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grigory [225]
3 years ago
5

A centrifuge in a medical laboratory rotates at an angular speed of 3,500 rev/min, clockwise (when viewed from above). When swit

ched off, it rotates through 46.0 revolutions before coming to rest. Assuming it is constant, what is the magnitude of angular acceleration (in rad/s2)?
Physics
1 answer:
Ratling [72]3 years ago
6 0

Answer:

The magnitude of angular acceleration is 232.38\ rad/s^2.

Explanation:

Given that,

Initial angular velocity, \omega_i=3500\ rev/min=366.5\ rad/s

When it switched off, it comes o rest, \omega_f=0

Number of revolution, \theta=46=289.02\ rad

We need to find the magnitude of angular acceleration. It can be calculated using third equation of rotational kinematics as :

\omega_f^2-\omega_i^2=2\alpha \theta\\\\\alpha =\dfrac{-\omega_i^2}{2\theta}\\\\\alpha =\dfrac{-(366.51)^2}{2\times 289.02}\\\\\alpha =-232.38\ rad/s^2  

So, the magnitude of angular acceleration is 232.38\ rad/s^2. Hence, this is the required solution.

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I uploaded the answer to a file hosting. Here's link:

tinyurl.com/wtjfavyw

7 0
3 years ago
What is the magnitude of the acceleration of a speck of clay on the edge of a potter's wheel turning at 45 rpm (revolutions per
Brilliant_brown [7]

Answer:

a_{cp}=7.77m/s^2

Explanation:

The equation for centripetal acceleration is a_{cp}=\frac{v^2}{r}.

We know the wheel turns at 45 rpm, which means 0.75 revolutions per second (dividing by 60), so our frequency is f=0.75Hz, which is the inverse of the period T.

Our velocity is the relation between the distance traveled and the time taken, so is the relation between the circumference C=2\pi r and the period T, then we have:

v=\frac{C}{T}=2\pi r f

Putting all together:

a_{cp}=\frac{(2\pi r f)^2}{r}=4 \pi^2f^2r=4 \pi^2(0.75Hz)^2(0.35m)=7.77m/s^2

4 0
3 years ago
You are trying to overhear a juicy conversation, but from your distance of 20.0 m , it sounds like only an average whisper of 30
12345 [234]

Answer:

r₂ = 0.2 m

Explanation:

given,

distance = 20 m

sound of average whisper = 30 dB

distance moved closer = ?

new frequency = 80 dB

using formula

\beta = 10 log(\dfrac{I_1}{I_0})

   I₀ = 10⁻¹² W/m²

now,

30 = 10 log(\dfrac{I_1}{10^{-12}})

\dfrac{I_1}{10^{-12}}= 10^3

I_1= 10^{-8}\ W/m^2

to hear the whisper sound = 80 dB

80 = 10 log(\dfrac{I_2}{10^{-12}})

\dfrac{I_2}{10^{-12}}= 10^8

I_2= 10^{-4}\ W/m^2

we know intensity of sound is inversely proportional to square of distances

\dfrac{I_1}{I_2}=\dfrac{r_2^2}{r_1^2}

\dfrac{10^{-8}}{10^{-4}}=\dfrac{r_2^2}{20^2}

10^{-4}=\dfrac{r_2^2}{20^2}

  r₂ = 0.2 m

6 0
3 years ago
A medieval instrument used to determine the position of the sun:
Stels [109]

Answer: astrolabe

Explanation:

4 0
2 years ago
if the resistance of a car headlight is 15 ohm and the current through it is 0.60, what is the voltage across the headlight?
Strike441 [17]

Answer:

9 volts (assuming 0.60 is in Amperes)

Explanation:

Recall that Ohms law can be expressed as

V = IR, where

V = voltage,

I = current (given as 0.6. I'm going to assume that the units is Amperes because it is not given)

R = resistance (given as 15 ohm)

substituting the above values into the formula

V = IR

V = (0.6)(15)

V = 9 Volts

4 0
3 years ago
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