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3 years ago

Describe what happens to the electric field lines when two objects with unlike charges are brought near each other.

Physics

2 answers:

VladimirAG [237]3 years ago

8 0

Answer:

Unlike Charges:

The unlike charges attracts each other, as we can say that the two atoms or point charges which posses different properties attracts one another.

Explanation:

The electric field lines:

The electric fields shows the direction, the nature, and the magnitude of the charges been placed there in a given space. As the field lines from the positive point charge(+) goes in the outward direction and the negative charges(-) has the field lines coming towards it or they are directed towards inside the test charge.

As, the different charges interact lines when interact with one another, the lines become more curved and they slide over the two charges. Now, that is also due to the number of charges attracted toward each present on the surface of the two charges.

harina [27]3 years ago

4 0

Hello.

The answer is:

It creates a spark.

Then thespark can sometimes start a fire or other serious problems.

Have a nice day

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1. write the meaning of the following terms:electrostatic,neutral, positively charged, negatively charged, coulomb,microcoulomb,

dybincka [34]

ELECTROSTATIC:

relating to stationary electric charges or fields as opposed to electric currents.

NEUTRAL:

nor negative nor positive/having no charge

POSITIVELY CHARGED:

positive charge occurs when the number of protons exceeds the number of electrons

NEGATIVELY CHARGED:

negative charge occurs when the number of electrons exceeds the number of protons.

COULOMB:

SI unit for electric charge. One coulomb is equal to the amount of charge from a current of one ampere flowing for one second.

MICROCOULOMB:

a unit of electrical charge equal to one millionth of a coulomb.

NANOCOULOMB:

Nanocoulombs are a unit of charge 1,000,000,000 times smaller than Coulomb.

CONSERVATION OF CHARGE:

constancy of the total electric charge in the universe or in any specific chemical or nuclear reaction

QUANTISATION OF CHARGE:

Charge quantization is the principle that the charge of any object is an integer multiple of the elementary charge.

5 0

2 years ago

A car is traveling at 39.7 mi/h on a horizontal highway. The acceleration of gravity is 9.8 m/s 2 . If the coefficient of fricti

777dan777 [17]

Answer:

The minimum distance in which the car will stop is

x=167.38m

Explanation:

$39.7\frac{mi}{h}*\frac{1km}{0.621371mi}*\frac{1000m}{1km}*\frac{1h}{3600s}=17.747\frac{m}{s}$

∑F=m*a

∑F=u*m*g

The force of friction is the same value but in different direction of the force moving the car so it can stop so

$F=m*a\\a=\frac{F}{m}\\a=\frac{u*m*g}{m}\\a=u*g\\a=0.096*-9.8\frac{m}{s^{2} }$

$a=-0.9408 \frac{m}{s^{2}}$

$v_{f}^{2}=v_{o}^{2}+2*a*(x_{f}-x_{o})\\v_{f}=0 \\x_{o}=0\\0=v_{o}^{2}+2*a*x_{f}\\x_{f}=\frac{v_{o}^{2}}{2*a} \\x_{f}=\frac{(-17.747\frac{m}{s})^{2}}{2*(-0.9408)} \\x_{f}=167.38m$

4 0

3 years ago

gAn optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is from the

igomit [66]

Answer:

d = 68.5 x 10⁻⁶ m = 68.5 μm

Explanation:

The complete question is as follows:

An optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is 1.70m from the slits. If the slits are illuminated with coherent light of wavelength 633 nm, how far apart should the slits be?

The answer can be given by using the formula derived from Young's Double Slit Experiment:

$y = \frac{\lambda L}{d}\\\\d =\frac{\lambda L}{y}\\\\$

where,

d = slit separation = ?

λ = wavelength = 633 nm = 6.33 x 10⁻⁷ m

L = distance from screen (detector) = 1.7 m

y = distance between bright fringes = 15.7 mm = 0.0157 m

Therefore,

$d = \frac{(6.33\ x\ 10^{-7}\ m)(1.7\ m)}{0.0157\ m}\\\\$

d = 68.5 x 10⁻⁶ m = 68.5 μm

7 0

3 years ago

1. A soccer ball is kicked horizontally off a cliff with an initial speed of 8 m/s and lands 16 m from the base of

lana66690 [7]

Answer:

Height of cliff = S = 20 m (Approx)

Explanation:

Given:

Initial velocity = 8 m/s

Distance s = 16 m

Starting acceleration (a) = 0

Computation:

s = ut + 1/2a(t)²

16 = 8t

t = 2 sec

Height of cliff = S

Gravitational acceleration = 10 m/s

S = 1/2a(t)²

S = 1/2(10)(2)²

Height of cliff = S = 20 m (Approx)

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3 years ago

The cycle of the moon through its phases or the synodic month is___.

ikadub [295]

It is 29 and a half days long

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3 years ago

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