkaskok [199]

Answer:

a = - 9.8 j ^ m/s²

Explanation:

This is a projectile launch problem, they give us the initial velocity in the two components

v₀ₓ = 17.1 m / s

$v_{oy}$ = 14.7 m / s

They indicate that the only acceleration that exists is the acceleration of gravity, which acts in the direction towards the center of the Earth, in general in a coordinate system it coincides with the direction of the y axis.

a = - g j ^

a = - 9.8 j ^ m /s²

6 0

2 years ago

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

1 year ago

Why does range of motion need to be measured separately for each joint

telo118 [61]

Because each joint is capable of a different measure of its own range of motion

6 0

3 years ago