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3 years ago

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A pendulum has 576 J of potential energy at the highest point of its swing. How much kinetic energy will it have at the bottom o

f its swing?

1 answer:

Natalka [10]3 years ago

3 0

Ideally, 576 J because energy is conserved.
In the real world, a tiny tiny tiny tiny bit less than 576 J ,
because we live in a world with friction and air resistance.

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A boy reaches out of a window and tosses a ball straight up with a speed of 10 m/s. The ball is 20 m above the ground as he rele

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Explanation:

It is given that,

Speed of the ball, v = 10 m/s

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(a) Let H is the maximum height reached by the ball. It can be calculated using the conservation of energy as :

$\dfrac{1}{2}mv^2=mgh'$

$h'=\dfrac{v^2}{2g}$

$h'=\dfrac{(10)^2}{2\times 9.8}$

h' = 5.1 m

The maximum height above ground,

H = 5.1 + 20

H = 25.1 meters

So, the maximum height reached by the ball is 25.1 meters.

(b) The ball's speed as it passes the window on its way down is same as the initial speed i.e. 10 m/s.

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3 years ago

Three equal point charges, each with charge 2.00 μC , are placed at the vertices of an equilateral triangle whose sides are of l

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Answer:

Explanation:

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9 x 10⁹ ( 2x2 x10⁻¹²/ .25 + 2x2 x10⁻¹²/ .25 + 2x2 x10⁻¹²/ .25 )

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3 years ago

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Explanation:

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3 years ago

A wheel rotating about a fixed axis with a constant angular acceleration of 2.0 rad/s2 starts from rest at t = 0. The wheel has

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Answer:

The total linear acceleration is approximately 0.246 meters per square second.

Explanation:

The total linear acceleration ( $a$ ) consist in two components, <em>radial</em> ( $a_{r}$ ) and <em>tangential</em> ( $a_{t}$ ), in meters per square second:

$a_{r} = \omega^{2}\cdot r$ (1)

$a_{t} = \alpha \cdot r$ (2)

Since both components are orthogonal to each other, the total linear acceleration is determined by Pythagorean Theorem:

$a = \sqrt{a_{r}^{2}+a_{t}^{2}}$ (3)

Where:

$r$ - Radius of the wheel, in meters.

$\omega$ - Angular speed, in radians per second.

$\alpha$ - Angular acceleration, in radians per square second.

Given that wheel accelerates uniformly, we use the following kinematic equation:

$\omega = \omega_{o}+ \alpha\cdot t$ (4)

Where:

$\omega_{o}$ - Initial angular speed, in radians per second.

$t$ - Time, in seconds.

If we know that $r = 0.1\,m$ , $\alpha = 2\,\frac{rad}{s^{2}}$ , $\omega_{o} = 0\,\frac{rad}{s}$ and $t = 0.60\,s$ , then the total linear acceleration is:

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$a_{t} = 0.2\,\frac{m}{s^{2}}$

$a = \sqrt{a_{r}^{2}+a_{t}^{2}}$

$a \approx 0.246\,\frac{m}{s^{2}}$

The total linear acceleration is approximately 0.246 meters per square second.

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