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3 years ago

How does a real image differ from a virtual image?

Physics

2 answers:

Snezhnost [94]3 years ago

6 0

Answer:

one contains datta that can take up hard drive or disc space weras a picture takes up space on walls or photo albums

Explanation:

goldfiish [28.3K]3 years ago

3 0

Real image is formed when the rays of light after reflection or refraction actually meet at some point whereas a virtual image is formed when the rays of light after reflection or refraction appear to meet at a point

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A compound is made up of two or more of what?

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For convection to occur,

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3 years ago

1. I drop a penny from the top of the tower at the front of Fort Collins High School and it takes 1.85 seconds to hit the ground

Svetradugi [14.3K]

You have three known variables:

Acceleration - $9.8m/s^2$
Time - $1.85s$
Initial Velocity - $0m/s$

For the first part of your question:

$v = u + at

v = (0)+(9.8)(1.1)

v = 10.78 m/s v=11 m/s (2 significant figures)$

For the second part of your question:

$v=u+at

v=(0)+(9.8)(1.85)
v=18.13 m/s

$

This still needs to be converted to m/h:

$18.13m/s = 18.13\times 3600 metres/h=65628 metres/hour

65628 metres/hour = 65628\div1600 mi/h = 40.7925 mi/h

= 41 mi/hr (2 significant figures)
$

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3 years ago

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A string with a mass density of 3 * 10^-3 kg/m is under a tension of 380 N and is fixed at both ends. One of its resonance frequ

Delvig [45]

Answer:

(a) the fundamental frequency of this string is 65 Hz

(b) the harmonics of the given frequencies are third and fourth respectively.

Explanation:

Given;

mass density of the string, μ = 3 x 10⁻³ kg/m

tension of the string, T = 380 N

resonating frequencies, 195 Hz and 260 N

For the given resonant frequencies;

$195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } ---(1)\\\\260 = \frac{n+1}{2l} \sqrt{\frac{T}{\mu} } ---(2)\\\\divide \ (2) \ by (1)\\\\\frac{260}{195} = \frac{n+1 }{n} \\\\260n = 195(n+1)\\\\260 n = 195 n + 195\\\\260n - 195n = 195\\\\65n = 195\\\\n = \frac{195}{65} \\\\n = 3$

$195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } \\\\195 = \frac{3}{2l}\sqrt{\frac{380}{3\times 10^{-3}} } \\\\l = \frac{3}{2\times 195}\sqrt{\frac{380}{3\times 10^{-3}} }\\\\l = 2.74 \ m$

(a) the fundamental frequency is calculated as;

$f_o = \frac{1}{2l} \sqrt{\frac{T}{\mu} } \\\\f_o = \frac{1}{2\times 2.74} \sqrt{\frac{380}{3\times 10^{-3} } }\\\\f_o = 65 \ Hz$

(b) harmonics of the given frequencies;

the first harmonic (n = 1) = f₀ = 65 Hz

the second harmonic (n = 2) = 2f₀ = 130 Hz

the third harmonic (n = 3) = 3f₀ = 195 Hz

the fourth harmonic (n = 4) = 4f₀ = 260 Hz

Thus, the harmonics of the given frequencies are third and fourth respectively.

7 0

3 years ago