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Alla [95]
1 year ago
8

A point charge q1 is held stationary at the origin. A second charge q2 is placed at point a, and the electric potential energy o

f the pair of charges is +5.4×10−8J. When the second charge is moved to point b, the electric force on the charge does −1.9×10−8J of work.
What is the electric potential energy of the pair of charges when the second charge is at point b?
Physics
1 answer:
Brrunno [24]1 year ago
7 0

The electric potential energy of the pair of charges when the second charge is at point b is 7.3 x 10⁻⁸ J.

<h3>Electric potential energy</h3>

When work is done on a positive test charge to move it from one location to another, potential energy increases and electric potential increases.

The electric potential energy between the charges when the second charge is at point b is calculated as follows;

ΔU = -w

Ui - Uf = w

Uf = Ui - w

where;

Uf is the final potential energy

Ui is the initial potential energy

w is the work done by the force

Uf = 5.4 x 10⁻⁸ J - (-1.9 x 10⁻⁸J)

Uf = 5.4 x 10⁻⁸ J + 1.9 x 10⁻⁸ J

Uf = 7.3 x 10⁻⁸ J

Thus, the electric potential energy of the pair of charges when the second charge is at point b is 7.3 x 10⁻⁸ J.

Learn more about electric potential energy here: brainly.com/question/14306881

#SPJ1

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A) lithium and beryllium

Explanation:

From the given row on the periodic table, only lithium and beryllium will conduct electricity.

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  • The presence of free mobile electrons and in some, ions allows them to carry electric currents.

Metals are generally known to be good conductors of heat and electricity. This is because, metals have a large pool of electrons i.e free mobile electrons. They are electropositive with a large size and readily release their electrons for conduction.

Lithium and Beryllium are in the metallic block on the periodic table.

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Answer:

D

Explanation:

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Multiple-Concept Example 13 presents useful background for this problem. The cheetah is one of the fastest accelerating animals,
Andre45 [30]

Answer:

9241.6 W or 12.39318 hp

Explanation:

u = Initial velocity = 0

v = Final velocity

m = Mass

t = Time taken

Energy

KE=\frac{1}{2}m(v^2-u^2)\\\Rightarrow KE=\frac{1}{2}108(30.4^2-0^2)\\\Rightarrow KE=49904.64\ Joules

Power

P=\frac{KE}{t}\\\Rightarrow P=\frac{49904.64}{5.4}\\\Rightarrow P=9241.6\ W

Converting to hp

1\ W=\frac{1}{745.7}\ hp

\\\Rightarrow 9241.6\ W=\frac{9241.6}{745.7}\ hp=12.39318\ hp

The power developed by the cheetah is 9241.6 W or 12.39318 hp

7 0
3 years ago
A modern compact fluorescent lamp contains 1.4 mg of mercury (Hg). If each mercury atom in the lamp were to emit a single photon
Reika [66]

Answer:

A. 1.64 J

Explanation:

First of all, we need to find how many moles correspond to 1.4 mg of mercury. We have:

n=\frac{m}{M_m}

where

n is the number of moles

m = 1.4 mg = 0.0014 g is the mass of mercury

Mm = 200.6 g/mol is the molar mass of mercury

Substituting, we find

n=\frac{0.0014 g}{200.6 g/mol}=7.0\cdot 10^{-6} mol

Now we have to find the number of atoms contained in this sample of mercury, which is given by:

N=n N_A

where

n is the number of moles

N_A=6.022\cdot 10^{23} mol^{-1} is the Avogadro number

Substituting,

N=(7.0\cdot 10^{-6} mol)(6.022\cdot 10^{23} mol^{-1})=4.22\cdot 10^{18} atoms

The energy emitted by each atom (the energy of one photon) is

E_1 = \frac{hc}{\lambda}

where

h is the Planck constant

c is the speed of light

\lambda=508 nm=5.08\cdot 10^{-7}nm is the wavelength

Substituting,

E_1 = \frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{5.08\cdot 10^{-7} m}=3.92\cdot 10^{-19} J

And so, the total energy emitted by the sample is

E=nE_1 = (4.22\cdot 10^{18} )(3.92\cdot 10^{-19}J)=1.64 J

4 0
3 years ago
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makkiz [27]

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minimum wavelength will occurs at Gamma rays

and maximum wavelength at Radio waves

the list of increasing order of wavelength is as following

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so least to maximum order is

1. Gamma rays

2. X rays

3 Ultraviolet

4 Visible light

5 Infrared waves

6 Radio waves

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