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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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At the beginning of a unit on forces, Ms. Alton is leading a class discussion asking her students
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Answer:

(iv), (v), (vi) would be incorrect.

Explanation:

(iv) Force isn't transferred from one colliding object to another, but momentum can be.

(v) An object doesn't stop immediately a force stops acting on it. Think of a thrown ball.

(vi) For an object not to move, it means that the net force on the object is zero, and not necessarily that there are no forces acting on the object. For example, an object could be pushed on one side, and be pushed on the other side with an equal force in the opposite direction. The forces would cancel each other and the net force would be zero.

The rest should be correct.

6 0
3 years ago
4.5 cm3 of water is boiled at atmospheric pressure to become 4048.3 cm3 of steam, also at atmospheric pressure. Calculate the wo
Mrrafil [7]

1. 408.4 J

The work done by a gas is given by:

W=p\Delta V

where

p is the gas pressure

\Delta V is the change in volume of the gas

In this problem,

p=1.01\cdot 10^5 Pa (atmospheric pressure)

\Delta V=4048.3 cm^3 - 4.5 cm^3 =4043.8 cm^3 = 4043.8 \cdot 10^{-6}m^3 is the change in volume

So, the work done is

W=(1.01\cdot 10^5 Pa)(4043.8 \cdot 10^{-6}m^3)=408.4 J

2. 10170 J

The amount of heat added to the water to completely boil it is equal to the latent heat of vaporization:

Q = m \lambda_v

where

m is the mass of the water

\lambda_v = 2.26\cdot 10^6 J/kg is the specific latent heat of vaporization

The initial volume of water is

V_i = 4.5 cm^3 = 4.5\cdot 10^{-6}m^3

and the water density is

\rho = 1000 kg/m^3

So the water mass is

m=\rho V_i = (1000 kg/m^3)(4.5\cdot 10^{-6}m^3)=4.5\cdot 10^{-3}kg

So, the amount of heat added to the water is

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4 years ago
A resistance of 4 ohm is offered by a conductor when a potential difference of 6 V l'd applied across it. Calculate the current
klemol [59]

Answer:

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Explanation:

The formula is V=IR

so now we can derive out the formula which is

I =<u>V</u>

<u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>R

I = <u>6</u>

4

I = 1.5 Ampere

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3 years ago
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defon

Answer: d. availability of places to play

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just olya [345]

The particles released during alpha decay are called alpha particles. Alpha particles have greater mass and charge than other emitted particles.

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Since alpha particles have more mass and charge, they are more ionising and lose more energy at a faster rate and can be blocked easily.

Learn more about alpha decay emissions here:

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