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Ber [7]
3 years ago
15

Positive and negative charger objects affect which type of field?

Physics
1 answer:
Ilya [14]3 years ago
6 0

Answer:-

<h2>Electric force</h2>

The interaction between two like-charged objects is repulsive. ... Positively charged objects and neutral objects attract each other; and negatively charged objects and neutral objects attract each other.

Explanation:

Hope it is helpful...

You might be interested in
Assuming typical speeds of 8.5 km/s and 5.5 km/s for P and S waves, respectively, how far away did the earthquake occur if a par
taurus [48]

Answer:

The earthquake occurred at a distance of 1122 km

Explanation:

Given;

speed of the P wave, v₁ = 8.5 km/s

speed of the S wave, v₂ =  5.5 km/s

The distance traveled by both waves is the same and it is given as;

Δx = v₁t₁ = v₂t₂

let the time taken by the wave with greater speed = t₁

then, the time taken by the wave with smaller speed, t₂ = t₁ + 1.2 min, since it is slower.

v₁t₁ = v₂t₂

v₁t₁ = v₂(t₁ + 1.2 min)

v₁t₁ = v₂(t₁ + 72 s)

v₁t₁ = v₂t₁ + 72v₂

v₁t₁ - v₂t₁ = 72v₂

t₁(v₁ - v₂) = 72v₂

t_1 = \frac{72v_2}{v_1-v_2}\\\\t_1 =   \frac{72*5.5}{8.5-5.5}\\\\t_1 = 132 \ s

The distance traveled is given by;

Δx = v₁t₁

Δx = (8.5)(132)

Δx = 1122 km

Therefore, the earthquake occurred at a distance of 1122 km

4 0
3 years ago
How are you guys today? I'm just wasting points' to talk.
scoundrel [369]
I’m honestly bored & tiredddd
5 0
3 years ago
Read 2 more answers
A power plant produces 1000 MW to supply a city 40 km away. Current flows from the power plant on a single wire of resistance 0.
nikitadnepr [17]

Answer:

Current = 8696 A

Fraction of power lost = \dfrac{80}{529} = 0.151

Explanation:

Electric power is given by

P=IV

where I is the current and V is the voltage.

I=\dfrac{P}{V}

Using values from the question,

I=\dfrac{1000\times10^6 \text{ W}}{115\times10^3\text{ V}} = 8696 \text{ A}

The power loss is given by

P_\text{loss} = I^2R

where R is the resistance of the wire. From the question, the wire has a resistance of 0.050\Omega per km. Since resistance is proportional to length, the resistance of the wire is

R = 0.050\times40 = 2\Omega

Hence,

P_\text{loss} = \left(\dfrac{200000}{23}\right)^2\times2

The fraction lost = \dfrac{P_\text{loss}}{P}=\left(\dfrac{200000}{23}\right)^2\times2\div (1000\times10^6)=\dfrac{80}{529}=0.151

3 0
3 years ago
If measurements if a gas are 75L and 300 full-scale and then the gas is measured a second time and found to be 50L, describe wha
swat32
The answer is a bit confusing, so I did some research and fount the original question. This is the complete question:

"<span>If measurements of a gas are 75L and 300 kilopascals and then the gas is measured a second time and found to be 50L, describe what had to happen to the pressure (if temperature remained constant). Include which law supports this observation".

From that:

1) Data:

V1 = 75 liter
P1 = 300 kPa

V2 = 50 liter
P2 = x

T = constant.

2) Analysis (physical law)

Boyle's law states that the volume of a fixed amount of gas, at constant temperarature, is inversely related with the pressure.

In mathematical form that statement becomes into this equation:

PV = constant (at constant T)

=> P1V1 = P2V2.

Then, the decreasing of the volumen (compression) is accompanied by an increase of the pressure.

3) So, using Boyle's law with the data given:

300 kPa * 75 liter = x * 50 liter

=> x = 300 kPa * 75 liter / 50 liter = 450 kPa.

Answer: following Boyle's law, when the gas passed from 75 liter to 50 liter the pressure had to increase from 300 kPa to 450 kPa.

</span>
3 0
3 years ago
The speed of a certain electron is 995 km s−1 . If the uncertainty in its momentum is to be reduced to 0.0010 per cent, what unc
Tpy6a [65]

Answer:

The uncertainty in the location that must be tolerated is 1.163 * 10^{-5} m

Explanation:

From the uncertainty Principle,

Δ_{y} Δ_{p} = \frac{h}{2\pi }

The momentum P_{y} = (mass of electron)(speed of electron)

                                = (9.109 * 10^{-31}kg)(995 * 10^{3} m/s)

                                = 9.0638 * 10^{-25}kgm/s

If the uncertainty is reduced to a 0.0010%, then momentum

                              = 9.068 * 10^{-30}kgm/s

Thus the uncertainty in the position would be:

                              Δ_{y} = \frac{h}{2\pi } * \frac{1}{9.068 * 10^{-30} }

                              Δ_{y} \geq  1.163 * 10^{-5}m

5 0
4 years ago
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