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Finger [1]
3 years ago
6

Find the linear speed of the bottom of a test tube in a centrifuge if the centripetal acceleration there is 5.4×104 times the ac

celeration of gravity. The distance from the axis of rotation to the bottom of the test tube is 7.2 cm .
Engineering
1 answer:
Luda [366]3 years ago
4 0

Answer: v= 195.2 m/s

Explanation:

The centripetal acceleration acts on the bottom of the test tube trying to take it closer to the axis of rotation, and it is defined by the following expression:

ac = v² / r

We know that ac = 5.4. 10⁴ . g = 5.4.10⁴.9.8 m/s²

Expressing r in m ⇒ r = 0.072 m.

So, the only unknown that remains is v, which is the linear speed of the bottom of the test tube, which we want to find out:

v = √ac. r = √5.4.10⁴.9.8 m/s². 0.072 m = 195.2 m/s

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(a) Aluminum foil used for storing food weighs about 0.3 grams per square inch. How many atoms of aluminum are contained in one
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Answer:

note:

solution is attached due to error in mathematical equation. please find the attachment

3 0
3 years ago
What type of test can show a chemical engineer if a material remains in a system and accumulates or if it moves right through?
UkoKoshka [18]

A chemical engineer can clearly see from this kind of test if a substance stays in a system and builds up or if it just passes through.

<h3>What is a chemical engineer?</h3>
  • Processes for manufacturing chemicals are created and designed by chemical engineers.
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To learn more about chemical engineer, refer to:

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7 0
2 years ago
I = 48 mA, R = 1125 2. What is Vs ?
german
54 volts

Ohms law. E= I x R
6 0
3 years ago
Block A hangs by a cord from spring balance D and is submerged in a liquid C contained in beaker B. The mass of the beaker is 1.
nikitadnepr [17]

Answer:

a)  m_e= 3.05 Kg

b)  \rho=1072.3kg/m^3

c)  m_e= 3.05 Kg

Explanation:

From the question we are told that:

Beaker Mass m_b=1.20

Liquid Mass m_l=1.85

Balance D:

Mass m_d=3.10

Balance E:

Mass m_e=7.50

Volume v=4.15*10^{-3}m^3

a)

Generally the equation for Liquid's density is mathematically given by

m_e=m_b+m_l+(\rho*v)

\rho=\frac{7.50-(1.2+1.85)}{4.15*10^{-3}}

\rho=1072.3kg/m^3

b)

Generally the equation for D's Reading at A pulled is mathematically given by

m_d = mass of block - mass of liquid displaced

m_d=m- (\rho *v )

m=3.10+ (1072.30 *4.15*10^{-3}m^3 )

m=18.10kg

c)

Generally the equation for E's Reading at A pulled is mathematically given by

m_e=m_b+m_l

m_e = 1.20 + 1.85

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6 0
3 years ago
Determine the design moment strength for a W21x73 steel beam with a simple span of 18 ft when lateral bracing for the compressio
SVETLANKA909090 [29]

This question is incomplete, the complete question is;

Determine the design moment strength (ϕMn) for a W21x73 steel beam with a simple span of 18 ft when lateral bracing for the compression flange is provided at the ends only (i.e., Lb = 18 ft). Report the result in kip-ft.

Use Fy=50 ksi and assume Cb=1.0 (if needed).

Answer: the design moment strength for the W21x73 steel beam is 566.25 f-ft

Explanation:

Given that;

section  W 21 x 73 steel beam;

now from the steel table table for this section;

Zx = Sx = 151 in³

also given that; fy = 50 ksi and Cb = 1.0

QMn = 0.9 × Fy × Zx

so we substitute

QMn = 0.9 × 50 × 151

QMn = 6795 k-inch

we know that;

12inch equals 1 foot

so

QMn = 6795 k-inch / 12

QMn = 566.25 f-ft

Therefore the design moment strength for the W21x73 steel beam is 566.25 f-ft

7 0
3 years ago
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