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3 years ago

5

A century ago, swords were made from various metals, especially steel. The property that makes steel a good choice for sword mak

ing is _____.

2 answers:

tigry1 [53]3 years ago

8 0

The answer should be hardness and strength
~emily

OLEGan [10]3 years ago

8 0

That it was harder than most other metals as well as not likely to shatter?

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A seesaw has an irregularly distributed mass of 30 kg, a length of 3.0 m, and a fulcrum beneath its midpoint. It is balanced whe

gladu [14]

Answer:

x = 0.9 m

Explanation:

For this exercise we must use the rotational equilibrium relation, we will assume that the counterclockwise rotations are positive

∑ τ = 0

60 1.5 - 78 1.5 + 30 x = 0

where x is measured from the left side of the fulcrum

90 - 117 + 30 x = 0

x = 27/30

x = 0.9 m

In summary the center of mass is on the side of the lightest weight x = 0.9 m

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3 years ago

Which of the following substances is covalently bonded? A. KF B. H2S C. CaCl2 D. SO2

poizon [28]

The answer is c for that question

6 0

3 years ago

Read 2 more answers

A box of mass 72 kg is at rest on a horizontal frictionless surface. A constant horizontal force F then acts on the box and acce

zvonat [6]

Answer:

$162\ N$

Explanation:

$We\ are\ given\ that,\\Mass\ of\ the\ box=72\ kg\\Distance/Displacement\ travelled\ by\ the\ box\ during\ the\ application\ of\\ Force=13\ m\\Time\ taken\ for\ it\ to\ displace=3.4 \seconds\\Now,\\As\ we\ know\ that,\\Force=Mass*Acceleration\\In\ order\ to\ know\ force\ we\ need\ to\ know\ the\ acceleration.\ Lets\ find\\ that\ out.\\We\ would\ use\ Newton's\ Third\ Equation\ Of\ Motion\ and\ Newton's\\ First\ Equation\ Of\ Motion:\\2as=v^2-u^2\\v=u+at$

$Hence,\\First\ lets\ consider\ Newtons\ First\ Equation\ Of\ Motion:\\v=u+at\\v-u=at\\Hence\ now\ lets\ move\ on\ to\ Newton's\ Third\ Law\ Of\ Motion:\\2as=v^2-u^2\\2as=(v+u)(v-u)\\Substituting\ (v-u)=at,\\2as=at(v+u) \\Hence,\ as\ the\ body\ moves\ from\ rest,\ u=0\\So,\\2as=at*v\\Cancelling\ a\ at\ both\ the\ sides\ we\ get,\\2s=vt\\Hence,\\Lets\ plug\ in\ the\ values\ of\ Displacement\ and\ Time,\ Shall\ we?\\Hence,\\2*13=3.4v\\26=3.4v\\\frac{26}{3.4}=v\\v \approx 7.647\ m/s$

$Now,\\As\ we\ know\ that,\\a=\frac{v-u}{t} [Equation\ for\ acceleration]\\Hence,\\a=\frac{v}{t} [As\ u=0]\\Hence,\\Acceleration=\frac{7.647}{3.4} \\Acceleration \approx 2.25\ m/s^2\\$

$Hence,\\Now,\\As\ by\ using\ expression\ for\ Force,\\Force= Mass*Acceleration\\Here,\\Force\ exerted\ on\ the\ box=72*2.25= 162 N$

3 0

2 years ago

UN COLET POSTAL ESTE IMPINS DE O MASA ORIZONTALA CU O FORTA F=40N.COLETU SE DEPLASEAZA PE O DISTANTA DE 1,5M CE LUCRU MECANIC EF

Wittaler [7]

Answer:

A POSTAL PACKAGE IS PUSHED BY A HORIZONTAL TABLE WITH A FORCE F = 40N. PACKAGE MOVES ON A DISTANCE OF 1.5M WHICH MECHANICAL WORK PERFORMS THIS FORCE

PT TEST

5 0

3 years ago

A block of mass 5 kg is placed on a rough table having a coefficient of static friction μs = 0.2.What is the force required to m

just olya [345]

Answer:

<em>We need to (at least) apply a force of 9.8 N to move the block</em>

Explanation:

<u>Second Newton's Law</u>

If a net force $F_n$ different from zero is applied to an object of mass m, then it will move at an acceleration a, given by

$F_n=ma$

If we apply a force F to an object placed on a rough surface, the only way to make it move is to beat the friction force which is given by

$F_r=\mu F_N$

Where $\mu$ is the static friction coefficient and $F_N$ is the normal force exerted by the table to the object. Since there is no motion in the vertical direction the normal force equals the weight of the object:

$F_N=mg=5\ kg\ 9.8\ m/s^2=49\ N$

The friction force is

$F_r=0.2 (49)=9.8\ N$

Thus, we need to (at least) apply a force of 9.8 N to move the block

5 0

3 years ago