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tankabanditka [31]
3 years ago
9

An archerfish, peering from just below the water surface, sees a grasshopper standing on a tree branch that's just above the wat

er. The archerfish spits a drop of water at the grasshopper and knocks it into the water. The grasshopper's initial position is 0.45 m above the water surface and 0.25 m horizontally away from the fish's mouth. The launch angle of the drop is 63° with respect to the water surface. Anticipating that the grasshopper would do this, the archerfish decides to spit the water at a higher speed but at the same launch angle (63°). At what speed should the archerfish spit the water for the drop to hit the grasshopper?
Physics
1 answer:
Mariana [72]3 years ago
4 0

Answer:

Explanation:

let the velocity of throw is u .

Time to reach horizontally .25 m is equal to reach .45 m vertically

t = .25 / u cos63

= .55 / u

For vertical motion

- h = - ut + 1/2 gt²

-.45 = -usin63 x .55 / u + .5 x9.8 (.55 / u )²

-.45 = - .49 + 1.48 / u²

.04 = 1.48 / u²

u² = 37

u = 6.08  m /s

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In the areas called pauses such as the tropopause, the temperature line is vertical. What does this indicate about the temperatu
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What will happen to an object's wavelength as the object moves towards you?
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There are 5510 lines per centimeter in a grating that is used with light whose wavelegth is 467 nm. A flat observation screen is
Mademuasel [1]

Answer:

1.696 nm

Explanation:

For a diffraction grating, dsinθ = mλ where d = number of lines per metre of grating = 5510 lines per cm = 551000 lines per metre and λ = wavelength of light = 467 nm = 467 × 10⁻⁹ m. For a principal maximum, m = 1. So,

dsinθ = mλ = (1)λ = λ

dsinθ = λ

sinθ = λ/d.

Also tanθ = w/D where w = distance of center of screen to principal maximum and D = distance of grating to screen = 1.03 m

From trig ratios 1 + cot²θ = cosec²θ

1 + (1/tan²θ) = 1/(sin²θ)

substituting the values of sinθ and tanθ we have

1 + (D/w)² = (d/λ)²

(D/w)² = (d/λ)² - 1

(w/D)² = 1/[(d/λ)² - 1]

(w/D) = 1/√[(d/λ)² - 1]

w = D/√[(d/λ)² - 1] = 1.03 m/√[(551000/467 × 10⁻⁹ )² - 1] = 1.03 m/√[(1179.87 × 10⁹ )² - 1] = 1.03 m/1179.87 × 10⁹  = 0.000848 × 10⁻⁹ = 0.848 × 10⁻¹² m = 0.848 nm.

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So for both principal maxima to be on the screen, its minimum width must be 2w = 2 × 0.848 nm = 1.696 nm

So, the minimum width of the screen must be 1.696 nm

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3 years ago
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