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3 years ago

9

Find the force between 2C and -1C separated by a distance 1m. Indicate if it is an attractive force (negative) or a repulsive fo

rce (positivie).

1 answer:

telo118 [61]3 years ago

6 0

Answer:

Explanation:

Its definitely an Attractive force since the two charges are Unlike.

From Coulombs Law

F=kq1q2/R²

Given

K=9x10^9

R=1m

q1=2C

q2=-1C

F=(9x10^9 x 2 x -1)/1²

F= - 1.8x10^10N. (Attractive).

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A 1.2 nf parallel-plate capacitor has an air gap between its plates. Its capacitance increases by 3.0 nf when the gap is filled

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Answer:

2.5

Explanation:

The capacitance of a parallel-plate capacitor filled with dielectric is given by

$C=kC_0$

where

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$C_0$ is the capacitance of the capacitor without dielectric

In this problem,

$C_0=1.2 nF$ is the capacitance of the capacitor in air

$C=3.0 nF$ is the capacitance with the dielectric inserted

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2 years ago

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2 years ago

A uniform rod rotates in a horizontal plane about a vertical axis through one end. The rod is 3.46 m long, weighs 12.8 N, and ro

Mkey [24]

Answer:

a. Rotational inertia: 5.21kgm²

b. Magnitude of it's angular momentum: 123.32kgm²/s

Explanation:

Length of the rod = 3.46m

Weight of the rod = 12.8 N

Angular velocity of the rod= 226 rev/min

a. Rotational Inertia (I) about its axis

The formula for rotational inertia =

I = (1/12×m×L²) + m × ( L ÷ 2)²

Where L = length of the rod

m = mass of the rod

Mass of the rod is calculated by dividing the weight of the rod with the acceleration due to gravity.

Acceleration due to gravity = 9.81m/s²

Mass of the rod = 12.8N/ 9.81m/s²

Mass of the rod = 1.305kg

Rotational Inertia =

(1/12× 1.305 × 3.46²)+ 1.305 ( 3.46÷2)²

Rotational Inertia = 1.3019115 + 3.9057345

Rotational Inertia = 5.207646kgm²

Approximately = 5.21kgm²

b. The magnitude of the rod's angular momentum about the rotational axis is calculated as

Rotational Inertia about its axis × angular speed of the rod.

Angular speed of the rod is calculated as= (Angular velocity of the rod × 2π)/60

= (226×2π) /60

= 23.67 rad/s

Rotational Inertia = 5.21kgm²

The magnitude of the rod's angular momentum about the rotational axis

= 5.21kgm²× 23.67 rad/s

= 123.3207kgm²/s

Approximately = 123.32kgm²/s

7 0

3 years ago