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3 years ago

Which of the following describes the relationship of conductance to a conductor?

1 answer:

4 0

The answer is The conductance of a conductor is inversely proportional to the cross-sectional area of the conductor. Conductance is directly related to the ease offered by any material to the passage of electric current. Conductance is the opposite of resistance. The higher the conductance, the lower the resistance and vice versa, the greater the resistance, the less conductance, so both are inversely proportional

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Aleonysh [2.5K]

Answer:

it's 1.57... radians and 90°

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2 years ago

An object moving at 30 m/s takes 5 sec. to come to a stop. What is the object’s acceleration

Klio2033 [76]

Answer:

6 m/s²

Explanation:

a=v/t

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3 years ago

Match the layers of the atmosphere with the description that best matches it according to the graph titled Temperature Profile.

Finger [1]

The correct matches are as follows:

Troposphere
A) Layer closest to the Earth where all weather occurs

Mesosphere
C) temperature remains constant as elevation increases

Mesopause
D) temperature remains at a constant zero degrees Celsius as elevation increases

Stratosphere
E) begins between 50-60km and decreases in temperature as elevation increases 

Stratopause
 B) temperature increases as elevation increases

Hope this answers the question. Have a nice day.

3 0

3 years ago

Read 2 more answers

A charge q1 = +5.00 nC is placed at the origin of an xy-coordinate system, and a charge q2 = -2.00 nC is placed on the positive

Ivahew [28]

Answer:

The x- and y-components of the total force exerted is

$F_{31 +32} = (8.64i - 5.52 j) *10^{-5}$

The magnitude of the force is

$|F_{31 +32}| = 10.25 *10^{-5} N$

The direction of the force is

$\theta =327.43 ^o$ Clockwise from x-axis

Explanation:

From the question we are told that

The magnitude of the first charge is $q_1 = +5.00nC = 5.00*10^{-9}C$

The magnitude of the second charge is $q_2 = -2.00nC = -2.00*10^{-9}C$

The position of the second charge from the first one is $d_{12} = 4.00i \ cm = \frac{4.00i}{100} = 4.00i *10^{-2} m$

The magnitude of the third charge is $q_3 = +6.00nC = 6.00*10^{-9}C$

The position of the third charge from the first one is $\= d_{31} = (4i + 3j) cm = \frac{ (4i + 3j)}{100} = (4i + 3j) *10^{-2}m$

$|d_{31}| =(\sqrt{4 ^2 + 3^2}) *10^{-2} m$

$|d_{31}| =5 *10^{-2} m$

The position of the third charge from the second one is

$\= d_{32} = 3j cm = 3j *10^{-2}m$

$|d_{32}| =(\sqrt{ 3^2}) *10^{-2} m$

$|d_{32}| =3 *10^{-2} m$

The force acting on the third charge due to the first and second charge is mathematically represented as

$F_{31 +32} = \frac{kq_3 q_1}{|d_{31}| ^3} *\= d_{31} + \frac{kq_3 q_2}{|d_{32}| ^3} *\= d_{32}$

Substituting values

$F_{31 +32} = \frac{9 *10^9 * 6 *10^{-9} * 5*10^{-9} }{(5*10^{-2}) ^3} * (4i + 3j ) *10^{-2} \\ \ + \ \ \ \ \ \ \ \ \ \frac{9 *10^9 * 6 *10^{-9} * -2*10^{-9} }{(5*10^{-2}) ^3} * (4i + 3j ) *10^{-2}$