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Klio2033 [76]
4 years ago
11

The electric potential, when measured at a point equidistant from two particles that have charges equal in magnitude but of oppo

site sign, isA) equal to the net electric field B) smaller than zero C) equal to zero D) equal to the averages of the two distances times the charge E) larger than zero
Physics
1 answer:
Reika [66]4 years ago
4 0

Answer:

C) equal to zero

Explanation:

Electric potential is calculated by multiplying constant and charge, then dividing it by distance. The location that we want to measure is equidistant from two particles, mean that the distance from both particles is the same(r2=r1). The charges of the particle have equal strength of magnitude but the opposite sign(q2=-q1). The resultant will be:V = kq/r

ΔV= V1 + V2= kq1/r1 + kq2/r2

ΔV= V1 + V2= kq1/r1 + k(-q1)/(r)1

ΔV= kq1/r1 - kq1/r1

ΔV=0

The electric potential equal to zero

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A data table shows the height of a person on his birthday each year for ten years. What is the the dependent variable? Provide e
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dep var = height

it depends on time ... year

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What is the voltage in a circuit if the current is 6 Amps and resistance is 6 ohms
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U=R•I=6•6=36 V

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4 years ago
if 162.4 j of heat are added to a gas contained within a cylinder and the gas expands, doing 87.3 j of work, what is the change
labwork [276]

The changes in internal energy of the system <u>75.1 J.</u>

The internal energy of a thermodynamic system is the total energy contained in it. It is the energy required to create or prepare a system in a given internal state and includes contributions of potential energy and internal kinetic energy.

Calculation:

Internal energy =  162.4 j - work done

                        =  162.4 j - 87.3 j

                         =<u> 75.1 J</u>

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Internal energy, in thermodynamics, is the property or state function that defines the energy of matter in the absence of capillary action or external electric, magnetic, or other fields.

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Learn more about internal energy here:-brainly.com/question/15735187

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6 0
1 year ago
A driver enters a one-lane tunnel at 34.4 m/s. The driver then observes a slow-moving van 154 m ahead travelling (in the same di
navik [9.2K]

Answer:

Ans. B) 22 m/s (the closest to what I have which was 20.16 m/s)

Explanation:

Hi, well, first, we have to find the equations for both, the driver and the van. The first one is moving with constant acceleration (a=-2m/s^2) and the van has no acceletation. Let´s write down both formulas so we can solve this problem.

X(van)=5.65t+154

X(driver)=34.4t+\frac{(-2)t^{2} }{2}

or by rearanging the drivers equation.

X(driver)=34.4t+t^{2}

Now that we have this, let´s equal both equations so we can tell the moment in which both cars crashed.

X(van)=X(driver)

5.65t+154=34.4t-t^{2}

0=t^{2} -(34.4-5.65)t+1540=t^{2} -28.75t+154

To solve this equation we use the following formulas

t=\frac{-b +\sqrt{b^{2}-4ac } }{2a}

t=\frac{-b +\sqrt{b^{2}-4ac } }{2a}

Where a=1; b=-28.75; c=154

So we get:

t=\frac{28.75 +\sqrt{(-28.75)^{2}-4(1)(154) } }{2(1)}=21.63st=\frac{28.75 -\sqrt{(-28.75)^{2}-4(1)(154) } }{2(1)}=7.12s

At this point, both answers could seem possible, but let´s find the speed of the driver and see if one of them seems ilogic.

V(driver)=V_{0} +at}

V(driver)=34.4\frac{m}{s} -2\frac{m}{s^{2} } *(7.12s)=20.16\frac{m}{s}V(driver)=34.4\frac{m}{s} -2\frac{m}{s^{2} } *(21.63s)=-8.86\frac{m}{s}

This means that 21.63s will outcome into a negative speed, for that reason we will not use the value of 21.63s, we use 7.12s and if so, the speed of the driver when he/she hits the van is 20.16m/s, which is closer to answer  A).

Best of luck

8 0
3 years ago
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