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3 years ago

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You are given two positively charged particles, of equal magnitude, separated by a distance, "d". What will happen to the force

field between the two particles when "d" is doubled?

2 answers:

11111nata11111 [884]3 years ago

7 0

Force becomes 1/4th

<u>Explanation</u>:

Force between two positively charged particles is given by the equation

F = k×q1×q2/r² So, force is inversely proportional to square of r.

Here, k is a constant, q1 and q2 are the charges which are equal and r is the distance between them.

So if the distance is doubled, r becomes 2r and r² becomes (2r)² = 4r²
Force, F then becomes 1/4th of the initial force.

amm18123 years ago

3 0

Answer:

c

Explanation:

because i look up the answer

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yuradex [85]

I think it B I’m not sure

7 0

3 years ago

Read 2 more answers

Find the values of the root mean square translational speed v, molecules in gaseous diatomic oxygen (O2), gaseous carbon dioxide

tiny-mole [99]

Answer:

For diatomic oxygen:V=539.06 m/s

For carbon dia oxide:V=459.71 m/s

For dia atomic hydrogen:V=2156.25 m/s

Explanation:

As we know that

Root mean square velocity V

$V=\sqrt{\dfrac{3RT}{M}}$

Where

R is the gas constant

$R=8.31\ \frac{kg.m^2}{s^2.mol.K}$

T is the temperature (K).

M is the molecular weight.

For diatomic oxygen:

M=32 g/mol

T=273+100 = 373 K

$R=8.31\ \frac{kg.m^2}{s^2.mol.K}$

$V=\sqrt{\dfrac{3RT}{M}}$

$V=\sqrt{\dfrac{3\times 8.31\times 373}{32\times 10^{-3}}}$

V=539.06 m/s

For carbon dia oxide

M=44 g/mol

T=273+100 = 373 K

$R=8.31\ \frac{kg.m^2}{s^2.mol.K}$

$V=\sqrt{\dfrac{3RT}{M}}$

$V=\sqrt{\dfrac{3\times 8.31\times 373}{44\times 10^{-3}}}$

V=459.71 m/s

For dia atomic hydrogen:

M= 2 g/mol

T=273+100 = 373 K

$R=8.31\ \frac{kg.m^2}{s^2.mol.K}$

$V=\sqrt{\dfrac{3RT}{M}}$

$V=\sqrt{\dfrac{3\times 8.31\times 373}{2\times 10^{-3}}}$

V=2156.25 m/s

3 0

4 years ago

If a car accelerated from 5 m/s to 25 m/s in 10 seconds what is it's acceleration?

Alex_Xolod [135]

Answer:

a= (25-5) /10=2m/s^2

............

6 0

3 years ago

A constant electric field of magnitude E = 148 V/m points in the positive x-direction. How much work (in J) does it take to move

DochEvi [55]

Answer:

$W=-2.1405\times 10^9\,J$

Explanation:

Given:

electric field, $E=148\,V.m^{-1}$

charge, $Q=-13\,\mu C=-13\times 10^{-6}\,C$

initial position coordinates, $p1 =(-18,-131)$

final position coordinates, $p2 =(107,76)$

We find the distance through which the charge has been moved:

$d=\sqrt{(x1-x2)^2+(y1-y2)^2}$

Where we have (x1,y1) & (x2,y2) as the initial and final coordinates of the points.

$d=\sqrt{(107-(-81))^2+(76-(-131))^2}$

$d= 279.63\,m$

Now we need the angle through which displacement is made with respect to the direction of electric field.

$tan\,\theta= \frac{y2-y1}{x2-x1}$

$\theta= tan^{-1}[\frac{76-(-131)}{107-(-81)} ]$

$\theta= 47.75^{\circ}$

Now from the relation between the change in potential difference:

$\Delta V= E.d.cos\,\theta$

$\Delta V= 148\times 279.63\times cos\,47.75^{\circ}$

$\Delta V= 27826.06 V$

∵The change in voltage is defined as the work done per unit charge.

∴ $\Delta V=\frac{W}{Q}$

$W=\frac{\Delta V}{Q}$

Putting the respective values

$W=\frac{27826.06 }{-13\times 10^{-6}}$

$W=-2.1405\times 10^9\,J$

3 0

3 years ago

Using local acceleration of gravity data from the Internet, de- termine the weight, in N, of a person whose mass is 80 kg living

Anna007 [38]

Answer

mass of a person is 80 kg

a) Mexico City, Mexico

acceleration due to gravity in mexico = 9.776 m/s²

weight = 80 × 9.776 = 782.08 N

b) Cape Town, South Africa

acceleration due to gravity in cape town = 9.796 m/s²

weight = 80 × 9.796 = 783.68 N

c) Tokyo, Japan

acceleration due to gravity in cape town = 9.798 m/s²

weight = 80 × 9.798 = 783.84 N

d) Chicago, IL

acceleration due to gravity in cape town = 9.803 m/s²

weight = 80 × 9.803 = 784.24 N

e) Copenhagen, Denmark

acceleration due to gravity in cape town = 9.815 m/s²

weight = 80 × 9.815 = 785.2 N

5 0

4 years ago