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Brums [2.3K]
3 years ago
7

How high has a 45 g golf ball been driven if it has a potential energy of 2.06 J?

Physics
2 answers:
Mice21 [21]3 years ago
7 0
PE=mgh
2.06 J = 2.06 kg·m²/s² = (0.045 kg)·(9.8 m/s²)·h
where h is the height the ball has reached
Solve for h:
h = 2.06/(0.045·9.8) = 4.67 to the nearest hundredth
h = 4.67 m

nydimaria [60]3 years ago
4 0
Here, U = mgh

We have: m = 45 g = 0.045 Kg
U = 2.06 J, g = 9.8 m/s², h = ?

Substitute the values in the formula:
2.06 = 0.045 × 9.8 × h
2.06 = 0.441 × h
h = 2.06 / 0.441
h = 4.67 m

In short, your final answer would be 4.67 meters

Hope this helps!

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Arm ab has a constant angular velocity of 16 rad/s counterclockwise. At the instant when theta = 60
geniusboy [140]

The <em>linear</em> acceleration of collar D when <em>θ = 60°</em> is - 693.867 inches per square second.

<h3>How to determine the angular velocity of a collar</h3>

In this question we have a system formed by three elements, the element AB experiments a <em>pure</em> rotation at <em>constant</em> velocity, the element BD has a <em>general plane</em> motion, which is a combination of rotation and traslation, and the ruff experiments a <em>pure</em> translation.

To determine the <em>linear</em> acceleration of the collar (a_{D}), in inches per square second, we need to determine first all <em>linear</em> and <em>angular</em> velocities (v_{D}, \omega_{BD}), in inches per second and radians per second, respectively, and later all <em>linear</em> and <em>angular</em> accelerations (a_{D}, \alpha_{BD}), the latter in radians per square second.

By definitions of <em>relative</em> velocity and <em>relative</em> acceleration we build the following two systems of <em>linear</em> equations:

<h3>Velocities</h3>

v_{D} + \omega_{BD}\cdot r_{BD}\cdot \sin \gamma = -\omega_{AB}\cdot r_{AB}\cdot \sin \theta   (1)

\omega_{BD}\cdot r_{BD}\cdot \cos \gamma = -\omega_{AB}\cdot r_{AB}\cdot \cos \theta   (2)

<h3>Accelerations</h3>

a_{D}+\alpha_{BD}\cdot \sin \gamma = -\omega_{AB}^{2}\cdot r_{AB}\cdot \cos \theta -\alpha_{AB}\cdot r_{AB}\cdot \sin \theta - \omega_{BD}^{2}\cdot r_{BD}\cdot \cos \gamma   (3)

-\alpha_{BD}\cdot r_{BD}\cdot \cos \gamma = - \omega_{AB}^{2}\cdot r_{AB}\cdot \sin \theta + \alpha_{AB}\cdot r_{AB}\cdot \cos \theta - \omega_{BD}^{2}\cdot r_{BD}\cdot \sin \gamma   (4)

If we know that \theta = 60^{\circ}, \gamma = 19.889^{\circ}, r_{BD} = 10\,in, \omega_{AB} = 16\,\frac{rad}{s}, r_{AB} = 3\,in and \alpha_{AB} = 0\,\frac{rad}{s^{2}}, then the solution of the systems of linear equations are, respectively:

<h3>Velocities</h3>

v_{D}+3.402\cdot \omega_{BD} = -41.569   (1)

9.404\cdot \omega_{BD} = -24   (2)

v_{D} = -32.887\,\frac{in}{s}, \omega_{BD} = -2.552\,\frac{rad}{s}

<h3>Accelerations</h3>

a_{D}+3.402\cdot \alpha_{BD} = -445.242   (3)

-9.404\cdot \alpha_{BD} = -687.264   (4)

a_{D} = -693.867\,\frac{in}{s^{2}}, \alpha_{BD} = 73.082\,\frac{rad}{s^{2}}

The <em>linear</em> acceleration of collar D when <em>θ = 60°</em> is - 693.867 inches per square second. \blacksquare

<h3>Remark</h3>

The statement is incomplete and figure is missing, complete form is introduced below:

<em>Arm AB has a constant angular velocity of 16 radians per second counterclockwise. At the instant when θ = 60°, determine the acceleration of collar D.</em>

To learn more on kinematics, we kindly invite to check this verified question: brainly.com/question/27126557

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Please help me with the correct answer​
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