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Kazeer [188]
2 years ago
14

What is the magnitude of the electric field at the dot in the figure?

Physics
1 answer:
Zielflug [23.3K]2 years ago
8 0

The magnitude of the electric field at the dot is :  10⁴ v/m

Given that there are three equipotential lines with equal spacing,we will apply the the relationship between P.D and electric field

<h3>Determine the magnitude of the electric field at the dot </h3>

change in voltage  = E .d

100 - 0 = E * ( 1 * 10⁻² m ) ----- ( 1 )

From equation ( 1 )

The magnitude of E = 100 v / ( 1 * 10⁻² m )

                                 = 10⁴ v/m

Hence we can conclude that The magnitude of the electric field at the dot is :  10⁴ v/m

Learn more about electric field : brainly.com/question/14372859

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Assume that the average mass of each of the approximately 1 billion people in China is 55 kg.Assume that they all gather in one
Harman [31]

Answer: 5.9(10)^{-8} m

Explanation:

The equation to calculate the center of mass C_{M} of a particle system is:

C_{M}=\frac{m_{1}r_{1}+m_{1}r_{1}+...+m_{n}r_{n}}{m_{1}+m_{2}+...+m_{n}}

In this case we can arrange for one dimension, assuming the geometric center of the Earth and the ladder are on a line, and assuming original center of mass located at the Earth's geometric center:

C_{M}=\frac{m_{E}(0 m) + m_{p} r_{E-p}}{m_{E}+m_{p}}

Where:

m_{E}=5.9(10)^{24} kg is the mass of the Earth

m_{p}=55(10)^{9} kg is the mass of 1 billion people

r_{E}=6371000 m is the radius of the Earth

r_{E-p}=6371000 m- 2m=6370998 m is the distance between the center of the Earth and the position of the people (2 m above the Earth's surface)

C_{M}=\frac{m_{p}55(10)^{9} kg (6370998 m)}{5.9(10)^{24} kg+55(10)^{9} kg}

C_{M}=5.9(10)^{-8} m This is the displacement of Earth's center of mass from the original center.

8 0
3 years ago
The ratio of output power to input power, in percent, is called
kow [346]

That ratio is called"efficiency".  It doesn't need to be a percent. 
It can just as well be a fraction or a decimal number.

3 0
2 years ago
The noise floor, also known as additive white Gaussian noise (AWGN), is a continuous noise level that appears over a wide spectr
harkovskaia [24]

Answer:

correct option is a. True

Explanation:

solution

the noise floor is AWGN ( additive white Gaussian noise )  

and when viewed in the frequency domain, it is the continuous noise level  

because as they have a  uniform power over all the frequency.

 

so that it is additive white Gaussian noise  

as we can say given statement is True  

correct option a true  

4 0
3 years ago
For a transverse wave, what is a wavefront?
vichka [17]

Explanation:

A wavefront is the long edge that moves, for example, the crest or the trough. Each point on the wavefront emits a semicircular wave that moves at the propagation speed v. These are drawn at a time t later, so that they have moved a distance s = vt.

4 0
2 years ago
Read 2 more answers
Justin Bieber is thrown horizontally at 10.0 m/s from the top of a cliff 122.5 m high.
sveticcg [70]

===>  Distance fallen from rest in free fall =

                                         (1/2) (acceleration) (time²)

                               (122.5 m) = (1/2) (9.8 m/s²) (time²)

Divide each side by (4.9 m/s²):   (122.5 m / 4.9 m/s²)  =  time²

                                                           (122.5/4.9) s²  =  time²

Take the square root of each side:    5.0 seconds


===> (Accelerating at 9.8 m/s², he will be dropping at
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when he goes 'splat'.  We'll need this number for the last part.)


===> With no air resistance, the horizontal component of velocity
doesn't change.

Horizontal distance = (10 m/s) x (5.0 s)  =  50 meters .

===>  Impact velocity =  (10 m/s horizontally) + (49 m/s vertically)

                                 = √(10² + 49²)  =  50.01 m/s  arctan(10/49)

                                 =    50.01 m/s   at  11.5° from straight down,
                                                           
away from the base of the cliff.  

7 0
3 years ago
Read 2 more answers
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