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3 years ago

The illuminance due to a 60.0-w lightbulb at 3.0m is 9.35 lx. What is the total luminous flux of the bulb

1 answer:

3 0

The equation for luminous flux is given as P = 4 $\pi$ $r^{2}$ E
where P is the luminous flux, r is the distance and E is the illumination. The unit for P is lumen, E is lux and r is in meters. Substituting the given to the equation:

P = 4 $\pi$ $r^{2}$ E

P= 4 $\pi$ $(3)^{2}$ (9.35) = 1057.46 lumens (lm)

The total luminous flux is equal to 1057.46 lumens (lm).

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An object is floating in equilibrium on the surface of a liquid. The object is then removed and placed in another container, fil

Anestetic [448]

Answer:

The fraction of its volume inside liquid is increased .

Explanation:

According to principle pf floatation , an object floats on the surface of water

when the weight of liquid displaced by it becomes equal to weight of the object . weight of the liquid depends upon the density of the liquid .

In the second case , when the body is dipped into liquid of lesser density , in order to balance the weight of body , more volume of liquid will be displaced so that weight of displaced liquid becomes equal to object's weight . So the body floats with greater depth inside liquid . The fraction of its volume inside liquid is increased .

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3 years ago

Two charged particles separated by a distance of = 3 and experienced electrostatic forces of = 60 . What would be this force if

klemol [59]

Answer: 539.4 N

Explanation:

Let's begin by explaining that Coulomb's Law establishes the following:

"The electrostatic force $F_{E}$ between two point charges $q_{1}$ and $q_{2}$ is proportional to the product of the charges and inversely proportional to the square of the distance $d$ that separates them, and has the direction of the line that joins them"

What is written above is expressed mathematically as follows:

$F_{E}= K\frac{q_{1}.q_{2}}{d^{2}}$ (1)

Where:

$F_{E}=60 N$ is the electrostatic force

$K=8.99(10)^{9} Nm^{2}/C^{2}$ is the Coulomb's constant

$q_{1}$ and $q_{2}$ are the electric charges

$d=3 m$ is the separation distance between the charges

Then:

$60 N= 8.99(10)^{9} Nm^{2}/C^{2}\frac{q_{1}.q_{2}}{(3 m)^{2}}$ (2)

Isolating $q_{1}$ and $q_{2}$ :

$q_{1}q_{2}=6(10)^{-8} C^{2}$ (3)

Now, if we keep the same charges but we decrease the distance to $d_{1}=1 m$ , (1) is rewritten as:

$F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{6(10)^{-8} C^{2}}{(1 m)^{2}}$ (4)

Then, the new electrostatic force will be:

$F_{E}= 539.4 N$ (5) As we can see, the electrostatic force is increased when we decrease the distance between the charges.

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3 years ago

A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of v0 = 15.0 m/s. The cliff i

ankoles [38]

Answer:

(a) x0 = 0m and y0 = 49.0m

(b) Vox = 15.0m/s Voy = 0m/s

(d) X = 15.0t and y = 49.0 - 4.9t²

(e) t = 3.16s

(f) Vf = 34.4m/s

Explanation:

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3 years ago

Why does it take significantly stronger magnetic and electric field strengths to move the beam of alpha particles compared with

wlad13 [49]

It takes significantly stronger magnetic and electric field strengths to move a beam of alpha particles compared with the beam of electrons(betaparticles) because the charge of an alpha particle is twice stronger than a beta particle. Therefore, more energy is needed to move the alpha particle.

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3 years ago

12345 [234]

The similarity is that they both are types of bonds in molecules.
Ionic bonds are between a metal and a nonmetal.
Covalent bonds are between two nonmetals.

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3 years ago