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shepuryov [24]
2 years ago
8

15. Three semicircles of radius 1 are constructed on diameter AB of a semicircle of radius 2. The centers of the small semicircl

es divide AB into four line segments of equal length. What is the area of the region that lies inside the large semicircle but outside the small semicircles?

Physics
1 answer:
AnnZ [28]2 years ago
5 0

Answer:

The remaining area = 4.201 m²

Explanation:

Given that

AB=D=  4 m  (R=2 m)

The area of AB semicircle

A=\pi R^2/2

A=2π

Thee area of small semicircle

a=5π/6 + 2√3/4  m²

a=5π/6 + √3/2  m²

So the remaining area = A-  a

                                     = 2π - (5π/6 + √3/2) m²

The remaining area =  2π - (5π/6 + √3/2) m²

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Which of the following best describes a capacitor?
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Answer:

B

Explanation:

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Due to this insulating layer, DC current can not flow through the capacitor.But it allows a voltage to be present across the plates in the form of an electrical charge.

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A noise level of 95 db is ______ than the lowest level at which hearing protection is required (85 db), and your exposure should
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The answer to this question is "Greater than". As per the OSHA or Organizational Health and Health Organization, a noise level of 95 DB or decibels is greater than the lowest level at which hearing protection is required in 85 decibels and the person's exposure should be limited only to six hours or less than of it. 
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The temperature and pressure at the surface of Mars during a Martian spring day were determined to be -50 °C and 900 Pa, respect
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Answer:

T = 273 + (-50) = 273 – 50 = 223 K

R = 188.82 J / kg K for CO2

Density (Martian Atmosphere) = P / RT = 900 / 188.92 x 223 = 900 / 42129.16 = 0.0213 kg / m^{3}

T = 273 +18 = 291 K, R = 287 J / kg k (for air) P = 101.6 k Pa = 101600 Pa

Density (Earth Atmosphere) = P / RT = 101600 / 287 x 291 = 1.216 kg / m^{3}

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For a group class project, students are building model roller coasters. Each roller coaster needs to begin at the top of the fir
abruzzese [7]

Case A :

A .75 kg 65 N/m 1.2 m

m = mass of car = 0.75 kg

k = spring constant of the spring = 65 N/m

h = height of the hill = 1.2 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (65) (0.25)² + (0.75 x 9.8 x 1.2) = (0.5) (0.75) v²

v = 5.4 m/s



Case B :

B .60 kg 35 N/m .9 m

m = mass of car = 0.60 kg

k = spring constant of the spring = 35 N/m

h = height of the hill = 0.9 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (35) (0.25)² + (0.60 x 9.8 x 0.9) = (0.5) (0.60) v²

v = 4.6 m/s




Case C :

C .55 kg 40 N/m 1.1 m

m = mass of car = 0.55 kg

k = spring constant of the spring = 40 N/m

h = height of the hill = 1.1 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

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v = 5.1 m/s




Case D :

D .84 kg 32 N/m .95 m

m = mass of car = 0.84 kg

k = spring constant of the spring = 32 N/m

h = height of the hill = 0.95 m

x = compression of spring = 0.25 m

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Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

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v = 4.6 m/s


hence closest is in case C at 5.1 m/s




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Equation:
v = 4q \div ( {d}^{2} \pi) \: where \: q = flow \\ v = velocity \: (speed) \: and \:  \\ d = diameter \: of \: pipe \: or \: hose \\ and \: \pi = 3.142
we \: can \: only \: assume \:that \\  flow \: (q) \:stays \: same \: since \: it \\  isnt \: impeded \: by \:  anything \\ thus \:it  \: (q)\:  stays \: the \: same \:  \\ so \: 4q \: can \: be \: removed \: from \:  \\ the \: equation
then \: we \: can \: assume \: that \: only \\ v \: and \: d \: change \: leading \:us \: to >  >  \\ (v1 \times {d1}^{2} \pi) = (v2  \times   {d2}^{2}\pi)
both \: \pi \: will \: cancel \: each \: other \: out \:  \\ as \: constants \:since \: one \: is \: on \\ each \: side \: of \: the \:  =

(v1  \times   {d1}^{2}) = (v2 \ \times {d2}^{2}) \\ (7.0 \times   {0.052}^{2}) = (v2  \times   {0.021}^{2}) \\ divide \: both \: sides \: by \:  {0.021}^{2} \\ to \: solve \: for \: v2 >  >
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new velocity coming out of the hose then is
44 ft/sec
4 0
2 years ago
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