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2 years ago

Difference between friction and gravity

Physics

1 answer:

diamong [38]2 years ago

5 0

Explanation:

Friction

Resistance defines friction. Friction consists of the resistance of one object in relation to another object with which it is in contact. Thus, friction is the force that opposes sliding motion, explains the Cornell Center for Materials Research. An example of friction involves removing a stain from clothing. You place detergent on a stained shirt, then repeatedly slide part of the shirt against the stained section. The friction eliminates the stain from the shirt.

Gravity

Gravity is simply defined as what goes up must come down. Gravity is the natural force exerted between two objects, drawing them toward each other. Therefore, instead of an object such as an apple thrown in the air staying there or floating, it falls down. Weight is extremely important to gravity. Gravity always exerts a force equal to the weight of the object it is acting on. A cup remains on a table because the upward force of the table is equal to the weight of the cup, causing it to stay in place.

<h2>hope it helps</h2>

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2 years ago

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3 years ago

horizontal clothesline is tied between 2 poles, 14 meters apart. When a mass of 3 kilograms is tied to the middle of the clothes

lbvjy [14]

Answer:

The tension is $T = 103.96N$

Explanation:

The free body diagram of the question is shown on the first uploaded image From the question we are told that

The distance between the two poles is $D =14 m$

The mass tied between the two cloth line is $m = 3Kg$

The distance it sags is $d_s = 1m$

The objective of this solution is to obtain the magnitude of the tension on the ends of the clothesline

Now the sum of the forces on the y-axis is zero assuming that the whole system is at equilibrium

And this can be mathematically represented as

$\sum F_y = 0$

To obtain $\theta$ we apply SOHCAHTOH Rule

So $Tan \theta = \frac{opp}{adj}$

$\theta = tan^{-1} [\frac{opp}{adj} ]$

$= tan^{-1} [\frac{1}{7}]$

$=8.130^o$

$=> \ \ \ \ \ \ \ \ 2T sin\theta -mg =0$

$=> \ \ \ \ \ \ \ \ T =\frac{mg}{2 sin\theta}$

$=> \ \ \ \ \ \ \ \ T = \frac{3 * 9.8 }{2 sin \theta }$

$=> \ \ \ \ \ \ \ \ T =\frac{29.4}{2sin(8.130)}$

$=> \ \ \ \ \ \ \ \ T = 103.96N$

5 0

2 years ago

Read 2 more answers

Chapter 14, Problem 042 A flotation device is in the shape of a right cylinder, with a height of 0.588 m and a face area of 4.19

Alisiya [41]

Answer:

The workdone is $W = 9.28 * 10^{3} J$

Explanation:

From the question we are told that

The height of the cylinder is $h = 0.588\ m$

The face Area is $A = 4.19 \ m^2$

The density of the cylinder is $\rho = 0.346 * \rho_w$

Where $\rho_w$ is the density of freshwater which has a constant value

$\rho_w = 1000 kg/m^3$

Now

Let the final height of the device under the water be $= h_f$

Let the initial volume underwater be $= V_n$

Let the initial height under water be $= h_i$

Let the final volume under water be $= V_f$

According to the rule of floatation

The weight of the cylinder = Upward thrust

This is mathematically represented as

$\rho_c g V_n = \rho_w gV_f$

$\rho_c A h = \rho A h_f$

So $\frac{0.346 \rho_w}{\rho_w} = \frac{h_f}{h}$

=> $\frac{h_f}{h_c} = 0.346$

Now the work done is mathematically represented as

$W = \int\limits^{h_f}_{h} {\rho_w g A (-h)} \, dh$

$= \rho_w g A [\frac{h^2}{2} ] \left | h_f} \atop {h}} \right.$

$= \frac{g A \rho}{2} [h^2 - h_f^2]$

$= \frac{g A \rho}{2} (h^2) [1 - \frac{h_f^2}{h^2} ]$

Substituting values

$W = \frac{(9.8 ) (4.19) (10^3)}{2} (0.588)^2 (1 - 0.346)$

$W = 9.28 * 10^{3} J$

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2 years ago

two tugboats pull a disabled supertanker. each tug exerts a constant force of 2.20×106 n , one at an angle 19.0 ∘ west of north,

asambeis [7]

The amount of work the two tugboats completed on the supertanker W = 3.12 × 10^9 joules .

Work is a physics term used to describe the energy transfer that takes place when an object is moved over a distance by an external force, at least some . The component of the force acting along the path multiplied by the length of the path can be used to calculate work if the force is constant. Mathematically, this idea is expressed as W = fd, where W is the work and f is the force multiplied by d, the distance. Work is completed when the force is applied at an angle of with respect to the displacement. Performing work on a body involves moving it in its entirety from one location to another as well as other methods. However, the work is thought to be negative if the applied force is in opposition to the item's motion, indicating that energy is being pulled away from the object

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8 months ago

Difference between friction and gravity​

Difference between friction and gravity