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masha68 [24]
3 years ago
9

Below is a list of characteristics for electric and magnetic fields. Put an "E" in front of the statement if it applies only to

electric fields. Put an "M" in front of the statement if it applies only to magnetic fields. And put a "B" in front of the statement if it applies to both types of fields.
___ Always form closed loops
___ Point away from positive charges
___ The closer together the lines are, the stronger the field is.
___ End on south poles
___ Point in the direction a positive charge would move
___ Never cross
___ Field decreases as you move away from what's causing it
Physics
1 answer:
Rainbow [258]3 years ago
6 0

_B__ Always form closed loops

__E_ Point away from positive charges  

__B_ The closer together the lines are, the stronger the field is.

_M__ End on south poles

_E__ Point in the direction a positive charge would move

__B_ Never cross

__B_ Field decreases as you move away from what's causing it


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A car travels at a constant velocity of 70 mph for one hour. At the second hour, the car’s velocity was 60 mph. At the third hou
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When the velocity increases, then the acceleration will be positive and when the velocity decrease then the acceleration will be negative.

During the first hour, the velocity was 70 mph and during the seconds hour the velocity was 60 mph. Hence, the velocity decrease in the seconds hour. So, the acceleration will be negative during the second hour.

Now, during the third hour the velocity increases as it is 80 mph. Hence, the acceleration will be positive during the third hour.


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A hydrogen atom that has an electron in the n = 2 state absorbs a photon. What wavelength must the photon possess to send the el
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Answer:

486nm

Explanation:

in order for an electron to transit from one level to another, the wavelength emitted is given by Rydberg Equation which states that

\frac{1}{wavelength}=R.[\frac{1}{n_{f}^{2} } -\frac{1}{n_{i}^{2} }] \\n_{f}=2\\n_{i}=4\\R=Rydberg constant =1.097*10^{7}m^{-1}\\subtitiute \\\frac{1}{wavelength}=1.097*10^{7}[\frac{1}{2^{2} } -\frac{1}{4^{2}}]\\\frac{1}{wavelength}= 1.097*10^{7}*0.1875\\\frac{1}{wavelength}= 2.06*10^{6}\\wavelength=4.86*10{-7}m\\wavelength= 486nm\\

Hence the photon must possess a wavelength of 486nm in order to send the electron to the n=4 state

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3 years ago
Cm<br> 1. Estimate how much water is contained in a round, plastic washing bowl
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Answer:  6-16 Galloons.

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2 years ago
You are traveling in a car toward a hill at a speed of 36.4 mph. The car's horn emits sound waves of frequency 231 Hz, which mov
Marina CMI [18]

Answer:

<em>a. The frequency with which the waves strike the hill is 242.61 Hz</em>

<em>b. The frequency of the reflected sound wave is 254.23 Hz</em>

<em>c. The beat frequency produced by the direct and reflected sound is  </em>

<em>    11.62 Hz</em>

Explanation:

Part A

The car is the source of our sound, and the frequency of the sound wave it emits is given as 231 Hz. The speed of sound given can be used to determine the other frequencies, as expressed below;

f_{1} = f[\frac{v_{s} }{v_{s} -v} ] ..............................1

where f_{1} is the frequency of the wave as it strikes the hill;

f is the frequency of the produced by the horn of the car = 231 Hz;

v_{s} is the speed of sound = 340 m/s;

v is the speed of the car = 36.4 mph

Converting the speed of the car from mph to m/s we have ;

hint (1 mile = 1609 m, 1 hr = 3600 secs)

v = 36.4 mph *\frac{1609 m}{1 mile} *\frac{1 hr}{3600 secs}

v = 16.27 m/s

Substituting into equation 1 we have

f_{1} =  231 Hz (\frac{340 m/s}{340 m/s - 16.27 m/s})

f_{1}  = 242.61 Hz.

Therefore, the frequency which the wave strikes the hill is 242.61 Hz.

Part B

At this point, the hill is the stationary point while the driver is the observer moving towards the hill that is stationary. The frequency of the sound waves reflecting the driver can be obtained using equation 2;

f_{2} = f_{1} [\frac{v_{s}+v }{v_{s} } ]

where f_{2} is the frequency of the reflected sound;

f_{1}  is the frequency which the wave strikes the hill = 242.61 Hz;

v_{s} is the speed of sound = 340 m/s;

v is the speed of the car = 16.27 m/s.

Substituting our values into equation 1 we have;

f_{2} = 242.61 Hz [\frac{340 m/s+16.27 m/s }{340 m/s } ]

f_{2}  = 254.23 Hz.

Therefore, the frequency of the reflected sound is 254.23 Hz.

Part C

The beat frequency is the change in frequency between the frequency of the direct sound  and the reflected sound. This can be obtained as follows;

Δf = f_{2} -  f_{1}  

The parameters as specified in Part A and B;

Δf = 254.23 Hz - 242.61 Hz

Δf  = 11.62 Hz

Therefore the beat frequency produced by the direct and reflected sound is 11.62 Hz

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