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3 years ago

How is a family defined

Physics

1 answer:

natulia [17]3 years ago

3 0

Answer:Family

Explanation: Family is defined as people you love, people that you can count on. People that make you happy. Family isnt only people who are related to you by blood, we are all a family. Certain people are your family.

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Please help me anything will help

spin [16.1K]

Your answer is C) The speed of sound is higher in solids than in liquids.

6 0

2 years ago

The Cosmological Argument for the existence of God intellectually follows cause-and-effect back in time -- in effect asking "Whe

ankoles [38]

Answer:

First uncaused cause

Explanation:

Aristotle states that an infinite regression in the principle of causality is not possible. If the regression were infinite, then there would never be a first cause (mover), since this would need another mover to start its motion. Therefore, according to Aristotle, there must be an unmoved mover that moves other things, but is not itself moved by any prior action,

3 0

2 years ago

There is a 90 kg woman attached at the end of a bungee cord (k = 35 N/m) that is experiencing simple harmonic motion. How long d

Cloud [144]

Answer:

The value is $T =10.1 \ s$

Explanation:

From the question we are told that

The mass of the woman is $m = 90 \ kg$

The spring constant of the bungee cord is $k = 35 \ N/ m$

Generally the period of the oscillation (i,e time taken to complete on cycle ) is mathematically represented as

$T = 2 \pi * \sqrt{ \frac{m}{k} }$

=> $T = 2 * 3.142 * \sqrt{ \frac{90 }{ 35} }$

=> $T =10.1 \ s$

3 0

3 years ago

A solid disk of mass 2 kg and radius 2 m is given a horizontal push of 20N at a point .3 m above its center. a. What is the mini

Margaret [11]

Answer:

$\mu_s=1.0205$

Explanation:

Given:

mass of solid disk, $m=2\ kg$

radius of disk, $r=2\ m$

force of push applied to disk, $F=20\ N$

distance of application of force from the center, $s=0.3\ m$

<em>For the condition of no slip the force of static friction must be greater than the applied force so that there is no skidding between the contact surfaces at the contact point.</em>

$\therefore F$