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Kazeer [188]
3 years ago
5

What would be the diameter of the earth if it had the density of a neurton star

Physics
1 answer:
Marat540 [252]3 years ago
5 0
The entire mass of the Earth at neutron star density would fit into a sphere of 305 m in diameter
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Which of these statements best describes an effect of changes in Earth's magnetic field?
Sveta_85 [38]
The only statement that makes sense as to which one describes an effect of changes in Earth's magnetic field is the first one - <span>Southern lights are seen at high altitude places like Alaska.
Normally, Southern lights are seen around the South Pole, which means that it is quite odd for them to be seen in Alaska, which is near the North Pole.</span>
6 0
3 years ago
Read 2 more answers
Which of the following is a form of pollution created when vehicle exhaust interacts with sunlight?
SVEN [57.7K]

Photochemical smog is formed when primary air pollutants interact with sunlight.

Photochemical smog is the result of the reaction between pollutants like  nitrogen oxides (NO), sunlight and volatile organic compound (VOC) in the atmosphere. The sources of NO are car exhaust, coal power plants, factory emissions, etc. This type of smog is also known by the name Los Angeles smog.

Air pollutants are the particles present dissolved in the air, which when inhaled by the organisms can cause serious health issues. These pollutants are :ozone, particulate matter, gaseous oxides, etc. These pollutants majorly affect the respiratory system of the humans.

Therefore, photochemical smog is a form of pollution created when vehicle exhaust interacts with sunlight.

To know more about photochemical smog, here: brainly.com/question/15728274

#SPJ4

8 0
1 year ago
Recall from Chapter 1 that a watt is a unit of en- ergy per unit time, and one watt (W) is equal to one joule per second ( J·s–1
harkovskaia [24]

Answer:

Explanation:

The energy of a photon is given by the equation E_p=h f, where h is the <em>Planck constant</em> and f the frequency of the photon. Thus, N photons of frequency f will give an energy of E_N=N h f.

We also know that frequency and wavelength are related by f=\frac{c}{\lambda}, so we have E_N=\frac{N h c}{\lambda}, where c is the <em>speed of light</em>.

We will want the number of photons, so we can write

N=\frac{\lambda E_N}{h c}

We need to know then how much energy do we have to calculate N. The equation of power is P=E/t, so for the power we have and considering 1 second we can calculate the total energy, and then only consider the 4% of it which will produce light, or better said, the N photons, which means it will be E_N.

Putting this paragraph in equations:

E_N=(\frac{4}{100})E=0.04Pt=(0.04)(100W)(1s)=4J.

And then we can substitute everything in our equation for number of photons, in S.I. and getting the values of constants from tables:

N=\frac{\lambda E_N}{h c}=\frac{(520 \times10^{-9}m) (4J)}{(6.626\times10^{-34}Js) (299792458m/s)}=1.047 \times10^{19}

3 0
3 years ago
The electric field in a region of space increases from 0 to 2150 N/C in 5.00 s. What is the magnitude of the induced magnetic fi
Feliz [49]

To solve this problem we will use the Ampere-Maxwell law, which   describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:

\oint \vec{B}\vec{dl} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}

Where,

B= Magnetic Field

l = length

\mu_0 = Vacuum permeability

\epsilon_0 = Vacuum permittivity

Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to

B(2\pi r) = \mu_0 \epsilon_0 \frac{d(EA)}{dt}

Recall that the speed of light is equivalent to

c^2 = \frac{1}{\mu_0 \epsilon_0}

Then replacing,

B(2\pi r) = \frac{1}{C^2} (\pi r^2) \frac{d(E)}{dt}

B = \frac{r}{2C^2} \frac{dE}{dt}

Our values are given as

dE = 2150N/C

dt = 5s

C = 3*10^8m/s

D = 0.440m \rightarrow r = 0.220m

Replacing we have,

B = \frac{r}{2C^2} \frac{dE}{dt}

B = \frac{0.220}{2(3*10^8)^2} \frac{2150}{5}

B =5.25*10^{-16}T

Therefore the magnetic field around this circular area is B =5.25*10^{-16}T

3 0
3 years ago
A long solenoid that has 1,130 turns uniformly distributed over a length of 0.430 m produces a magnetic field of magnitude 1.00
sweet-ann [11.9K]

Given Information:

Number of turns = N = 1130 turns

Length of solenoid = L = 0.430 m

Magnetic field = B =  1.0x10⁻⁴ T

Required Information:

Current = I = ?

Answer:

I = 0.0302 A

Explanation:

The current flowing in the solenoid winding can be found using

I = BL/μ₀N

Where μ₀ is the permeability of free space, N is the number of turns, B is the magnetic field and L is the length of solenoid

I = 1.0x10⁻⁴*0.430/4πx10⁻⁷ *1130

I = 0.0302 A

or

I = 30.28 mA

6 0
3 years ago
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