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dem82 [27]
3 years ago
11

suppose that you look into a photometer's eyepiece and the fluorescent disks appear to be equal in intensity. If the distance be

tween the photometer to lamp 1 is 400mm, the distance between the photometer to lamp 2 is 200 mm, and the intensity of lamp 2 is known to be 15 candelas, what is the intensity to lamp 1?
Physics
1 answer:
ehidna [41]3 years ago
3 0
Use the Inverse square law, Intensity (I)<span> of a light </span>is inversely proportional to the square of the distance(d).

I=1/(d*d)

Let Intensity for lamp 1 is L1 distance be D1 so on, L2 D2 for Intensity for lamp 2 and its distance.

L1/L2=(D2*D2)/(D1*D1)

L1/15=(200*200)/(400*400)
L1=15*0.25
L1=3.75 <span>candela</span>
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That unit has the special name "<em>Watt</em>".

Your question is not necessarily a question. It's also a statement, You could write

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4 years ago
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matrenka [14]

Answer:

The answer to your question is

Explanation:

Data

mass = 0.5kg

T1 = 35

T2 = ?

Q = - 6.3 x 10⁴ J  = - 63000 J

Cp = 4184 J / kg°C

Formula

                        Q = mCp(T2 - T1)

                         T2 = T1 + Q/mCp    

Substitution

                       T2 = 35 - 63000/(0.5 x 4184)

                        T2 = 35 - 63000/2092

                        T2 = 35 - 30.1

                         T2 = 4.9 °C

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3 years ago
Physics double pivot question​
andriy [413]

Explanation:

Assuming the wall is frictionless, there are four forces acting on the ladder.

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Reaction force pushing up at the foot of the ladder (Rf).

Friction force pushing to the right at the foot of the ladder (Ff).

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Take the sum of the moments about the foot of the ladder.

∑τ = Iα

Rw (3.0 sin 60°) − mg (1.5 cos 60°) = 0

Rw (3.0 sin 60°) = mg (1.5 cos 60°)

Rw = mg / (2 tan 60°)

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Take the sum of the forces in the x direction.

∑F = ma

Ff − Rw = 0

Ff = Rw

Ff = 28 N

(c) State the reaction at the foot of the ladder.

Take the sum of the forces in the y direction.

∑F = ma

Rf − mg = 0

Rf = mg

Rf = 98 N

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