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2 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]2 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]2 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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choli [55]

I think its C or D.

Einstein's Nobel winning photoelectric equation.

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2 years ago

Plz help will give brainliest and ten points

Svetradugi [14.3K]

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2 years ago

A circuit breaker is rated for a current of 25 a rms at a voltage of 240 v rms. (a) what is the largest value of imax that the b

Mademuasel [1]

The largest value of current that the breaker can carry = Imax=35.4A

Explanation:

Rms value of current= Irms= 25 A

The rms current and the maximum current are related as

Imax= √2 Irms

Imax=√2 (25)

Imax=35.4 A

Thus the maximum current carried by the breaker= 35.4 A

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3 years ago

A fisherman is fishing from a bridge and is using a "42.0-N test line." In other words, the line will sustain a maximum force of

lara31 [8.8K]

Answer:

(a) 42 N

(b)36.7 N

Explanation:

Nomenclature

F= force test line (N)

W : fish weight (N)

Problem development

(a) Calculating of weight of the heaviest fish that can be pulled up vertically, when the line is reeled in at constant speed

We apply Newton's first law of equlibrio because the system moves at constant speed:

∑Fy =0

F-W= 0

42N -W =0

W = 42N

(b) Calculating of weight of the heaviest fish that can be pulled up vertically, when the line is reeled with an acceleration whose magnitude is 1.41 m/s²

We apply Newton's second law because the system moves at constant acceleration:

m= W/g , m= W/9.8 , m:fish mass , W: fish weight g:acceleration due to gravity

∑Fy =m*a

m= W/g , m= W/9.8 , m:fish mass , W: fish weight g:acceleration due to gravity

F-W= ( W/9.8 )*a

42-W= ( W/9.8 )*1.41

42= W+0.1439W

42=1.1439W

W= 42/1.1439

W= 36.7 N

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3 years ago

A baseball is thrown straight up from a building that is 25 meters tall with an initial velocity v = 10 m/s. How fast is it goin

Yanka [14]

Answer:-24,5m/s

Explanation: what we have here is a UALM with these gravity as acceleration (-9.8 m/s^2). The initial position is 25 m and initial speed is 10m/s.

Speed and gravity are increasing in the opposite direction, speed upwards and gravity downwards, while the position is also upwards, depending on your reference system.

The first thing I need to know is the maximum high it will reach.

Hmax=- S(0)^2/2g=

S= speed.

0= initial

G= gravity

Hm= 100/19,6= 5.1 m

So, the ball will go 5,1 m higher than the initial position, and from there it will fall free.

Then, I need to know how long it takes to fall. For that we use UALM equation:

X(t)= X(0) + S(0)*t + (A*t^2)/2.

X: position

S: speed

A: acceleration

T:time

0: initial

0 = 25m +10*t -(9.8 * t^2)/2

Solving the quadratic equation we get

T= 3,5 sec. ( Negative value for time is impossible)

So now we know that the ball to go up and then fall needs 3,5 sec.

Let's see how long it takes to go up:

30,1=25+10*t-4,9*t^2

0=-5,1+10*t-4,9*t^2

T= 1 sec. So it will take 1 sec to the ball to reach the maximum high and 0=speed and then it'll fall during the resting 2,5 sec

Finally, to know the speed just before it touches the ground, we use the following formula:

A= (St-S0)/t

-9.8m/s^2 = (St- 0m/s)/ 2,5s

-24,5 m/s= St

-24,5 m/s is the speed at 3,5 sec, which is the time just before falling

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2 years ago