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3 years ago

9

Which one of the following formulas is used to find the voltage of a circuit?

1 answer:

Sedbober [7]3 years ago

8 0

D , since Voltage is one joule per coulomb

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Which type of galaxy is too small for gravity to form it into an easily describable shape?

oksano4ka [1.4K]

If you think about it, irregular galaxies dont really have a describable shape, as you can tell by the name. So B.

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3 years ago

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( 8c5p79) A certain force gives mass m1 an acceleration of 13.5 m/s2 and mass m2 an acceleration of 3.5 m/s2. What acceleration

Talja [164]

Answer:

$4.725 m/s^{2}$

Explanation:

We know that from Newton's second law of motion, F=ma hence making acceleration the subject then $a=\frac {F}{m}$ where a is acceleration, F is force and m is mass

Also making mass the subject of the formula $m=\frac {F}{a}$

For $m1= \frac {F}{13.5}$ and $m2=\frac {F}{3.5}$ hence $F=(m2-m1)a= (\frac {F}{3.5}-\frac {F}{13.5})a=0.2116402116\\\frac {1}{a}=0.2116402116\\a=4.725 m/s^{2}$

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3 years ago

Which lists types of materials from most conductive to least conductive?

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Superconductor, conductor, semiconductor, insulator

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3 years ago

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3000 N is exerted for 4.0 seconds on a 9500 kg object.<br><br> What is the change in momentum?

WARRIOR [948]

Force is the change in momentum over a specific time. The change of momentum is therefore the force multiplied by the time that the force acts, so 3000x4.0=12000 N s=12000 kg m/s

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2 years ago

A tensile test specimen has a gage length = 50 mm and its cross-sectional area = 100 mm2. The specimen yields at 48,000 N, and t

Shalnov [3]

Answer:

a) yield strength

$\sigma_y = \dfrac{F_y}{A} = =\dfrac{48000}{100} = 480 MPa$

b) modulus of elasticity

strain calculation

$\varepsilon_0=\dfrac{L-L_0}{L_0}=\dfrac{50.23-50}{50} = 0.0046$

strain for offset yield point

$\varepsilon_{new} = \varepsilon_0 -0.002$

=0.0046-0.002 = 0.0026

now, modulus of elasticity

$E = \dfrac{\sigma_y}{\varepsilon_{new}}=\dfrac{480}{0.0026}$

= 184615.28 MPa = 184.615 GPa

c) tensile strength

$\sigma_u =\dfrac{F_{max}}{A}=\dfrac{87000}{100}=870MPa$

d) percentage elongation

$\% Elongation = \dfrac{L-L_0}{L_0}\times 100 = \dfrac{67.3-50}{50}\times 100 = 34.6\%$

e) percentage of area reduction

$\% Area\ reduction = \dfrac{A-A_f}{A}\times 100=\dfrac{100-53}{100}= 47 \%$

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3 years ago