All categories

3 years ago

Can A positively charged body attract another positively charged body

Physics

1 answer:

andriy [413]3 years ago

3 0

Like charges repel, unlike charges attract

Two protons will also tend to repel each other because they both have a positive charge. On the other hand, electrons and protons will be attracted to each other because of their unlike charges.

So I would say no, unless the two bodies are placed close to each other where one has much more charge than the other, then due to induction, force of attraction becomes more than the force of repulsion.

You might be interested in

What's the climate change over the past 200 years

Mandarinka [93]

Answer:

The Spanish philosopher George Santayana wrote, “those who cannot remember the past are condemned to repeat it.” When it comes to climate change, repeating the past is a luxury we can’t afford. If partisan politics continues to derail policy or if denial continues to win over science, it will mean irreversible changes to our planet. Future generations will look at ours as the one that didn’t have the courage to act, rather than the one that recognized the fierce urgency of the moment and met it head on.

With this in mind, we’ve created a climate change timeline highlighting the evolution of science, the intrusion of denial, and the sluggishness of policy over the past 200 years. Let’s learn from the mistakes of the past, so we can make tomorrow a brighter—but not hotter—future.

4 0

3 years ago

Read 2 more answers

A star is known to be moving at 8.46km/s toward the earth. If you observe the spectral line to be at 5.02nm, at what wavelength

anastassius [24]

Answer:

$\lambda_x=5.019858nm$

Explanation:

From the question we are told that

Speed of star $S=8.46km/s$

Distance of spectral line $\lambda_0= 5.02nm$

Generally the equation for wavelength with respect to spectral lines is mathematically given by

$\lambda=\lambda _0 *\frac{v}{c}$

where

$\lambda_0= length\ of\ spectral\ line$

$c=The\ speed\ of\ light$

$v= speed\ of\ moving\ object$

therefore

$\lambda=5.02*10^{-9} *\frac{8.46*10^{12}}{299 792 458*10^9}$

$\lambda=1.42*10^{-4} nm$

Generally the equation for new wavelength is mathematically given as

$\lambda_x=\lambda _0-\lambda$

$\lambda_x=5.02 nm-1.42*10^{-4} nm$

$\lambda_x=5.02-1.42*10^{-4}$

Therefore

$\lambda_x=5.019858nm$

5 0

4 years ago

A street light is at the top of a 13.0 ft. tall pole. A man 6.3 ft tall walks away from the pole with a speed of 3.5 feet/sec al

amm1812

Answer:

$\dfrac{dL}{dt}=5.82 \ ft/s$

Explanation:

given,

street light height = 13 ft

man height = 6.3 ft

speed of the man = 3.5 ft/sec

$\dfrac{H}{L} = \dfrac{h}{l}$

$\dfrac{H}{L} = \dfrac{h}{L-x}$

$\dfrac{L}{H} = \dfrac{L-x}{h}$

hL = H(L-x)

hL = HL-Hx

$L = \dfrac{Hx}{H-h}$

$L = \dfrac{13x}{13-6.3}$

L = 1.94 x

$\dfrac{dL}{dt}=\dfrac{dL}{dx}\dfrac{dx}{dt}$

$\dfrac{dL}{dt}=1.94\times 3$

$\dfrac{dL}{dt}=5.82 \ ft/s$

7 0

3 years ago

Problem: A lossless 50-Ω transmission line is terminated in a load with ZL = (50 + j25) Ω. Use the Smith chart to find the follo

Nadya [2.5K]

Answer: (a). ΓL = 0.246 < 75°

(b). S = 1.7

(c). Zin = (30-j)λ

(d). jreal = Arc Po = 0.105λ

(e). jmax = jreal = 0.105λ

Explanation:

attached is a document to help in understanding.

So we will begin with a step by step analysis of the problem.

from the diagram we have that ZL = (50 + j25) Ω.

where ZL = ZL / Z₀ = 50 + j25 / 50 = 1 + j0.5

so we mark this on the chart as point 'P'

(a) ΓL = mP/m 'P' < Θ L = 1.7/6.9 < 75°

ΓL = 0.246 < 75°

(b) This s-circle 's' is given thus s = r = 1.7 on the RHS of the chart

S = 1.7

ζin = Q = 0.6 - j0.02

therefore Zin = Z₀ζin = 50(0.6 - j0.02) = (30-j)λ

Zin = (30-j)λ

(d) here we are taking R as the diameter opposite of Q on the s=circle

so R = γin = 1.7 + j0.02

yin = yo (γin) = (1.7+j0.02) / 50 = (34 + j0.4)ms

yin = (34 + j0.4)ms

(e) move from 'p' on s-circle to 'o'

where maximum impedance = Znxl = Zos

which gives jreal = Arc Po = 0.105λ

(f) jmax = jreal = 0.105λ

cheers i hope this helps

3 0

3 years ago

Welcome to this IE. You may navigate to any page you've seen already using the IE Outline tab on the right. A particle beam is m

Genrish500 [490]

Answer:

the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m

Explanation:

Given the data in the question;

Kinetic energy of each proton that makes up the beam = 3.25 × 10⁻¹⁵ J

Mass of proton = 1.673 × 10⁻²⁷ kg

Charge of proton = 1.602 × 10⁻¹⁹ C

distance d = 2 m

we know that

Kinetic Energy = Charge of proton × Potential difference ΔV

Potential difference ΔV = Kinetic Energy / Charge of proton

we substitute

Potential difference ΔV = ( 3.25 × 10⁻¹⁵ ) / ( 1.602 × 10⁻¹⁹ )

Potential difference ΔV = 20287.14 V

Now, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m will be;

E = Potential difference ΔV / distance d

we substitute

E = 20287.14 V / 2 m

E = 10143.57 V/m or 1.01 × 10⁴ V/m

Therefore, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m

3 0

3 years ago

Can A positively charged body attract another positively charged body​

Can A positively charged body attract another positively charged body