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yan [13]
4 years ago
6

Green light shines through a 100-mm-diameter hole and is observed on a screen.If the hole diameter is increased by 20%, does the

circular spot of light on the screen decrease in diameter, increase in diameter, or stay the same
Physics
1 answer:
Ede4ka [16]4 years ago
3 0

Answer:

For this case we can use the model given by:

W = \frac{K \lambda L}{D}

Where K is a constant and we can see W represent the spot of light and if D increases then W needs to decrease since the other variables \lambda , L remains constant.

Conclusion: The circular spot of light on the screen increase in diameter.

Explanation:

For this case we know that the diameter is:

D = 100 mm

And the hole diameter is increased by 20%, so the final diameter would be:

D_f = 1.2*100=120 mm

And for this case if we use concepts from the theory of rays we know that if the diameter of the hole gets bigger then the hole wuld increase.

For this case we can use the model given by:

W = \frac{K \lambda L}{D}

Where K is a constant and we can see W represent the spot of light and if D increases then W needs to decrease since the other variables \lambda , L remains constant.

Conclusion: The circular spot of light on the screen increase in diameter.

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A team of eight dogs pulls a sled with waxed wood runners on wet snow (mush!). The dogs have average masses of 19.0 kg, and the
liberstina [14]

Answer

given,

number of dog = 8

mass of each dog= 19 Kg

mass of sled = 210 Kg

average force = 185 Nss

a) writing all the horizontal force

   force acting by dog - friction force = (M + 8m) a

   8 F_d - μ m g = (M + 8m) a

assuming coefficient of friction of snow be μ = 0.14

    8 x 185 - 0.14 x 210 x 9.8 = (210 + 8 x 19 ) x a

               a = 3.29 m/s²

b)  the kinetic friction of coefficient is less than static friction

 hence, we can suggest that acceleration of the sled will increase once the sled start to move.

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8 0
3 years ago
A hollow cylinder of mass 2.00 kgkg, inner radius 0.100 mm, and outer radius 0.200 mm is free to rotate without friction around
kipiarov [429]

Answer: 2.86 m

Explanation:

To solve this question, we will use the law of conservation of kinetic and potential energy, which is given by the equation,

ΔPE(i) + ΔKE(i) = ΔPE(f) + ΔKE(f)

In this question, it is safe to say there is no kinetic energy in the initial state, and neither is there potential energy in the end, so we have

mgh + 0 = 0 + KE(f)

To calculate the final kinetic energy, we must consider the energy contributed by the Inertia, so that we then have

mgh = 1/2mv² + 1/2Iw²

To get the inertia of the bodies, we use the formula

I = [m(R1² + R2²) / 2]

I = [2(0.2² + 0.1²) / 2]

I = 0.04 + 0.01

I = 0.05 kgm²

Also, the angular velocity is given by

w = v / R2

w = 4 / (1/5)

w = 20 rad/s

If we then substitute these values in the equation we have,

0.5 * 9.8 * h = (1/2 * 0.5 * 4²) + (1/2 * 0.05 * 20²)

4.9h = 4 + 10

4.9h = 14

h = 14 / 4.9

h = 2.86 m

8 0
3 years ago
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MrMuchimi
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