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scoray [572]
3 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]3 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]3 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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The amount of force required to stretch or compress the spring is known as the spring force. Its unit is Newton(N). Force is needed to stretch spring is 10.2 N.

<h3>What is spring force?</h3>

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2 years ago
An emf is induced in a conducting loop of wire 1.22 m long as its shape is changed from square to circular. Find the average mag
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Answer:

The induced emf in the loop is 7.35\times 10^{-4}\ V

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In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400
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Complete Question:

In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400 kg). Both cars then slide with locked wheels until the frictional force from the slick road (with a low ?k of 0.15) stops them, at distances dA = 6.1 m and dB = 4.4 m. What are the speeds of (a) car A and (b) car B at the start of the sliding, just after the collision? (c) Assuming that linear momentum is conserved during the collision, find the speed of car B just before the collision.

Answer:

a) Speed of car A at the start of sliding = 4.23 m/s

b) speed of car B at the start of sliding = 3.957 m/s

c) Speed of car B before the collision = 7.28 m/s

Explanation:

NB: The figure is not provided but all the parameters needed to solve the question have been given.

Let the frictional force acting on car A, f_{ra} = \mu mg\\............(1)

Since frictional force is a type of force, we are safe to say f_{ra} = ma.......(2)

Equating (1) and (2)

ma = \mu mg\\a = \mu g\\\mu = 0.15\\a = 0.15 * 9.8 = 1.47 m/s^{2}

a) Speed of A at the start of the sliding

d_{A} = 6.1 m\\Speed of A at the start of sliding, v_{A} = \sqrt{2ad_{A} }\\ v_{A} = \sqrt{2*1.47*6.1 } \\v_{A} = \sqrt{17.934 } \\v_{A} = 4.23 m/s

b) Speed of B at the start of the sliding

d_{A} = 4.4 m\\Speed of A at the start of sliding, v_{B} = \sqrt{2ad_{B} }\\ v_{B} = \sqrt{2*1.47*4.4 } \\v_{B} = \sqrt{12.936 } \\v_{B} = 3.957 m/s

Let the speed of car B before collision = v_{B1}

Momentum of car B before collision = m_{B} v_{B1}

Momentum after collision = m_{A} v_{A} + m_{B} v_{B2}

Applying the law of conservation of momentum:

m_{B} v_{B1}  = m_{A} v_{A} +m_{B} v_{B2}

m_{A} = 1100 kg\\m_{B} = 1400 kg

(1400*v_{B1} ) = (1100 * 4.23) + ( 1400 * 3.957)\\(1400*v_{B1} ) = 10192.8\\v_{B1} = 10192.8/1400\\v_{B1 = 7.28 m/s

3 0
4 years ago
Read 2 more answers
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