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Molodets [167]
4 years ago
6

Suppose a student rubs a Teflon rod with wool and then briefly touches it to an initially neutral aluminum rod suspended by insu

lating threads. What would the net charge on the Teflon rod be? What would the net charge on the aluminum rod be? After the student has touched the aluminum rod with the rubbed Teflon rod, as described above, the student then holds the Teflon rod near the aluminum rod, but does not allow them to touch. The student observes that when the rods are close enough tog
Engineering
1 answer:
Oliga [24]4 years ago
3 0

Answer:

Due to touch of teflon, its charge will reduce but will not go to zero. Some amount of its initial charge will be transferred to Aluminum rod. So, aluminum rod will have a non-zero negative charge.

Explanation:

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Atlantis benefited from lessons learned in the construction and testing of Enterprise, Columbia and Challenger. ... The Experience gained during the Orbiter assembly process also enabled Atlantis to be completed with a 49.5 percent reduction in man hours (compared to Columbia).

Explanation:

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What is NOT an example of professional networking?
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4 years ago
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3 0
4 years ago
A furnace wall is to be built of 20-cm firebrick and building (structural) brick of same thickness. The thermal conductivities o
Norma-Jean [14]

Answer:

q=2313.04W/m^2

T=690.86°C

Explanation:

Given that

Thickness t= 20 cm

Thermal conductivity of firebrick= 1.6 W/m.K

Thermal conductivity of structural brick= 0.7 W/m.K

Inner temperature of firebrick=980°C

Outer temperature of structural brick =30°C

We know that thermal resistance

R=\dfrac{t}{KA}

These are connect in series

R=\left(\dfrac{t}{KA}\right)_{fire}+\left(\dfrac{t}{KA}\right)_{struc}

R=\dfrac{0.2}{1.6A}+\dfrac{0.2}{0.7A}\ K/W

R=\dfrac{23}{56A}\ K/W

Heat transfer

Q=\dfrac{\Delta T}{R}

Q=56A\times \dfrac{980-30}{23}\ W

So heat flux

q=2313.04W/m^2

Lets temperature between interface is T

Now by equating heat in both bricks

\dfrac{980-T}{\dfrac{0.2}{1.6A}}=\dfrac{T-30}{\dfrac{0.2}{0.7A}}

So T=690.86°C

6 0
3 years ago
A large class with 1,000 students took a quiz consisting of ten questions. To get an A, students needed to get 9 or 10 questions
VMariaS [17]

Answer:

a. 0.11

b. 110 students

c. 50 students

d. 0.46

e. 460 students

f. 540 students

g. 0.96

Explanation:

(See attachment below)

a. Probability that a student got an A

To get an A, the student needs to get 9 or 10 questions right.

That means we want P(X≥9);

P(X>9) = P(9)+P(10)

= 0.06+0.05=0.11

b. How many students got an A on the quiz

Total students = 1000

Probability of getting A = 0.11 ---- Calculated from (a)

Number of students = 0.11 * 1000

Number of students = 110 students

So,the number of students that got A is 110

c. How many students did not miss a single question

For a student not to miss a single question, then that student scores a total of 10 out of possible 10

P(10) = 0.05

Total Students = 1000

Number of Students = 0.05 * 1000

Number of Students = 50 students

We see that 5

d. Probability that a student pass the quiz

To pass, a student needed to get at least 6 questions right.

So we want P(X>=6);

P(X>=) =P(6)+P(7)+P(8)+P(9)+P(10)

=0.08+0.12+0.15+0.06+0.05=0.46

So, the probability of a student passing the quiz is 0.46

e. Number of students that pass the quiz

Total students = 1000

Probability of passing the quiz = 0.46 ----- Calculated from (d)

Number of students = 0.46 * 1000

Number of students = 460 students

So,the number of students that passed the test is 460

f. Number of students that failed the quiz

Total students = 1000

Total students that passed = 460 ----- Calculated from (e)

Number of students that failed = 1000 - 460

Number of students that failed = 540

So,the number of students that failed is 540

g. Probability that a student got at least one question right

This means that we want to solve for P(X>=1)

Using the complement rule,

P(X>=1) = 1 - P(X<1)

P(X>=1) = 1 - P(X=0)

P(X>=1) = 1 - 0.04

P(X>=1) = 0.96

7 0
3 years ago
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