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3 years ago

A fusible link should be how many wire sizes smaller than the actual circuit wire?

Engineering

1 answer:

V125BC [204]3 years ago

5 0

Answer:

.75mm

Explanation:

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A glass bottle washing facility uses a well agitated hot water bath at 50°C with an open top that is placed on the ground. The b

DochEvi [55]

Answer:

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Explanation:

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3 years ago

State the number of terms for each of the following algebraic expression 2x+1

harina [27]

Answer:

Expressions are made up of terms.

A term is a product of factors.

Coefficient is the numerical factor in the term

Before moving to terms like monomials, binomials, and polynomials, like and unlike terms are discussed.

When terms have the same algebraic factors, they are like terms.

When terms have different algebraic factors, they are unlike terms.

Explanation:

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3 years ago

Where can you go to find the system requirements?

SSSSS [86.1K]

Answer:c

Explanation:

yea, jus go to the systems help center then youll be confronted with a talks. person n they'll walk u threw it.

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1 year ago

Test de evaluare

ivolga24 [154]

C why it’s c bc they I just got it right

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3 years ago

Assume that a person skiing high in the mountains at an altitude of h = 14200 ft takes in the same volume of air with each breat

goldfiish [28.3K]

Answer:

The ratio of the mass of oxygen inhaled for each breath at this high altitude compared to that at sea level is 64.7 %

Explanation:

Mass of air at sea level is given by;

$m_o = \rho_o V_o$

Mass of air at 14,200 ft altitude is given by;

$m_{14.2} = \rho _{14.2} V_{14.2}$

The ratio of the mass of oxygen at high altitude to that at sea level is given by;

$\frac{m_{14.2}}{m_o} = \frac{\rho _{14.2} V_{14.2}}{\rho _oV_{o}}\\\\ Assume \ that \ the \ air \ composition \ (V_{14.2} = V_o)\\\\\frac{m_{14.2}}{m_o} = \frac{\rho _{14.2} }{\rho _o}\\\\$

density of air at sea level, $\rho _o = 0.002378 \ slug/ft^3$

density of air at 10,000 ft = 0.001756 slug/ft³

density of air at 14,200 ft = x

density of air at 15,000 ft, = 0.001496 slug/ft³

Interpolate between 10,000 ft and 15,000 ft

$\frac{14,200 - 10,000}{15,000-10,000} = \frac{X - 0.001756}{0.001496 -0.001756}\\\\ 0.84(-0.00026) = X - 0.001756\\\\-0.0002184 = X - 0.001756\\\\X = 0.001756 - 0.0002184\\\\ X = 0.001538 \ slug/ft^3$

The ratio of the mass of oxygen at 14,200 ft to that at sea level is given by;

$\frac{m_{14.2}}{m_o} = \frac{\rho _{14.2}}{\rho_o} \\\\\frac{m_{14.2}}{m_o} =\frac{0.001538}{0.002378} = 0.647 = 64.7 \%$

Therefore, the ratio of the mass of oxygen inhaled for each breath at this high altitude compared to that at sea level is 64.7 %

6 0

3 years ago