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3 years ago

_____ friction occurs when a object rolls over a surface

2 answers:

6 0

<h2>Answer:</h2><h2><em><u>-Rolling friction occurs</u></em></h2><h3><em><u>Thank you for asking this great question need any other help please let me know by commenting below I'd be glad to help. </u></em></h3><h3><em><u>I'd also greatly appreciate you if you mark me as brainliest and click that thanks button.( optional )</u></em></h3><h3><em><u>Your brainly friend ( lauralit1 )</u></em></h3><h2><em><u /></em></h2>

Andrew [12]3 years ago

6 0

<u>Answer:</u>

<u><em>Rolling friction</em></u><em> occurs when a object rolls overs a surface.</em>

<u>Explanation:</u>

Friction is the resisting force which occurs during contact of two objects leading to preventing its motion. The loss due to friction is released in the form of heat energy.

When two objects come in contact with each other for any kind of motion, the electromagnetic attraction of the charges exhibited by the two objects leads to resisting the motion of the object with the release of heat energy.

In this case, when a object is rolling over a surface, the roughness of the surface will contribute to the friction between the surface and the rolling object leading to variation of the speed of the rolling object. This kind of friction experienced by a rolling object is termed as rolling friction.

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3 years ago

The deck of a bridge is suspended 235 feet above a river. If a pebble falls off the side of the bridge, the height, in feet, of

Aleks04 [339]

Answer:

(a) 1, average velocity = -65.6 m/s

2, average velocity = -64.8 m/s

3, average velocity = -64.16 m/s

(b) The instantaneous velocity is -96 m/s

Explanation:

(a)

Average velocity is given by;

$y(t_2,t_1) = \frac{y(t_2) - y(t_1)}{t_2-t_1}$

(1)

$y(2.1,2) = \frac{(235-16*2.1^2) - (235-16*2^2)}{2.1-2}\\\\ y(2.1,2) = -65.6 \ m/s$

(2)

$y(2.05,2) = \frac{(235-16*2.05^2) - (235-16*2^2)}{2.05-2}\\\\ y(2.05,2) = -64.8 \ m/s$

(3)

$y(2.01,2) = \frac{(235-16*2.01^2) - (235-16*2^2)}{2.01-2}\\\\ y(2.01,2) = -64.16 \ m/s$

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3 years ago

A 0.335 kg mass is attached to a spring and executes simple harmonic motion with a period of 0.38 s. The total energy of the sys

kramer

Answer:

$k=91.54 \frac{N}{m}$

Explanation:

The angular speed is defined as the angle traveled in one revolution over the time taken to travel it, that is, the period. Therefore, it is given by:

$\omega=\frac{2\pi}{T}\\\omega=\frac{2\pi}{0.38s}\\\omega=16.53\frac{rad}{s}$

The angular frequency of the simple harmonic motion of the mass-spring system is defined as follows:

$\omega=\sqrt{\frac{k}{m}}$

Here, k is the spring's constant and m is the mass of the body attached to the spring. Solving for k:

$\omega^2=\frac{k}{m}\\k=m\omega^2\\k=0.335kg(16.53\frac{rad}{s})^2\\k=91.54 \frac{N}{m}$

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3 years ago