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Alexandra [31]
3 years ago
9

You are drinking from a standard (10 oz) coffee cup when a fly starts buzzing around your hands. After finishing your beverage,

you quickly trap the fly inside the cup. You then notice another fly is buzzing nearby and you deftly manage to trap the second fly in the cup without releasing the first fly. If you were able to continue trapping flies until the cup was full, estimate how many flies you could trap in the cup.
Physics
1 answer:
geniusboy [140]3 years ago
6 0
I would estimate over about 25 flies
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Heat transfers energy from a hot object to a cold object. Both objects are isolated from their surroundings. The change in entro
aniked [119]

To develop this problem we will start from the definition of entropy as a function of total heat, temperature. This definition is mathematically described as

S = \frac{Q}{T}

Here,

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\Delta S = \frac{Q}{T_{cold}}-\frac{Q}{T_{hot}}

From this relationship we can realize that the change in entropy by the second law of thermodynamics will be positive. Therefore the temperature in the hot body will be higher than that of the cold body, this implies that this term will be smaller than the first, and in other words it would imply that the magnitude of the entropy 'of the hot body' will always be less than the entropy 'cold body'

Change in entropy \Delta S_{hot} is smaller than \Delta S_{cold}

Therefore the correct answer is C. Will always have a smaller magnitude than the change in entropy of the cold object

5 0
2 years ago
Two identical 9.10-g metal spheres (small enough to be treated as particles) are hung from separate 300-mm strings attached to t
Musya8 [376]

Answer:

n = 1.266\times 10^{12}

Explanation:

Given data:

mass of sphere is 10 g

Angle between string and vertical axis is \theta = 13 degree

thickness of string  300 mm = 0.3 m

sin\theta =\frac{2}{0.3 m}

r =0.3 sin 13 = 0.067 m

Fe = \frac{ kq_1 q-2}{d^2}

Fe = \frac{kq^2}{(2r)^2} = mg tan\theta

q^2 =  mg tan\theta \frac{(2r)^2}{k}

    = 0.0091 \times 9.8 tan13 \times \frac{(2\times 0.067)^2}{9\times 10^9}

q^2 = 4.10\times 10^{-14}

q = 2.026 \times 10^{-7} C

q = ne

n = \frac{1.6\times 10^{-19}}{2.02\times 10^{-7}}

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3 0
2 years ago
Calculate the kinetic energy in joules of a 1500 kg automobile moving at 19 m/s
Svetlanka [38]
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5 0
3 years ago
A block of mass 0.1 kg is attached to a spring of spring constant 21 N/m on a frictionless track. The block moves in simple harm
bogdanovich [222]

Answer:

A) 2.75 m/s  B) 0.1911 m    C) 0.109 s

Explanation:

mass of block = M =0.1 kg

spring constant = k = 21 N/m

amplitude = A = 0.19 m

mass of bullet = m = 1.45 g = 0.00145 kg

velocity of bullet = vᵇ = 68 m/s

as we know:

Angular frequency of S.H.M = ω₀ = \sqrt\frac{k}{M}

                                                       = \sqrt\frac{21}{0.1}

                                                       = 14.49 rad/sec

<h3>A) Speed of the block immediately before the collision:</h3>

displacement of Simple Harmonic  Motion is given as:

                                x = A sin (\omega t + \phi)\\

Differentiating this to find speed of the block immediately before the collision:

                    v=\frac{dx}{dt}= A\omega_{o} cos (\omega_{o}t =\phi}\\

As bullet strikes at equilibrium position so,

                                  φ = 0

                                   t= 2nπ

                             ⇒ cos (ω₀t + φ) = 1

                             ⇒ v= A\omega_{o}

                                       v=(.19)(14.49)\\v= 2.75 ms^{-1}

<h3>B) If the simple harmonic motion after the collision is described by x = B sin(ωt + φ), new amplitude B:</h3>

S.H.M after collision is given as :

                              x= Bsin(\omega t + \phi)

To find B, consider law of conservation of energy

K.E = P.E\\K.E= \frac{1}{2}(m+M)v^{2}  \\P.E = \frac{1}{2} kB^{2}

\frac{m+M}{k} v^{2} = B^{2} \\B =\sqrt\frac{m+M}{k} v\\B = \sqrt\frac{.00145+0.1}{21} (2.75)\\B = .1911m

<h3>C) Time taken by the block to reach maximum amplitude after the collision:</h3>

Time period S.H.M is given as:

T=2\pi \sqrt\frac{m}{k}\\ for given case\\m= m=M\\then\\T=2\pi \sqrt\frac{m+M}{k}

Collision occurred at equilibrium position so time taken by block to reach maximum amplitude is equal to one fourth of total time period

T=\frac{\pi }{2}\sqrt\frac{m+M}{k} \\T=0.109 sec

5 0
3 years ago
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