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aniked [119]
2 years ago
14

A spider spins a web with silk threads of density 1300 kg/m3 and diameter 3.0 μm . a typical tension in the radial threads of su

ch a web is 7.0 mn. suppose a fly hits this web. part a which will reach the spider first: the very slight sound of the impact or the disturbance traveling along the radial thread of the web?
Physics
1 answer:
Tema [17]2 years ago
5 0

Answer:

Explanation:

The velocity of a wave in a string is equal to:

v = √(T / (m/L))

where T is the tension and m/L is the mass per length.

To find the mass per length, we need to find the cross-sectional area of the thread.

A = πr² = π/4 d²

A = π (3.0×10⁻⁶ m)²

A = 2.83×10⁻¹¹ m²

So the mass per length is:

m/L = ρA

m/L = (1300 kg/m³) (2.83×10⁻¹¹ m²)

m/L = 3.68×10⁻⁸ kg/m

So the wave velocity is:

v = √(T / (m/L))

v = √(7.0×10⁻³ N / (3.68×10⁻⁸ kg/m))

v ≈ 440 m/s

The speed of sound in air at sea level is around 340 m/s.  So the spider will feel the vibration in the thread before it hears the sound.

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You wad up a piece of paper and throw it into the wastebasket. How far will
vitfil [10]

The range of the piece of paper is C) 1.4 m

Explanation:

The motion of the piece of paper is the motion of a projectile, which consists of two separate motions:

- A uniform motion along the horizontal direction, with constant velocity

- A uniformly accelerated motion along the vertical direction, with constant acceleration (the acceleration of gravity, g=9.8 m/s^2)

From the equation of motion, it is possible to find an expression for the range (the total horizontal distance covered) of a projectile, which is given by:

d=\frac{u^2 sin 2\theta}{g}

where

u is the initial velocity

\theta is the angle of projection

g is the acceleration of gravity

For the piece of paper in this problem,

u = 4.3 m/s

\theta=65^{\circ}

Substituting,

d=\frac{(4.3)^2 sin(2\cdot 65^{\circ})}{9.8}=1.45 m \sim 1.4 m

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

6 0
2 years ago
Read 2 more answers
Which type of listening response includes the use of head nods, facial expressions, and short utterances such as "uh-huh" that s
BARSIC [14]
This type of listening response is called back-channel signal. This allows the speaker to know that the listener is attentive or willing to engage a conversation between them. It is shown through short utterances, facial expressions, head nods and others. 
4 0
2 years ago
In a simple model of the hydrogen atom, the electron moves in a circular orbit of radius 0.053nm around a stationary proton. par
Alenkasestr [34]
<span>Here the force that is applied between the electron and proton is centripetal, so equate the two forces to determine the velocity.
 We know charge of the electron which for both Q1 and Q2, e = 1.60 x 10^-19 C
 The Coulombs Constant k = 9.0 x 10^9
 Radius r = 0.053 x 10^-9m = 5.3 x 10^-11 m
  Mass of the Electron = 9.11 x 10^-31
  F = k x Q1 x Q2 / r^2 = m x v^2 / r(centripetal force)
  ke^2 / r^2 = m x v^2 / r => v^2 = ke^2 / m x r
 v^2 = ((1.60 x 10^-19)^2 x 9.0 x 10^9) / (9.11 x 10^-31 x 5.3 x 10^-11 )
 v^2 = 4.77 x 10^12 = 2.18 x 10^6 m/s
 Since one orbit is the distance,
  one orbit = circumference = 2 x pi x r; distance s = v x t.
 v x t = 2 x pi x r => t = (2 x 3.14 x 5.3 x 10^-11) / (2.18 x 10^6)
  t = 33.3 x 10^-11 / 2.18 x 10^6 = 15.27 x 10^-17 s
 Revolutions per sec = 1 / t = 1 / 15.27 x 10^-17 = 6.54 x 10^15 Hz</span>
6 0
2 years ago
The structural diversity of carbon-based molecules is determined by which properties?
Leokris [45]

Explanation:

The structural diversity of carbon-based molecules is determined by following properties:

1. the ability of those bonds to rotate freely,

2.the ability of carbon to form four covalent bonds,

3.the orientation of those bonds in the form of a tetrahedron.

5 0
2 years ago
A vessel, divided into two parts by a partition, contains 4 mol of nitrogen gas at 75°C and 30 bar on one side and 2.5 mol of ar
Elena L [17]

Answer:

assume nitrogen is an ideal gas with cv=5R/2

assume argon is an ideal gas with cv=3R/2

n1=4moles

n2=2.5 moles

t1=75°C   <em>in kelvin</em> t1=75+273

t1=348K

T2=130°C  <em>in kelvin</em> t2=130+273

t2=403K

u=пCVΔT

U(N₂)+U(Argon)=0

<em>putting values:</em>

=>4x(5R/2)x(Tfinal-348)=2.5x(3R/2)x(T final-403)

<em>by simplifying:</em>

Tfinal=363K

6 0
2 years ago
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