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3 years ago

What does it mean if electric field lines are close together versus farther apart ?

Physics

1 answer:

hoa [83]3 years ago

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Answer:

As a result, if the field lines are close together (that is, the field line density is greater), this indicates that the magnitude of the field is large at that point. If the field lines are far apart at the cross-section, this indicates the magnitude of the field is small. (Figure) shows the idea.

Explanation:

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2 years ago

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Answer:

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Explanation:

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2 years ago

What region of the atom determines the volume of the atom

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4 years ago

A 83 kg man is resting on the roof top of his house. If he has a potential energy of

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3 years ago

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a disk with a radius of 0.1 m is spinning about its center with a constant angular speed of 10 rad/sec. What are the speed and m

padilas [110]

The speed of the bug at the rim of the circular disk is $\boxed{1\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}}$ and the acceleration of the bug is $\boxed{10\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right.\kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}}$ .

Further Explanation:

Given:

The angular velocity of the disk is $10\,{{{\text{rad}}} \mathord{\left/{\vphantom {{{\text{rad}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}$ .

The radius of the circular disk is $0.1\,{\text{m}}$ .

Concept:

The linear speed of the bug present at the rim of the circular disk is given by.

$v = r \times \omega$

Here, $v$ is the linear speed, $r$ is the radius of the disk and $\omega$ is the angular speed of the disk.

Substitute $0.1\,{\text{m}}$ for $r$ and $10\,{{{\text{rad}}} \mathord{\left/{\vphantom {{{\text{rad}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}$ for $\omega$ in above expression.

$\begin{aligned}v &= 0.1\,{\text{m}} \times 10\,{{{\text{rad}}} \mathord{\left/{\vphantom {{{\text{rad}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}} \\&= 1\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}} \\\end{aligned}$

The magnitude of the centripetal acceleration of the bug cling to the rim of the disk is given as.

${a_c} = \dfrac{{{v^2}}}{r}$

Substitute $1\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}$ for $v$ and $0.1\,{\text{m}}$ for $r$ in above equation.

$\begin{aligned}a_c&=\dfrac{(1\text{m/s})^2}{0.1\text{m}}\\&=\frac{1}{0.1}\text{m/s}^2\\&=10\,\text{m/s}^2\end{aligned}$

Thus, the speed of the bug at the rim of the circular disk is $\boxed{1\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}}$ and the acceleration of the bug is $\boxed{10\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right.\kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}}$ .

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Answer Details:

Grade: College

Chapter: Uniform Circular motion

Subject: Physics

Keywords: Circular disk, circular motion, angular speed, linear speed, bug clinging, rim of the disk, acceleration, magnitude, constant, rad/s.

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3 years ago

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