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3 years ago

8

If a rigid body rotates about a fixed axis passing through its center of mass, the body's linear momentum is

1 answer:

dexar [7]3 years ago

6 0

Answer:

The linear momentum is zero

Explanation:

Because

When a rigid body is rotating about a fixed axis passing through point O, the body’s linear momentum given as L = mvG

But VG= 0 so

Linear momentum is zero

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An airplane went from 120 m/s to 180 m/s in 4.0 seconds. What was its acceleration?

scoundrel [369]

<span>15 m/s^2 The first thing to calculate is the difference between the final and initial velocities. So 180 m/s - 120 m/s = 60 m/s So the plane changed velocity by a total of 60 m/s. Now divide that change in velocity by the amount of time taken to cause that change in velocity, giving 60 m/s / 4.0 s = 15.0 m/s^2 Since you only have 2 significaant figures, round the result to 2 significant figures giving 15 m/s^2</span>

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3 years ago

Calculate the electric field at the center of a square

pantera1 [17]

Answer:

$E_y=1175510.2\ N.C^{-1}$

The Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

Explanation:

Given:

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charge on one corner of the square, $q_1=+45\times 10^{-6}\ C$

charge on the remaining 3 corners of the square, $q_2=q_3=q_4=-27\times 10^{-6}\ C$

<u>Distance of the center from each corners</u> $=\frac{1}{2} \times diagonals$

$diagonal=\sqrt{52.5^2+52.5^2}$

$diagonal=74.25\cm=0.7425\ m$

∴Distance of center from corners, $b=0.3712\ m$

Now, electric field due to charges is given as:

$E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}$

<u>For charge $q_1$ we have the field lines emerging out of the charge since it is positively charged:</u>

$E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}$

$E_1=2938775.5\ N.C^{-1}$

<u>Force by each of the charges at the remaining corners:</u>

$E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}$

$E_2=E_3=E_4=1763265.3\ N.C^{-1}$

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$E_y=E_1-E_4$

$E_y=1175510.2\ N.C^{-1}$

<u>Now, net electric field in the horizontal direction:</u>

$E_y=E_2-E_3$

$E_y=0\ N.C^{-1}$

So the Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

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