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2 years ago

How are the concepts of quantitative and qualitative related?

Physics

1 answer:

Alexeev081 [22]2 years ago

4 0

Answer:

c) The concepts of quantitative and qualitative are not related at all.

Explanation:

Quantitative properties are those which can be compared by its magnitude only

so it is always in relative terms like we can compare them by which one is larger and which one is smaller

Qualitative terms are those which will compare the quality of two like which one is better in terms of given property

so here we can say that quantitative and qualitative terms is not comparable

hence correct answer will be

c) The concepts of quantitative and qualitative are not related at all.

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A LASIK vision-correction system uses a laser that emits 10-ns-long pulses of light, each with 3.0 mJ of energy. The laser beam

lyudmila [28]

P = energy/t = 0.0025/1E-8 = 250000 W
I(ave) = P/A = 250000/(pi*0.425E-3^2) = 4.4056732E11 W/m^2
I(peak) = 2I(ave) = 8.8113463E11 W/m^2
Electric field E = sqrt(I(peak)*Z0) = 1.8219499E7 V/m, where
free-space impedance Z0 = sqrt(µ0/e0) = 376.73031 ohms

7 0

3 years ago

Calculate the magnitude of the electric field inside the solid at a distance of 9.50 cm from the center of the cavity. Express y

WITCHER [35]

Question:

A point charge of -2.14uC is located in the center of a spherical cavity of radius 6.55cm inside an insulating spherical charged solid. The charge density in the solid is 7.35×10−4 C/m^3.

a) Calculate the magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity.

Express your answer using two significant figures.

Answer:

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity $3.65\times 10^5N/C$

Explanation:

A point charge ,q = $-2.14\times 10^{-6} C$ is located in the center of a spherical cavity of radius , $r =6.55\times 10^{-2}$ m inside an insulating spherical charged solid.

The charge density in the solid , d = $7.35 \times 10^{-4}C/m^3.$

Distance from the center of the cavity,R = $9.5\times 10^{-2 }m$

Volume of shell of charge= V = $(\frac{4\pi}{3})[ R^3 - r^3 ]$

Charge on the shell ,Q = $V \times d'$

$Q =(\frac{4\pi}{3})[ R^3 - r^3 ] \times d$

$Q = 4.1888*\times 10^{-4 }[8.57375 - 2.81011 ]\times 7.35\times 10^{-4}$

$Q = 4.1888\times 10^{-4} [5.76364 ] \times 7.35 \times 10^{-4}$

$Q =2.4143 \times 10^{-4} \times 7.35 \times 10^ { -4}$

$Q =1.7745 \times 10^{-6 }C$

Electric field at $9.5\times 10^{-2}$ m due to shell $E1 = \frac{k Q}{R^2}$

E1 = $\frac{ 9 \times 10^9\times 1.7745\times 10^{-6 }}{ 90.25\times 10^{-4}}$

$E1 =1.769\times 10^6 N/C$

Electric field at $9.5\times 10^{-2}$ due to 'q' at center $E2 = \frac{kq}{R^2}$

E2 = $\frac{ - 9 \times 10^9\times 2.14\times 10^{-6 }}{ 90.25\times 10^{-4}}$

$E2 =2.134\times 10^6 N/C$

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity

= E2- E1

$=[ 2.134 - 1.769 ]\times 10^6$

$= 3.65\times 10^5 N/C$

8 0

3 years ago

A net force of 345 N accelerates a boy on a sled at 3.2 m/s^2 . What is combined mass of the sled

Daniel [21]

Answer:

Mass, m = 26.54kg

Explanation:

Net force can be defined as the vector sum of all the forces acting on a body or an object i.e the sum of all forces acting simultaneously on a body or an object.

Mathematically, net force is given by the formula;

$Fnet = Fapp + Fg$

Where;

Fnet is the net force

Fapp is the applied force

Fg is the force due to gravitation

Given the following data;

Net force, Fnet = 345

Acceleration, a = 3.2m/s²

To find mass;

Fnet = Fapp + Fg

Fnet = ma + mg

Fnet = m(a+g)

m = Fnet/(a+g)

We know that acceleration due to gravity, g = 9.8m/s²

Substituting into the equation, we have;

m = 345/(3.2 + 9.8)

m = 345/13

Mass, m = 26.54kg

6 0

2 years ago

What is the basic unit of each type of elements

Lynna [10]

A Atom is the basic unit of each type of element

8 0

3 years ago

A satellite is revolving the earth 4km above the surface.find the orbital velocity of the satellite (R =6400km,g=9.8m/s^2)

Karo-lina-s [1.5K]

Answer:

v ≈ 7900 m/s

Explanation:

centripetal force will equal gravity force

mv²/R = mg

v²/R = g

v² = Rg

v = √(Rg)

v = √(6.4e6(9.8))

v = 7.91959...e+3

v ≈ 7900 m/s

of course, at those velocities and that deep into the atmosphere, the satellite would quickly burn up, slow down, and cause tremendous damage to buildings etc. with the sonic boom shock wave. It would also have to avoid a lot of mountains as 4000 m is not that high.

3 0

3 years ago