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Archy [21]
3 years ago
14

You can comfortably hold your fingers close beside a candle flame, but not very close above the flame. why? challenge (optional)

: why do candle flames quickly snuff out in gravity-free regions?
Physics
2 answers:
Inessa05 [86]3 years ago
6 0
The candle flame releases hot gases, which directly go in upwards directions. Due to which the air near the flame of the candle is very hot and dense. The particles along with vapour move up. And since the sideways, the air is not very dense and hot, we are able to hold the candle. In anti-gravity region, there will be no density differences and also, the convection process wont occur. So, the candle quickly snuffs off.
vekshin13 years ago
3 0

Candle flame gases go up and not out, so your hand is okay with being next to the flame and not above it. You can also see the gases above the flame if you look. :)

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An aluminum bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. The hole is 40mm in diameter and 10
s2008m [1.1K]

Answer:

<em>1.228 x </em>10^{-6}<em> mm </em>

<em></em>

Explanation:

diameter of aluminium bar D = 40 mm  

diameter of hole d = 30 mm

compressive Load F = 180 kN = 180 x 10^{3} N

modulus of elasticity E = 85 GN/m^2  = 85 x 10^{9} Pa

length of bar L = 600 mm

length of hole = 100 mm

true length of bar = 600 - 100 = 500 mm

area of the bar A = \frac{\pi D^{2} }{4} =  \frac{3.142* 40^{2} }{4} = 1256.8 mm^2

area of hole a = \frac{\pi(D^{2} - d^{2}) }{4} = \frac{3.142*(40^{2} - 30^{2})}{4} = 549.85 mm^2

Total contraction of the bar = \frac{F*L}{AE} + \frac{Fl}{aE}

total contraction = \frac{F}{E} * (\frac{L}{A} +\frac{l}{a})

==> \frac{180*10^{3}}{85*10^{9}} *( \frac{500}{1256.8} + \frac{100}{549.85}) = <em>1.228 x </em>10^{-6}<em> mm </em>

6 0
3 years ago
A group of runners complete a 26.2 mile marathon in 3.4 hours. The distance between the start and finish lines is 12.2 miles. Wh
NNADVOKAT [17]

26.2/3.4 would be the average velocity for the run.

7.7 miles/hr

8 0
3 years ago
Read 2 more answers
If a force of 32000N exerted pressure of 160N/m² , find the area on which the force acts.​
Alex17521 [72]

Answer: 200m^2

Explanation:

160N=32000N/x

x*160N=32000

x=200m^2

4 0
1 year ago
A 500-kg roller coaster car travels with some initial velocity along a track that is 5 m above the ground. The car goes down a s
LUCKY_DIMON [66]

Answer:

12m/s

Explanation:

4 0
3 years ago
You stand on a merry-go-round which is spinning at f = 0:25 revolutions per second. You are R = 200 cm from the center. (a) Find
ivanzaharov [21]

Answer:

(a) ω = 1.57 rad/s

(b) ac = 4.92 m/s²

(c) μs = 0.5

Explanation:

(a)

The angular speed of the merry go-round can be found as follows:

ω = 2πf

where,

ω = angular speed = ?

f = frequency = 0.25 rev/s

Therefore,

ω = (2π)(0.25 rev/s)

<u>ω = 1.57 rad/s </u>

(b)

The centripetal acceleration can be found as:

ac = v²/R

but,

v = Rω

Therefore,

ac = (Rω)²/R

ac = Rω²

therefore,

ac = (2 m)(1.57 rad/s)²

<u>ac = 4.92 m/s² </u>

(c)

In order to avoid slipping the centripetal force must not exceed the frictional force between shoes and floor:

Centripetal Force = Frictional Force

m*ac = μs*R = μs*W

m*ac = μs*mg

ac = μs*g

μs = ac/g

μs = (4.92 m/s²)/(9.8 m/s²)

<u>μs = 0.5</u>

7 0
2 years ago
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