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Marat540 [252]
3 years ago
11

What is required for an electromagnet to produce a magnetic field that is

Physics
2 answers:
rewona [7]3 years ago
5 0

Answer:if you have different ans your may be a soleno with a current running though it.

Explanation:

Ape.x

jek_recluse [69]3 years ago
3 0

Answer:

A. A coil of wire with current running through it

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A 20 kg object is floating in space. What is its mass?
trapecia [35]

It's still 20 kg. Mass doesn't change according to gravity, only weight changes.

5 0
4 years ago
You are on your balcony and notice some bad squirrels digging in your garden directly below. You start tossing your pistachios a
Rashid [163]

Answer:

0.43 s

Explanation:

We have the following parameters:

Initial velocity, u = 7.4 m/s

Acceleration of gravity, g = 9.8 m/s^2

Distance, s = 43 in + 10 ft = 1.092 m + 3.048 m = 4.14 m

Time, t = ?

Using the equation of motion s=ut +\frac{1}{2}gt^2, we have

4.14 = 7.4t + 0.5\times9.8t^2

4.9t^2 + 7.4t - 4.14 =0

Using the quadratic formula \dfrac{-b\pm\sqrt{b^2-4ac}}{2a} where a = 4.9, b = 7.4 and c = - 4.14, and solving for the positive value of t only, we have

t = 0.43 s

3 0
4 years ago
Electromagnetism consists of what two processes?
kherson [118]

Answer: it’s A and B

Explanation: everyone else on this post was giving you the wrong answer.

5 0
3 years ago
Read 2 more answers
What are the characteristics of the radiation emitted by a blackbody? According to Wien's Law, how many times hotter is an objec
jasenka [17]

Answer:

a) What are the characteristics of the radiation emitted by a blackbody?

The total emitted energy per unit of time and per unit of area depends in its temperature (Stefan-Boltzmann law).

The peak of emission for the spectrum will be displaced to shorter wavelengths as the temperature increase (Wien’s displacement law).

The spectral density energy is related with the temperature and the wavelength (Planck’s law).

b) According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wave length of 450 nm, than a object whose spectrum peaks in the red, at 700 nm?

The object with the blackbody emission spectrum peak in the blue is 1.55 times hotter than the object with the blackbody emission spectrum peak in the red.

Explanation:

A blackbody is an ideal body that absorbs all the thermal radiation that hits its surface, thus becoming an excellent emitter, as these bodies express themselves without light radiation, and therefore they look black.

The radiation of a blackbody depends only on its temperature, thus being independent of its shape, material and internal constitution.

If it is study the behavior of the total energy emitted from a blackbody at different temperatures, it can be seen how as the temperature increases the energy will also increase, this energy emitted by the blackbody is known as spectral radiance and the result of the behavior described previously is Stefan's law:

E = \sigma T^{4}  (1)

Where \sigma is the Stefan-Boltzmann constant and T is the temperature.

The Wien’s displacement law establish how the peak of emission of the spectrum will be displace to shorter wavelengths as the temperature increase (inversely proportional):

\lambda max = \frac{2.898x10^{-3} m. K}{T}   (2)

Planck’s law relate the temperature with the spectral energy density (shape) of the spectrum:

E_{\lambda} = {{8 \pi h c}\over{{\lambda}^5}{(e^{({hc}/{\lambda \kappa T})}-1)}}}  (3)

b) According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wavelength of 450 nm, than a object whose spectrum peaks in the red, at 700 nm?

It is need it to known the temperature of both objects before doing the comparison. That can be done by means of the Wien’s displacement law.

Equation (2) can be rewrite in terms of T:

T = \frac{2.898x10^{-3} m. K}{\lambda max}   (4)

Case for the object with the blackbody emission spectrum peak in the blue:

Before replacing all the values in equation (4), \lambda max (450 nm) will be express in meters:

450 nm . \frac{1m}{1x10^{9} nm}  ⇒ 4.5x10^{-7}m

T = \frac{2.898x10^{-3} m. K}{4.5x10^{-7}m}

T = 6440 K

Case for the object with the blackbody emission spectrum peak in the red:

Following the same approach above:

700 nm . \frac{1m}{1x10^{9} nm}  ⇒ 7x10^{-7}m

T = \frac{2.898x10^{-3} m. K}{7x10^{-7}m}

T = 4140 K

Comparison:

\frac{6440 K}{4140 K} = 1.55

The object with the blackbody emission spectrum peak in the blue is 1.55 times hotter than the object with the blackbody emission spectrum peak in the red.

4 0
3 years ago
The amount of heat energy required to raise the temperature of a unit mass of a material one degree is
charle [14.2K]
The amount of heat energy required to raise the temperature of a unit mass of a material to one degree is called D. its heat capacity.

The relationship of the heat when applied to the object and the change in temperature of the object when heat is being applied is directly proportional to each other. This means that when heat is applied to the object, the temperature of the object increases and when heat is not applied to the object, the temperature of the object decreases.
5 0
3 years ago
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