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3 years ago

21. A positively charged rod will have the strongest attraction for

Physics

1 answer:

Ilya [14]3 years ago

8 0

It will have the strongest attraction for a conductor.

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Explain why earth's gravity pulls you toward the ground

pogonyaev

Answer:

Earth's gravity comes from all its mass. All its mass makes a consolidated gravitational draw on all the mass in your body.

Explanation:

5 0

3 years ago

The conductive tissues of the upper leg can be modeled as a 40-cm-long, 12-cm-diameter cylinder of muscle and fat. The resistivi

ahrayia [7]

To calculate and solve the problem it is necessary to apply the concepts related to resistance and resistivity.

The equation that is responsible for relating the two variables is:

$R = \rho \frac{L}{A}$

Where,

R= Resistance of the conductor

$\rho =$ Resistivity of the conductor material

L = Length

A = Cross-sectional area of conductor

With the previous values the area of the muscle (Real Muscle-82%)is,

$A_m = (0.82)\pi r^2 = (0.82)\pi (12/2*10^{-2})^2$

$A_m = 9.274*10^{-3}m^2$

Using the equation from Resistance we have that at the muscle the value is:

$R_m = \rho \frac{L}{A}$

$R_m = \frac{13*(0.4)}{9.274*10^{-3}}$

$R_m = 560.70\Omega$

At the same time we can make the same process to calculate the resistance of the fat, then

$A_m = (0.18)\pi r^2 = (0.18)\pi (12/2*10^{-2})^2$

$A_m = 2.0357*10^{-3}m^2$

The resistance of the fat would be,

$R_f = \rho \frac{L}{A}$

$R_f = \frac{25*(0.4)}{2.0357*10^{-3}}$

$R_f = 4912.3\Omega$

Then the total resistance in a set as the previously writen, i.e, in parallel is:

$R=\frac{R_mR_f}{R_m+R_f}$

$R = \frac{(560.70)(4912.3)}{4912.3+560.70}$

$R = 502.62\Omega$

We can here apply Ohm's law, then

$I= \frac{V}{R}$

$I = \frac{1.5}{502.62}$

$I = 2.984*10^{-3}A$

$I = 2.984mA$

3 0

4 years ago

A(n) ________ describes the shape of a planet’s orbital path. astronomical unit

sleet_krkn [62]

It is an ellipse.

Hope this helps!

8 0

3 years ago

Read 2 more answers

Mental processes refers to

hjlf

Internal,covert processes

5 0

3 years ago

a car accelerates from 2 m/s to 28m/s at a constant rate of 3 m/s^2. How far does it travel while accelerating?

xxTIMURxx [149]

Answer:

Distance, S = 130m

Explanation:

Given the following data;

Initial velocity = 2m/s

Final velocity = 28m/s

Acceleration = 3m/s²

To find the distance, we would use the third equation of motion.

V² = U² + 2aS

Substituting into the equation, we have;

28² = 2² + 2*3*S

784 = 4 + 6S

6S = 784 - 4

6S = 780

S = 780/6

Distance, S = 130m

3 0

3 years ago