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scoray [572]
2 years ago
11

You are working in a biology lab and learning to use a new ultracentrifuge for blood tests. The specifications for the centrifug

e say that a red blood cell rotating in the ultracentrifuge moves at 470 m/s and has a radial acceleration of 150,000 g's (that is, 150,000 times 9.8 m/s2). The radius of the centrifuge is 0.15 m. You wonder if this claim is correct.
Physics
2 answers:
Sloan [31]2 years ago
7 0

Answer:

Yes, correct

Explanation:

velocity, v = 470 m/s

radius, r = 0.15 m

The radial acceleration is the centripetal acceleration which always acts towards the centre of the circular centrifuge.

The formula for the centripetal acceleration is given by

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

a =\frac{470^{2}}{0.15}

a = 1472666.667

a = 150272.1 g

According to the question, we can get the acceleration as mentioned. So the claim is correct.

Salsk061 [2.6K]2 years ago
6 0

Answer:

no, the claim is not correct

Explanation:

You are working in a biology lab and learning to use a new ultracentrifuge for blood tests. The specifications for the centrifuge say that a red blood cell rotating in the ultracentrifuge moves at 470 m/s and has a radial acceleration of 150,000 g's (that is, 150,000 times 9.8 m/s2). The radius of the centrifuge is 0.15 m. You wonder if this claim is correct.

centripetal force is the force that is needed to keep an object undergoing a circular motion in a circular path .

a centripetal force is responsible for centripetal acceleration

a=v^2/r

a=470^2/0.15

a=1472666.66m/s^2

for the radial acceleration

150000*9.8

1470000m/s^s

the  two differ by

da=1472666.66m/s^2-1470000m/s^2

da=2666.66m/s^2

therefore the claim is not correct

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A lab cart is loaded with different masses and moved at various velocities
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A lab cart is loaded with different masses and moved at various constant velocities? the anser should be

1.0m/s → 4kg
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3 years ago
A 0.400-kg ice puck, moving east with a speed of 5.86 m/s , has a head-on collision with a 0.900-kg puck initially at rest.
andreev551 [17]

Answer:

a) The final speed of the 0.400-kg puck after the collision is 2.254 meters per second, b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards, c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

Explanation:

a) Since collision is perfectly elastic and there are no external forces exerted on pucks system, the phenomenon must be modelled after the Principles of Momentum and Energy Conservation. Changes in gravitational potential energy can be neglected. That is:

Momentum

m_{1}\cdot v_{1,o} + m_{2}\cdot v_{2,o} = m_{1}\cdot v_{1,f} + m_{2}\cdot v_{2,f}

Energy

\frac{1}{2}\cdot (m_{1}\cdot v_{1,o}^{2}+ m_{2}\cdot v_{2,o}^{2})=\frac{1}{2}\cdot (m_{1}\cdot v_{1,f}^{2}+ m_{2}\cdot v_{2,f}^{2})

m_{1}\cdot v_{1,o}^{2} + m_{2}\cdot v_{2,o}^{2} = m_{1}\cdot v_{1,f}^{2} + m_{2}\cdot v_{2,f}^{2}

Where:

m_{1}, m_{2} - Masses of the 0.400-kg and 0.900-kg pucks, measured in kilograms.

v_{1,o}, v_{2,o} - Initial speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

v_{1}, v_{2} - Final speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

If m_{1} = 0.400\,kg, m_{2} = 0.900\,kg, v_{1,o} = +5.86\,\frac{m}{s}, v_{2,o} = 0\,\frac{m}{s}, the system of equation is simplified as follows:

2.344\,\frac{kg\cdot m}{s} = 0.4\cdot v_{1,f} + 0.9\cdot v_{2,f}

13.736\,J = 0.4\cdot v_{1,f}^{2}+0.9\cdot v_{2,f}^{2}

Let is clear v_{1,f} in first equation:

0.4\cdot v_{1,f} = 2.344 - 0.9\cdot v_{2,f}

v_{1,f} = 5.86-2.25\cdot v_{2,f}

Now, the same variable is substituted in second equation and resulting expression is simplified and solved afterwards:

13.736 = 0.4\cdot (5.86-2.25\cdot v_{2,f})^{2}+0.9\cdot v_{2,f}^{2}

13.736 = 0.4\cdot (34.340-26.37\cdot v_{2,f}+5.063\cdot v_{2,f}^{2})+0.9\cdot v_{2,f}^{2}

13.736 = 13.736-10.548\cdot v_{2,f} +2.925\cdot v_{2,f}^{2}

2.925\cdot v_{2,f}^{2}-10.548\cdot v_{2,f} = 0

2.925\cdot v_{2,f}\cdot (v_{2,f}-3.606) = 0

There are two solutions:

v_{2,f} = 0\,\frac{m}{s} or v_{2,f} = 3.606\,\frac{m}{s}

The first root coincides with the conditions before collision and the second one represents a physically reasonable solution.

Now, the final speed of the 0.400-kg puck is: (v_{2,f} = 3.606\,\frac{m}{s})

v_{1,f} = 5.86-2.25\cdot (3.606)

v_{1,f} = -2.254\,\frac{m}{s}

The final speed of the 0.400-kg puck after the collision is 2.254 meters per second.

b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards.

c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

3 0
3 years ago
Assume you have an ice cube and also a small rock that is the same size and shape as the ice cube. Predict what would happen if
Step2247 [10]

Answer: The ice cube would float on top of the water and the rock would sink to the bottom.

Explanation: The ice cube has a smaller density than the rock which allows the ice cube to float but makes the rock sink to the bottom of the glass of water.

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Part K Find the y-component of the stone's velocity when it is 8.00 m below your hand
Romashka-Z-Leto [24]

The y-component of the stone's velocity when it is 8 m below the hand is 14.86 m / s

v² = u² + 2 a s

s = Displacement

u = Initial velocity

a = Acceleration

u = 8 m / s

s = 8 m

v² = 8² + 2 * 9.8 * 8

v² = 64 + 156.8

v = √ 220.8

v = 14.86 m / s

The equation used to solve the problem is an equation of motion. These equations are designed to locate an object in motion using components such as velocity, displacement, acceleration and time.

Therefore, the y-component of the stone's velocity is 14.86 m / s

To know more about Equations of motion

brainly.com/question/5955789

#SPJ1

4 0
1 year ago
Which of the following best describes the human population from early times to the present?
Makovka662 [10]

Answer:

We have a not significant increase of the population until 1700s or 1800s and then a significant increase growth from these years to the present.

Explanation:

From the figure attached we see the evolution of the human population since early times (1050).

We see that from 1050 until 1750-1850 we have an increase slowly with a low value for the increase per year.

But after these years (1750-1850) we see a considerable increase of the population, like an exponential model.

So then we can conclude in general terms this:

We have a not significant increase of the population until 1700s or 1800s and then a significant increase growth from these years to the present.

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