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Gnoma [55]
3 years ago
15

4. How long would it take for a water balloon to fall 39 m if we dropped it starting from rest down an

Physics
1 answer:
Paraphin [41]3 years ago
8 0

Answer:

Around 2.8212 sec

Explanation:

Given the eqn x=1/2at^2+vot

your vo=0

39=1/2(-9.8)t^2

=7.95=t^2

=2.82sec

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Are all elements naturally forming?
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1 through 92 occur naturally on Earth

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What is Snells Law? ...?
butalik [34]
Snell's law<span> (also known as </span>Snell<span>–Descartes </span>law<span> and the </span>law<span> of refraction) is a formula used to describe the relationship between the angles of incidence and refraction, when referring to light or other waves passing through a boundary between two different isotropic media, such as water, glass, or air.


I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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3 0
3 years ago
A soap bubble, when illuminated at normal incidence with light of 463 nm, appears to be especially reflective. If the index of r
Sedbober [7]

Answer:

the minimum thickness the soap film can be if it is surrounded by air is 85.74 nm

Explanation:

Given the data in the question;

wavelength of light; λ = 463 nm = 463 × 10⁻⁹ m

Index of refraction; n = 1.35

Now, the thinnest thickness of the soap film can be determined from the following expression;

t_{min = ( λ / 4n )

so we simply substitute in our given values;

t_{min = ( 463 × 10⁻⁹ m ) / 4(1.35)

t_{min = ( 463 × 10⁻⁹ m ) / 5.4

t_{min = ( 463 × 10⁻⁹ m ) / 4(1.35)

t_{min = 8.574 × 10⁻⁸ m

t_{min = 85.74 × 10⁻⁹ m

t_{min = 85.74 nm

Therefore, the minimum thickness the soap film can be if it is surrounded by air is 85.74 nm

5 0
2 years ago
5–111. A box having a weight of 8 lb is moving around in a circle of radius rA = 2 ft with a speed of (vA)1 = 5 ft&gt;s while co
Elis [28]

Answer:

a) vB = 10.77 ft/s

b) W = 11.30 lb*ft

Explanation:

a) W = 8 lb   ⇒  m = W/g = 8 lb/32.2 ft/s² = 0.2484 slug

vA <em>lin</em> = 5 ft/s

rA = 2 ft

v <em>rad</em> = 4 ft/s

vB = ?

rB = 1 ft

W = ?

We can apply The law of conservation of angular momentum

L<em>in</em> = L<em>fin</em>

m*vA*rA =  m*vB*rB    ⇒    vB = vA*rA / rB

⇒   vB = (5 ft/s)*(2 ft) / (1 ft) = 10 ft/s  (tangential speed)

then we get

vB = √(vB tang² + vB rad²)   ⇒   vB = √((10 ft/s)² + (4 ft/s)²)

⇒   vB = 10.77 ft/s

b) W = ΔK = K<em>B</em> - K<em>A</em> = 0.5*m*vB² - 0.5*m*vA²

⇒     W = 0.5*m*(vB² - vA²) = 0.5*0.2484 slug*((10.77 ft/s)²-(5 ft/s)²)

⇒     W = 11.30 lb*ft

6 0
2 years ago
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