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Arte-miy333 [17]
3 years ago
15

A centrifuge is a device used to separate materials by their masses. A sample in a centrifuge is rotated at high speeds along a

circular path a distance of 10.0 cm from the center of the centrifuge. A centrifuge constantly accelerates a 1.00-gram sample from rest to a speed of 1,000 revolutions per minute in 1.00 seconds. Eventually, the sample reaches a speed 10.5 m/s. What is the magnitude and direction of the centripetal force acting on the sample
Physics
1 answer:
german3 years ago
8 0

Answer:

 F = 1.047 10⁻² N

Explanation:

Let's use kinematics to find the angular acceleration

             w = w₀ + α t

as for rest w₀ = 0

             w = α t

             α = w / t

let's reduce the magnitudes to the SI system

              w = 1000 rev / min (2π rad/ 1 rev) (1 min/ 60s) = 104.72 rad / s

              m = 1.00 g (1 kg / 1000 g) = 1,000 10⁻³ kg

              r = 10.0 cm (1 m / 100 cm) = 0.100 m

let's calculate

              α = 104.72 / 1

              α = 104.72 rad / s²

angular and linear variables are related

               a = α  r

               a = 104.72 0.100

               a = 10.47 m / s²

finally we substitute in Newton's second law

               F = 1 10⁻³ 10.47

               F = 1.047 10⁻² N

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8 0
3 years ago
Andy is waiting at the signal. As soon as the light turns green, he accelerates his car at a uniform rate of 8.00 meters/second2
lbvjy [14]
-- The car starts from rest, and goes 8 m/s faster every second.

-- After 30 seconds, it's going (30 x 8) = 240 m/s.

-- Its average speed during that 30 sec is  (1/2) (0 + 240) = 120 m/s

-- Distance covered in 30 sec at an average speed of 120 m/s 

                                                                           = <span> 3,600 meters .</span>
___________________________________

The formula that has all of this in it is the formula for 
distance covered when accelerating from rest:

       Distance = (1/2) · (acceleration) · (time)²

                       = (1/2) ·      (8 m/s²)     · (30 sec)²

                       =      (4 m/s²)          ·      (900 sec²)

                       =            3600 meters.

_________________________________

When you translate these numbers into units for which
we have an intuitive feeling, you find that this problem is 
quite bogus, but entertaining nonetheless.

When the light turns green, Andy mashes the pedal to the metal
and covers almost 2.25 miles in 30 seconds. 

How does he do that ?

By accelerating at 8 m/s².  That's about 0.82 G  !

He does zero to 60 mph in 3.4 seconds, and at the end
of the 30 seconds, he's moving at 534 mph !  

He doesn't need to worry about getting a speeding ticket.
Police cars and helicopters can't go that fast, and his local
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5 0
3 years ago
Which best explains how the diffraction pattern observed in young’s experiment supports the wave theory of light?.
KIM [24]

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6 0
2 years ago
Determine the angular speed, in rad/s of:
ryzh [129]

Answer:

Explanation:

A. The earth about its axis:

The earth makes one revolution in 24 hours. to know the number of revolutions per second it makes, we need to convert hours to seconds and the revolution to rad.

\frac{1 rev}{24hours}\times \frac{1 hr}{3600s}\times \frac{2\pi}{1}= 7.27 \times10^-5 rad/s

B. The minute hand of the clock makes one revolution in 60 minutes

To convert this to rad per second, we have

\frac{1 rev}{60mins}\times \frac{1 min}{60s}\times \frac{2\pi}{1}= 1.745 \times10^-3rad/s

C. The hour hand of  a clock completes one revolution in 12 hours

\frac{1 rev}{12hours}\times \frac{1 hr}{3600s}\times \frac{2\pi}{1}= 1.454 \times10^-4 rad/s

D. an egg beater turning at 300 rpm.

\frac{300 rev}{1minute}\times \frac{1 min}{60s}\times \frac{2\pi}{1}= 7.27 \times10^-5 rad/s=0.0218rad/s

6 0
3 years ago
A popular car stereo has four speakers, each rated at 60 W. In answering the following questions, assume that the speakers produ
Fantom [35]

Answer:

Explanation:

Intensity of sound = sound energy emitted by source / 4 π d² , where d is distance of the source .

A )

Intensity of sound at 1 m distance = 60 /4 π d²

d = 1 m

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B )

Intensity of sound at 1.5 m distance = 60 /4 π d²

d = 1.5  m

Intensity of sound at 1 m distance = 60 /(4 π 1.5²)

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C )

Intensity of sound due to 4 speakers at 1.5 m distance = 4 x 60 /4 π d²

d = 1.5  m

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D )

Intensity of sound due to .06 W speaker must be 10⁻¹² W s ⁻² . Let the distance be d .

.06 /4 π d² = 10⁻¹²

d² = .06 /4 π 10⁻¹²

d = 6.9 x 10⁴ m .

7 0
3 years ago
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