Do you think energy can be “used up?” Why or why not?

Goryan [66]

Because the Earth's axis is not "straight up and down" as we move
around the sun.

So when we're on one side of the sun, the top pole leans slightly toward
the sun. During that time the sun shines more directly on the top half
of the Earth, and less directly on the bottom half. The people on the
top half see the sun higher in the sky, and their weather is warmer,
while the people on the bottom half see the sun lower in the sky, and
their weather is cooler.

Then, when we're on the other side of the sun, the top pole leans slightly
away from the sun. During that time the sun shines more directly on the
bottom half of the Earth, and less directly on the top half. The people on
the bottom half see the sun higher in the sky, and their weather is warmer,
while the people on the top half see the sun lower in the sky, and their
weather is cooler.

The Earth makes the complete trip around the sun in one year, so the
people on the Earth go through this cycle of higher/lower sun and
warmer/cooler weather every year.

8 0

2 years ago

Rama's weight is 4okg: She is carrying a load of 20kg up to a height of 20 meters.what work does she do?Also mention its type of

Maslowich

Answer:

rama is doing

Explanation:

work done=f×d×g

=60×20×9.8

=11760j

she is doing work against gravity

mark me

8 0

2 years ago

Example of optical object

Alik [6]

Answer:

Telescope,cameras

Hope it helped

5 0

2 years ago

Read 2 more answers

A 10 kg bowling ball sits at the top of a 10 m hill and then slides down its icy hillside.

Nataly_w [17]

We need more informtion

5 0

3 years ago

Read 2 more answers

Aloop of wire of area 71 cm^2 is placed with its plane parallel to a 16 mt magnetic field. the loop is then rotated so that its

kkurt [141]

Answer:

Approximately 1.62 × 10⁻⁴ V.

Explanation:

The average EMF in the coil is equal to

$\displaystyle \frac{\text{Final Magnetic Flux} - \text{Initial Magnetic Flux}}{2}$ ,

Why does this formula work?

By Faraday's Law of Induction, the EMF $\epsilon$ induced in a coil (one loop) is equal to the rate of change in the magnetic flux $\Phi$ through the coil.

$\displaystyle \epsilon(t) = \frac{d}{dt}(\Phi(t))$ .

Finding the average EMF in the coil is similar to finding the average velocity.

$\displaystyle \text{Average}\; \epsilon = \frac{1}{t}\int_0^t \epsilon(t)\cdot dt$ .

However, by the Fundamental Theorem of Calculus, integration reverts the action of differentiation. That is:

$\displaystyle \int_0^{t} \epsilon(t)\cdot dt = \int_0^{t} \frac{d}{dt}\Phi(t)\cdot dt = \Phi(t) - \Phi(0)$ .

Hence the equation

$\displaystyle \text{Average}\; \epsilon = \frac{1}{t}\int_0^t \epsilon(t)\cdot dt = \frac{\Phi(t)- \Phi(0)}{t}$ .

Note that information about the constant term in the original function will be lost. However, since this integral is a definite one, the constant term in $\Phi(t)$ won't matter.

Apply this formula to this question. Note that $\Phi$ , the magnetic flux through the coil, can be calculated with the equation

$\Phi = B \cdot A \cdot N \; \sin{\theta}$ .

For this question,

$B = \rm 16\; mT = 16\times 10^{-3}\; T$ is the strength of the magnetic field.
$A = \rm 71\; cm^{2} = 71\times \left(10^{-2}\right)^2 \; m^{2}$ is the area of the coil.
$N = 1$ is the number of loops in the coil.
$\theta$ is the angle between the field lines and the coil.
At $\rm 0\;s$ , the field lines are parallel to the coil, $\theta = 0^{\circ}$ .
At $\rm 0.7\; s$ , the field lines are perpendicular to the coil, $\displaystyle \theta = 90^{\circ}$ .