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____ [38]
4 years ago
12

A typical laboratory centrifuge rotates at 4100 rpm . test tubes have to be placed into a centrifuge very carefully because of t

he very large accelerations. part a what is the acceleration at the end of a test tube that is 10 cm from the axis of rotation? express your answer with the appropriate units.
Physics
1 answer:
alekssr [168]4 years ago
3 0

The test tube will be subject to centripetal acceleration. This acceleration is given by the following formula

(accel.) = (tangential velocity)^2 / (radius)

a = \frac{v^2}{r}

The velocity of the probe at a distance of 10 cm from the center of the centrifuge, can be calculated using the circumference of the circle:

v = 2\pi r\cdot \frac{rpm}{60} = \omega r

where omega denotes the angular velocity (radians per second). So, combining both:

v = \omega^2 r = (2\pi\cdot\frac{rpm}{60s})^2\cdot r = (2\pi\cdot\frac{4100}{60s})^2\cdot 0.1m = 18434.2 \frac{m}{s^2}

The test tube is subjected to an acceleration of 18434 m/s^2!

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3 years ago
A housefly walking across a clean surface can accumulate a significant positive or negative charge. In one experiment, the large
nalin [4]

Answer:

1681714.28571 N/C

Yes it could exist

Explanation:

m = Mass of housefly = 12 m

q = Charge = 70 pC

g = Acceleration due to gravity = 9.81 m/s²

E = Electric field

When an object accumulates charge it means that it is gaining electrons making it negatively charged. This is the concept of static electricity.

Here, the electric force and the graviational force will balance each other

F_e=W\\\Rightarrow qE=mg\\\Rightarrow E=\frac{mg}{q}\\\Rightarrow E=\frac{12\times 10^{-6}\times 9.81}{70\times 10^{-12}}\\\Rightarrow E=1681714.28571\ N/C

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5 0
3 years ago
8.How long is a day? A year?
prisoha [69]

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3 years ago
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A car has a weight of 25000 N and its brakes can apply a maximum force of 628 N to stop it. The car is initially moving at a spe
svlad2 [7]

Time required : 25.876 s

<h3>Further explanation</h3>

Given

A car weight = 25000 N

Brake force = 628 N

vo= 6.5 m/s

vt = 0(stop)

Required

time to stop

Solution

W = m . g

25000 = m.10

m=2500 kg

Newton's second law

F = m . a

628 = 2500 . a

\tt a=\dfrac{628}{2500}=0.2512~m/s^2

vt=vo-at(deceleration a=-)

0=6.5-0.2512.t

0.2512.t=6.5

t=25.876 s

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Ad libitum [116K]

Answer:

v= 4.8 m/s

Explanation:

m= 2.0

KE= 1/2 mv^2

v= ?

v= 1/2 (2.0)^2

v= 4.8 m/s

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3 years ago
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