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Crank
3 years ago
5

Point charge A is located at point A and point charge B is at point B. Points A and B are separated by a distance r. To determin

e the electric potential at the mid-point along a line between points A and B, which of the following mathematical approaches is correct? a. The difference in the absolute value (the sign of the charges does not enter into the calculation) of the two electric potentials is determined at a distance r/2 from each of the charges. b. The algebraic sum of the two electric potentials is determined at a distance r/2 from each of the charges, making sure to include the signs of the charges. c. The electric potential due to each charge is determined at a distance r/2 from each of the charges and an average is taken of the two values. d. The vector sum of the two electric potentials determines the total electric potential at a distance r/2 from each of the charges
Physics
1 answer:
fredd [130]3 years ago
3 0

Answer:

B. The algebraic sum of the two electric potentials is determined at a distance r/2 from each of the charges, making sure to include the signs of the charges.

Explanation:

Total electric potential is the sum of all the electric potential. And because electric potential is a scalar quantity you have to account for the signs.

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Answer:

See attached document

Explanation:

Entire process for deriving the asked expression dV across the bridge as function of dP is illustrated in the attachment below.

The document gives a step-by step process for arriving at the expression. However, manipulation of algebraic equations is skipped for the conciseness of the document.

It also gives the expression for the case when all resistors have different nominal values.

Download docx
6 0
3 years ago
a car accelerates from 4 meters/second to 16 meter/second in 4 seconds. The cars acceleration is how many meter/seconds.​
Allushta [10]

Answer:

=3 metre per second ^2

Explanation:

Formula for acceleration is

V-U÷T

In the given information

V=16

U=4

T=4

Acceleration =16-4/4

=3 metre per second ^2

4 0
3 years ago
Many Amtrak trains can travel at a top speed of 42.0 m/s. Assuming a train maintains that speed for several hours, how many kilo
777dan777 [17]

Answer:

605 km

Explanation:

Hello

the same units of measure should be used, then

Step 1

convert  42 m/s ⇒   km/h

1 km =1000 m

1 h = 36000 sec

42 \frac{m}{s}*\frac{1\ km}{1000\ m}=0.042\ \frac{km}{s}\\ 0.042\ \frac{km}{s}\\

0.042\ \frac{km}{s}*\frac{3600\ s}{1\ h} =151.2 \frac{km}{h}\\ \\Velocity =151.2\ \frac{km}{h}

Step 2

find kilometers traveled after 4  hours

V=\frac{s}{t}\\ \\

V,velocity

s, distance traveled

t. time

now, isolating s

V=\frac{s}{t} \\s=V * t\\

and replacing

s=V * t\\s=151.2\frac{km}{h}*4 hours\\ s=604.8 km\\

S=604.8 Km

Have a great day

4 0
3 years ago
Which change occurs when an atom undergoes alpha decay?
oee [108]
In an alpha decay, an atom emits an alpha particle. An alpha particle consists of 2 protons and 2 neutrons: this means that during this kind of decay, the original atom loses 2 protons and 2 neutrons from its nucleus.

This also means that the atomic number Z of the element (the atomic number is the number of protons in the nucleus) decreases by 2 units in the process, while the mass number A (the mass number is the sum of the number of protons and neutrons) decreases by 4 units. 
3 0
3 years ago
Read 2 more answers
A research submarine can withstand an external pressure of 62 megapascals (million pascals) all the while maintaining a comforta
Alex

Answer:

<em>The depth will be equal to</em> <em>6141.96 m</em>

<em></em>

Explanation:

pressure on the submarine P_{sea} = 62 MPa = 62 x 10^6 Pa

we also know that P_{sea} = ρgh

where

ρ is the density of sea water = 1029 kg/m^3

g is acceleration due to gravity = 9.81 m/s^2

h is the depth below the water that this pressure acts

substituting values, we have

P_{sea} = 1029 x 9.81 x h = 10094.49h

The gauge pressure within the submarine P_{g} = 101 kPa =  101000 Pa

this gauge pressure is balanced by the atmospheric pressure (proportional to 101325 Pa) that acts on the surface of the sea, so it cancels out.

Equating the pressure P_{sea}, we have

62 x 10^6 = 10094.49h

depth h = <em>6141.96 m</em>

7 0
3 years ago
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