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3 years ago

8

Which one of the following women won the American Mathematical Society Scatter Prize, a Clay Mathematics Institute Fellowship, a

nd the Blumenthal Award in 2013? Sally Ride Amalie Noether Maryam Mirzakhani Clara Ma

1 answer:

AVprozaik [17]3 years ago

3 0

Answer:

Option C, Maryam Mirzakhani, is the right answer.

Explanation:

Maryam Mirzakhani was a mathematician from Iran and a professor at Stanford University. Her main contribution can be seen in the "moduli spaces of Riemann surfaces" theory. For her contribution to mathematics, she has been awarded numerous awards. These awards include the "Ruth Lyttle Satter Prize in Mathematics, a Clay Mathematics Institute Fellowship, and the Blumenthal Award". She was the first Iranian women to receive the Fields Medal.

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2.14 (a) Using series/parallel resistance reductions, find the equivalent resistance between terminals A and B in the circuit of

olasank [31]

Answer:

There is no attachment

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3 years ago

7. What is the voltage across a 100 ohm circuit element that draws a current of 1 A?

Marianna [84]

Answer:

100 V

Explanation:

V = IR

100 x 1 = 100

100 volts

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3 years ago

The wall of drying oven is constructed by sandwiching insulation material of thermal conductivity k = 0.05 W/m°K between thin me

masha68 [24]

Answer:

86 mm

Explanation:

From the attached thermal circuit diagram, equation for i-nodes will be

$\frac {T_ \infty, i-T_{i}}{ R^{"}_{cv, i}} + \frac {T_{o}-T_{i}}{ R^{"}_{cd}} + q_{rad} = 0$ Equation 1

Similarly, the equation for outer node “o” will be

$\frac {T_{ i}-T_{o}}{ R^{"}_{cd}} + \frac {T_{\infty, o} -T_{o}}{ R^{"}_{cv, o}} = 0$ Equation 2

The conventive thermal resistance in i-node will be

$R^{"}_{cv, i}= \frac {1}{h_{i}}= \frac {1}{30}= 0.033 m^{2}K/w$ Equation 3

The conventive hermal resistance per unit area is

$R^{"}_{cv, o}= \frac {1}{h_{o}}= \frac {1}{10}= 0.100 m^{2}K/w$ Equation 4

The conductive thermal resistance per unit area is

$R^{"}_{cd}= \frac {L}{K}= \frac {L}{0.05} m^{2}K/w$ Equation 5

Since $q_{rad}$ is given as 100, $T_{o}$ is 40 $T_ \infty$ is 300 $T_{\infty, o}$ is 25

Substituting the values in equations 3,4 and 5 into equations 1 and 2 we obtain

$\frac {300-T_{i}}{0.033} +\frac {40-T_{i}}{L/0.05} +100=0$ Equation 6

$\frac {T_{ i}-40}{L/0.05}+ \frac {25-40}{0.100}=0$

$T_{i}-40= \frac {L}{0.05}*150$

$T_{i}-40=3000L$

$T_{i}=3000L+40$ Equation 7

From equation 6 we can substitute wherever there’s $T_{i}$ with 3000L+40 as seen in equation 7 hence we obtain

$\frac {300- (3000L+40)}{0.033} + \frac {40- (3000L+40)}{L/0.05}+100=0$

The above can be simplified to be

$\frac {260-3000L}{0.033}+ \frac {(-3000L)}{L/0.05}+100=0$

$\frac {260-3000L}{0.033}=50$

-3000L=1.665-260

$L= \frac {-258.33}{-3000}=0.086*10^{-3}m= 86mm$

Therefore, insulation thickness is 86mm

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3 years ago

A barometer reads a height of 78 cmHg. Express this atmospheric pressure to:

Yakvenalex [24]

A barometer reads a height of 78 cmHg. Express this atmospheric pressure to $\large\mathsf{\red{\underline{Pascal(pa)}}}$

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3 years ago

Read 2 more answers

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allsm [11]

Answer:

wow

Explanation:

that's so cute!!!!!!!!

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3 years ago