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4 years ago

Which statement would most likely be found in an advertisement from a cell phone provider

Physics

1 answer:

ki77a [65]4 years ago

3 0

Answer:B

Explanation:

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How does radiation transfer thermal energy from the Sun to Earth?

Svetradugi [14.3K]

Answer:

The thermal energy is carried by electromagnetic waves

Explanation:

There are three types of transfer of heat (thermal energy):

- Conduction: conduction occurs when two objects/two substances are in contact with each other. The heat is transferred from the hotter object to the colder object by the collisions between the molecules of the two mediums.

- Convection: convection occurs when a fluid is heated by an external source of heat. The part of the fluid closer to the heat source gets warmer, therefore it becomes less dense and it rises, and it is replaced by the colder part of the fluid, which is colder. Then, this part of fluid is heated as well, so it gets warmer, it rises, etc.. in a cycle.

- Radiation: radiation occurs when thermal energy is carried by electromagnetic waves. Since electromagnetic waves do not need a medium to propagate, this is the only method of heat transfer that can occur through a vacuum (so, in space as well).

Indeed, the Sun emits a lot of electromagnetic radiation, which travels through space and eventually reaches the Earth, heating it.

4 0

3 years ago

Read 2 more answers

A toaster oven indicates that it operates at 1500 W on a 110 V

Mamont248 [21]

$P=U.I => I=\frac{P}{U}=\frac{1500}{110}=\frac{150}{11}\\I=\frac{U}{R}=> R=\frac{U}{I} = \frac{110}{\frac{150}{11} }=8.06< ohm>$

The answer is: A. 8.06 ohm

ok done. Thank to me :>

8 0

3 years ago

A particle initially located at the origin has an acceleration of a=3jm/s^2 and an initial velocity of Vi=5im/s.

-BARSIC- [3]

Given the particle's acceleration is

$\vec a(t) = \left(3\dfrac{\rm m}{\mathrm s^2}\right)\vec\jmath$

with initial velocity

$\vec v(0) = \left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath$

and starting at the origin, so that

$\vec r(0) = \vec 0$

you can compute the velocity and position functions by applying the fundamental theorem of calculus:

$\vec v(t) = \vec v(0) + \displaystyle \int_0^t \vec a(u)\,\mathrm du$

$\vec r(t) = \vec r(0) + \displaystyle \int_0^t \vec v(u)\,\mathrm du$

We have

• velocity at time <em>t</em> :

$\vec v(t) = \left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath + \displaystyle \int_0^t \left(3\dfrac{\rm m}{\mathrm s^2}\right)\,\vec\jmath\,\mathrm du \\\\ \vec v(t) = \left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath + \left(3\dfrac{\rm m}{\mathrm s^2}\right)t\,\vec\jmath \\\\ \boxed{\vec v(t) = \left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath + \left(3\dfrac{\rm m}{\mathrm s^2}\right)t\,\vec\jmath}$

• position at time <em>t</em> :

$\vec r(t) = \displaystyle \int_0^t \left(\left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath + \left(3\dfrac{\rm m}{\mathrm s^2}\right)u\,\vec\jmath\right) \,\mathrm du \\\\ \boxed{\vec r(t) = \left(5\dfrac{\rm m}{\rm s}\right)t\,\vec\imath + \frac12 \left(3\frac{\rm m}{\mathrm s^2}\right)t^2\,\vec\jmath}$

8 0

3 years ago

If the bar magnet is flipped over and the south pole is brought near the hanging ball, the ball will be?

disa [49]

The ball may attracted to the magnet.

<h3>How can we understand that the hanging ball will be attracted to the magnet or not?</h3>

From the question, we understand that the ball is attracted by the north pole of the bar magnet, then the bar magnet flipped over and the south pole is brought near the hanging ball.
As we know, in this type of experiments of bar magnet most of the times the ball is made out of steel.
Steel is a magnetic material.
Magnetic materials gets attracted to the magnet at both the North and South pole.
This can be compared to how neutral objects also gets attracted to the positively and negatively charged rods through the Polarization force.

So, If the bar magnet is flipped over and the south pole is brought near the hanging ball, The ball will be attracted to the magnet.

Learn more about the bar magnet:

brainly.com/question/27943723

#SPJ4

7 0

2 years ago

An object of mass m is dropped from height h above a planet of mass M and radius R .

Margarita [4]

Answer:

$v = \sqrt{2GM(\frac{1}{R+h)}-\frac{1}{R})}$